Talk:Two envelopes problem/Arguments

Simple Solution to Hardest Problem

"Let the amount in the envelope you chose be A. Then by swapping, if you gain you gain A but if you lose you lose A/2. So the amount you might gain is strictly greater than the amount you might lose."

Yes, but if you swap again, as it's the FIRST envelope that has A dollars, then by switching again, you either gain A/2 or lose A. Therefore, you can't switch forever and gain forever. It operates under the false assumpation that whatever envelope you have has A, but that A is also a constant. You can't have both.

                 -t3h 1337 r0XX0r

Oh, just noticed it's irrelevance. Sorry, won't do this again... 1337 r0XX0r 15:39, 19 January 2006 (UTC)

Its irrelevance? I'm confused. The solution to the hardest problem seems pretty simple... you are switching the amount in the envelope depending on if you look at it like it has more, or less, from Y to 2Y. You could do the same thing with Y (you gain a Y if the other envelope has more) and 100Y (you lose 50 Y if your envelope had the most). ~Tever

• "To be rational I will thus end up swapping envelopes indefinitely" - This sentence is clearly false. After the first swap, I already know the contents of both the envelopes, and were I allowed a second swap, the decision whether on not to do it is simple. - Mike Rosoft 23:42, 24 March 2006 (UTC)

No, you never look into any envelopes in the original statement of the problem. INic 21:14, 27 March 2006 (UTC)

[[ The solution is simple. Everyone is trying to discern the probability of getting the greater amount, thus gaining the most out of the problem. However, if you were to not open either one, and continiously follow the pattern of swapping between envelopes, you would thus gain nothing. So, in order to gain the most, one would have to be content upon gaining at all, thus gaining something, and thus gaining the most out of the probablity, for without taking one, you gain nothing. ]] - Apocalyptic_Kisses, April 6th, 2006

You are right. But what rule of decision theory do you suggest should be added to allow for this behaviour? It seems to me to be a meta-rule rather than an ordinary rule. Something like "When decision theory leads to absurd results please abandon decision theory!" INic 17:43, 18 April 2006 (UTC)

I may have missed something here, but my own analysis is based on the fact that money exists in units and A or whatever the amount is in the envelope is therefore a discrete variable. Any amount which is an odd number of units can only be doubled, which means that for any reasonable range on monetary amounts, amounts with even numbers of units are more likely to be halved than doubled. If fact if you assume that there is a limit on the amount of cash available, any large even number can only be halved, and there are no large odd numbers. This means that across the distribution the losses from halving come out equal to the gains from doubling - because the larger numbers cannot be doubled. If you were to open the first envelope (which you might expect to tell you nothing), you would immediately swap any odd amount and stick with an even amount unless you were sufficiently confident it was a small amount. (Mark Bennet 26 Nov 07) —Preceding unsigned comment added by MDBennet (talk • contribs) 21:35, 26 November 2007 (UTC)

That money comes in discrete units is irrelevant to the problem itself as it is easy to restate the problem using a continuous reward. We can put gold instead of money in the envelopes for example. As gold doesn't come in discrete units (except at atomic levels but then you can't see it anyway) your reasoning based on odd/even amounts doesn't help us at all. We can assume that there is a limit to the amount of money/gold available (even though I don't know what the upper limit would be), but it's irrelevant to this problem too as the subject opening the first envelope doesn't know the limit anyway. It is also possible to restate the problem in such a way that no limit exists, so any solution based on a limit assumption will fail to shed light upon the real cause of the problem. iNic (talk) 23:33, 28 November 2007 (UTC)

What's the problem?

Can someone explain why this is such a "paradox"? It seems to me to be so much mathematical sleight of hand. Representing the payoffs as Y and 2Y or 2A and A/2 are all just misdirection. Just being allowed to play the game means you're getting paid Y. The decision involved is all about the other chunk. So you've got 1/2 chance of getting it right the first time and 1/2 of getting it wrong. Switching envelopes doesn't change that... 1/2*0 + 1/2*Y = Y/2, which, added to the guaranteed payoff of Y gives 3Y/2, which is the expectation that we had before we started playing. 68.83.216.237 03:59, 30 May 2006 (UTC)

The problem is not to find another way to calculate that doesn't lead to contradictions (that is easy), but to pinpoint the erroneous step in the presented reasoning leading to the contradiction. That includes to be able to say exactly why that step is not correct, and under what conditions it's not correct, so we can be absolutely sure we don't make this mistake in a more complicated situation where the fact that it's wrong isn't this obvious. So far none have managed to give an explanation that others haven't objected to. That some of the explanations are very mathematical in nature might indicate that at least some think that this is a subtle problem in need of a lot of mathematics to be fully understood. You are, of course, free to disagree! INic 21:17, 30 July 2006 (UTC)

Spot the flaw? Okay.

1. Denote by A the amount in the selected envelope.
2. The probability that A is the smaller amount is 1/2, and that it's the larger also 1/2

And there's the flaw.

The 50/50 pick isn't between a doubled value and the halved value. There are four situations here not two. Either we have an initial condition of A & 2A and you picked A with a 50% chance, or we had an initial condition A & 0.5A and you picked A with a 50% chance. But you have no way of knowing the relative probabilities of the two initial conditions: between A/2A and A/0.5A, so you can't calculate the combined expectation value.

Jon Hurwitz 212.159.88.189 (talk) 13:33, 3 April 2009 (UTC)

The problem is logically flawed

The solution of the paradox is that it is incorrect to assume that, with just two amounts of money available, you have a fixed amount of money in your hand and still the possibility of getting a larger and smaller amount by swapping (this isn't changed at all by opening the first envelope).

As an example, assume that there are two envelopes on the table, one with $50 and one with $100. If you choose the $50 envelope first, you gain $50 by swapping, and if you choose the $100 envelope first, you lose $50 by swapping, so on average you neither gain or lose anything.

The formula resulting in a gain of 5/4 of the original amount applies only if there are three amounts of money available (with the ratios 4/2/1) and you have the middle amount in the first place. So if you have $50 and there are $100 and $25 on the table you gain (0.5*$100 +0.5*$25 -$50) =$12.50.
If on the other hand you have initially $25, you gain (0.5*$100 +0.5*$50 -$25) =$50 ,
whereas if you have the $100 initially, you lose (0.5*$50 +0.5*$25 -$100) = -$62.50 ,
so on average you wouldn't gain or lose anything either with this 'three envelope' situation.

Thomas

The problem of measuring gain/loss

I suggest that the problem exists with the fact that the "gain" or "loss" of the envelope holder is being considered. Assuming A is the smaller amount in one of the envelopes, then the average amount of money to be holding after picking the first envelope at random is:

${1 \over 2}A+{1 \over 2}2A={3 \over 2}A$

Contrarily, assuming A is the larger amount in one of the envelopes, the average amount of money to be holding after picking the first envelope at random is:

${1 \over 2}A+{1 \over 2}{A \over 2}={3 \over 4}A$

In both cases, after having picked the first envelope, the envelope holder is unable to determine whether the amount in his/her envelope is the greater or lesser amount. If he/she were to assume that he/she had the average amount, in both cases, he/she would realise that he/she had the same amount to gain/lose in switching envelopes (${A \over 2}$ and ${A \over 4}$ respectively).
--TechnocratiK 16:56, 22 September 2006 (UTC)

Another proposed solution

Motion to include this in the main article... its proposed by me. All in favour say "I"... seriously though, please give me feedback (rmessenger@gmail.com if you like), and please read the whole thing:

First, take A to be the quantity of money in the envelope chosen first, and B to be that of the other envelope. Then take A and B to be multiples of a quantity Q. Even if the quantity in A is known, Q is not. In this situation, there are two possible outcomes:

(I) A = Q, B = 2Q
(II) A = 2Q, B = Q

The question is whether we would stand to benefit, on average, by taking the money in B instead of our random first choice A. Both situations (I) and (II) are equally likely. In situation (I), we would gain Q by choosing B. In situation (II), we would lose Q by choosing B. Thus, the average gain would be zero:

${1 \over 2}Q+{1 \over 2}(-Q)=0$

The average proportion of B to A is irrelevant in the determination of whether to choose A or B. The entire line of reasoning in the problem is a red herring. It is true that the average proportion of B to A is greater than one, but it does not follow from that determination that the better option is to choose B.

For Example: Assume one of the envelopes contains $100 and the other $50. The two possibilities are:

(I) A = $50, B = $100
(II) A = $100, B = $50

If you repeat the above event many, many times, each time recording the following:

${R_{i}}={{B_{i}} \over {A_{i}}},{A_{i}},{B_{i}}$

then ${\bar {R}}$ will approach ${5 \over 4}$, and both ${\bar {A}}$ and ${\bar {B}}$ will approach $75. The value of ${\bar {R}}$ is totally irrelevant. What is relevant is that ${\bar {A}}={\bar {B}}$. This means that it makes no different which envelope you choose. On average, you will end up with the same amount of money, which is the expected result.

Step 8 assumes that because ${\bar {R}}\neq 1$, one will gain on average by choosing B. This is simply and plainly false, and represents the source of the "paradox."

I like your reasoning! The only reason I deleted your addition of it was that it was original research, and that is strictly forbidden. However, it's not forbidden here at the talk page (but not encouraged either). I think you should elaborate on your idea some and try to get it published somewhere. What I lack in your explanation right now is a clear statement of the scope of your reasoning; how can I know in some other context that the expected value is of no use? I don't think you say it's always of no use, right? And another thing that is interesting with your solution is that you say it's correct to calculate the expected value as in step 7, only that it's of no use. But if it's of no use, in what sense is it correct? iNic 22:46, 13 November 2006 (UTC)

No, I was wrong about that. The expected value calculation is done wrong. This is because we are supposed to be comparing the expected values of two options. One option is to keep A, the other is to go with B. If we do two different EV calculations for both, we will see that we can expect the same value for each. Take Q to be the smallest value between the two envelopes. There are two equally likely possibilities for each option: we hold the greater amount, or we don't. If we hold the greater amount, we hold 2Q, if not, we hold Q; as such:

$E_{A}={1 \over 2}Q+{1 \over 2}2Q={3 \over 2}Q$

$E_{B}={1 \over 2}Q+{1 \over 2}2Q={3 \over 2}Q$

$E_{A}=E_{B}={3 \over 2}Q$

They are equal, so we can expect the same amount of money regardless of whether we keep A or go with B. Amen.

And, their equation for $E_{B}$ is wrong for one simple reason: the value of A is different in both of the two possibilities. As soon as the two envelopes are on the table, the value of Q never changes. Since we don't know Q from opening one envelope, A must take on two different values in the two terms of their EV equation. A is not a constant in the equation even if we know what it is. Here's their equation:

$E_{B}={1 \over 2}{1 \over 2}A+{1 \over 2}2A$

In the first term, A=Q, in the second, A=2Q. Since Q is constant, A must change. An EV equation must be written in terms of constants. Since theirs isn't, it's wrong.

Think of it this way: since there are two different possibilities, their are two different A's. You don't know which one you have:

Possibility 1: $A_{1}=2Q,B_{1}=Q$

Possibility 2: $A_{2}=Q,B_{2}=2Q$

So their EV equation would look like this:

$E_{B}={1 \over 2}{1 \over 2}A_{1}+{1 \over 2}2A_{2}$

OK, so this means that you now agree with all other authors that step 7 and not step 8 is the erroneous step in the first case? And you seem to claim that step 7 is the culprit even in the next case where we look in the envelope, as you say that "A is not a constant in the equation even if we know what it is." Does this mean that you disagree with the common opinion that it's step 2 that is the erroneous step there? And what about the next variant? Is the expected value calculation not valid there either? And how can i know, in some other context, that a number, say 1, is not really a constant? You still have to specify how we can avoid this paradox in general. As long as you haven't generalized your idea to a general rule you haven't really proposed any new solution, I'm afraid. iNic 02:26, 15 November 2006 (UTC)

Your right. But its simple: if you forget probability for a second, and imagine that as soon as the envelopes are on the table, you repeat the possible outcomes of the event millions of times. Each time the event takes place, there is some $A_{i}$ and $B_{i}$ which are the A and B values you got that time. These are different each time you do it. So as a simple rule, they can play no part in an expected value calculation; unless A actually was a constant. Here's where it gets interesting: It seems like the same situation, but its not. If we imagine A is a constant, and if the same event were repeated, A would not change, as is explained in steps 1-6. If you model the situation this way (I have done computer models for both), such that the experimenter, if you will (the computer), first chooses a value for A, then randomly selects a value for B as a function of A based on steps 1-6, then the expected value of B actually does end up equaling five fourths of the EV of A!

If instead you tell the experimenter to first choose a value to be the smallest value, then choose which envelope to put it in, and put two times that in the other: The result is as you would expect for the real life scenario. So while the possibility of B>A is the same for each instance of both situations... and the two possible values of $B_{i}$ with respect to $A_{i}$ are the same, 'tis not the same situation.

I can make it even clearer how there are two different situations being represented by separating out the probabilistic variable:

Starting with the true event: the experimenter chooses a smallest value, call it Q, and puts it in one envelope, and puts 2Q in the other. Then, lets say you flip a coin: if it's heads, $Z_{i}=0$, if tails, $Z_{i}=1$ for instance... now observe:

$A_{i}=(1+Z_{i})Q$

$B_{i}=(2-Z_{i})Q$

$B_{i}={{2-Z_{i}} \over {1+Z_{i}}}A_{i}$

As you can see, now we have an unchanging algebraic expression for all the variables. Its easy to see the results of the coin toss simply dictates which envelope you pick up. Please note that (referring to the last expression) $B_{i}$ is either (1/2)A or 2A with equal probability. Its also easy to write our expected value expression! There is only one (well understood) probabilistic event: a coin toss. We know the expected value of Z is exactly 1/2. Plug it in and we get what we would expect: $E_{B}=E_{A}=(3/2)Q$, which holds up regardless of whether Q is the average of a constantly changing value, or simply a constant through all instances. NOW, the "OTHER EVENT":

The experimenter first chooses a value for A, then chooses $B_{i}$ based on steps 1-6. If we accommodate the coin toss, we get:

$B_{i}=({1 \over 2}+{3 \over 2}Z_{i})A$

First, notice that if Z=0, $B_{i}=(1/2)A$, and if Z=1, $B_{i}=2A$, and since the probability of either is equal, we would expect that this is identical to the true event, but it's not! If we try to solve for the expected value of B, we plug in our known expected value for Z and what do we get? five fourths. What a surprise! So its easy to see that, while these too situations appear similar, the similarities break down when Z assumes value other than 0 or 1. It's like saying: I'm a function with roots at 2, 4 and 7.. what function am I? There can be many seemingly similar situations at first glance, with the primitive tools used to analyze this problem. But they behave differently.

So the moral: If it changes from instance to instance AND its expected value isn't already known, it CAN'T be used as a term in an expected value calculation! Just as you'd expect! We can't even use $B_{i}$ to find the expected value of B! because if an EV equation was written in terms of a variable that changes all the time, then the expected value would always be changing!

And specifically addressing the problem: Step one fine IF we remember that they have set A to equal the money in the envelope THIS TIME. Step 2 is testably TRUE. Steps 3-5 are stating the obvious. Step 6 is the problem. It is true within the parameters of step 1: that is, AT EACH INSTANCE, half of the time, B=2A, and the other half, B=(1/2)A. But this relates specifically to values of $B_{i}$ and $A_{i}$. As I stated, if we want to calculate expected value in this type of situation, It must be written in terms of ALREADY KNOWN expected values! We don't already know the expected values of either A or B, hence we can't squeeze blood from a turnip!

In the true situation, there are two truely random probabilistic variables: Q and Z. A and B are simply functions of Q and Z (Z being the binary decision, Q being the random number). In their situation, they took the random probabilistic variables to be A and Z, and let B be a function of those. This is different because it assumes that the probability distribution of A is totally random, when in fact it depends on that of the true independent variables (the ones that are actually chosen at random!!!) A was never chosen at random! only Q and Z are!

When determining the expected value of a variable: Express the variable at any instance in terms of the random decisions or numbers that compose it. This equation will always hold true when relating the expected value of the variable to the known expected values of the random decisions or numbers.

i.e. suppose I said I was giving away between 0-5 dollars to one of two people, AND between 10-15 dollars to the other. You're one of those people, and.. say, Joe is the other. so Y is the amount of money you just got, and J is the amount Joe got. Is it worth you're while to switch with Joe?

First, identify the variables: i am choosing between two people, so there is one binary decision Z (independent). There is the amount of money between 0-5 dollars, call it q (independent), and the other quantity between 0-5 dollars (to be added to 10), call it Q (independent). Y and J are dependent upon these as such:

$Y_{i}=Z_{i}q_{i}+(1-Z_{i})(10+Q_{i})$

$J_{i}=(1-Z_{i})q_{i}+Z_{i}(10+Q_{i})$

And write expected value expressions as follow:

$E_{Y}=E_{Z}E_{q}+(1-E_{Z})(10+E_{Q})$

$E_{J}=(1-E_{Z})E_{q}+E_{Z}(10+E_{Q})$

We know $E_{Q}=E_{q}=2.5;E_{Z}=1/2;$ As such:

$E_{Y}=E_{J}=\$7.50$

The point is, had you assumed your money $Y_{i}$ was an independent variable and somehow tried to define poor Joe's money $J_{i}$ in a given instance in terms of your own, you would have gotten it wrong! Neither J nor Y are independent variables, hence we know nothing of their expected values, SO: if we had an expression for J in terms of Y, we wouldn't be able to fill in the expected value of Y, because its dependent! We would be forced to do something tragic.. assume Y is a constant, and define Joe's money in terms of this magical made up constant! Therein lies the mistake! Note: When I give you your money, you know $Y_{i}$. This doesn't mean you know anything about the infinitely many other possible values of Y!

Why is it that this thorough and indisputable resolution is not in the article? It just doesn't make sense that the original "problem", which is an insult to intelligence, has its own article, and the rational argument debunking the "problem" is disallowed from that article. Instead the article remains, an inane problem with a bunch of inane pseudo-solutions, while the real rational solution will not be submitted for reasons of academic bureaucracy. What happened to you Wikipedia, you used to be cool... Denito 10:28, 28 June 2007 (UTC)

History of the problem

Someone correct me if I'm wrong here, but in the original problem

Two people, equally rich, meet to compare the contents of their wallets. Each is ignorant of the contents of the two wallets. The game is as follows: whoever has the least money receives the contents of the wallet of the other (in the case where the amounts are equal, nothing happens). One of the two men can reason: "Suppose that I have the amount A in my wallet. That's the maximum that I could lose. If I win (probability 0.5), the amount that I'll have in my possession at the end of the game will be more than 2A. Therefore the game is favourable to me." The other man can reason in exactly the same way. In fact, by symmetry, the game is fair. Where is the mistake in the reasoning of each man?

the mistake in reasoning is two-fold. The first mistake is that the probability of winning is not 0.5. If they both are equally rich, then they both have a total wealth of W. The amount in one wallet would be any amount A such that 0 <= A <= W. So the probablity that A is greater than the amount in the other wallet is A/W, which may or may not be 0.5.

Second, if A is greater than B, then A + B < 2A and the amount in his possession at the end of the game cannot be more than 2A.

Fryede 20:42, 11 December 2006 (UTC)

Three envelopes

I don't have a solution to propose but the formula (1/2 * 2A) + (1/2 * A/2) can also be interpreted as: there are three envelops, they contain the amount A/2, A and 2A, you know you are holding the envelop with the amount A, you are given the choice to keep it or swap with one of the other two envelops. So in this case if you choose to swap envelop, the average or expected value is indead 5/4A, but with this interpretation the above formula can be applied only for the first swap.H eristo 00:17, 5 March 2007 (UTC)

I agree with Heristo, in my opinion the paradox is originated because it consider only 2 envelopes, but the envelopes are 3 conteining:

X/2, X, 2X.

Only 2 are actual envelopes, and only the couples: {x/2, x} e {x, 2x}.

If we assume X = 100 e and we consider that we have the same probability to have one of the three envelopes (2 real and one virtual)

$	Probability	Value x Prob.
50	1/3	16.66
100	1/3	33.33
200	1/3	66.66

If we have the first possible couple {X/2, X}:

We will have:

$	Probability	Value x Prob.
50	1/3	16.66
100	1/3	33.33
Sum		50

And will be indifferent to change envelop.

If we have the second possible couple {X, 2X}:

We will have:

$	Probability	Value x Prob.
100	1/3	33.33
200	1/3	66.66
Sum		100

And again will be indifferent to change envelop.

In all possible scenarios will be indifferent to change the envelop.

Salvacam 14:23, 24 April 2007 (UTC)

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Comment

The fallacy or source of error may be in that the formula seems to allow 3 possible amounts for 2 envelopes. There is A, 2A and A/2.

I would suggest assigning only values A and 2A OR A and 0.5A. Using A and 2A: By selecting one envelope, I have a 50% chance of selecting amount A and 50% of selecting 2A. Without knowing what A may be, that gives me odds of winning 1.5A with my initial selection.

If I have selected the envelope holding A, then the other holds 2A. If I selected 2A, then the other holds A.

The two outcomes are equally likely, so the probability is that opposite envelope would contain:

0.5 * A + 0.5 * 2A, or 1.5A.

This is no more than what I am holding. I have no incentive to switch.

Srobidoux 18:19, 13 May 2007 (UTC) srobidoux

Commentary 1

In general, I agree with Chase's answer. However, I also believe there is another, which is more straightforward, and ties in with his.

This paradox results from a wishful thinking bias. It only addresses the "destined to win" sequence, while there is a whole other "destined to lose" possibility that is not being addressed. If there is an apparently correct rationale for the "destined to win" sequence (which this article provides) then there must also be an apparently correct rationale for the "destined to lose" sequence which suffers from the same fallacy in and of itself. Combining the two and balancing them against each other gives you the correct answer of "coinflip". --76.217.81.40 16:21, 29 May 2007 (UTC)

my comment

Suppose that all we know is one envelope contains more money and the other less money. Then there is no paradox, right? This double/half version is just a special case of that, so there should be no paradox there either.

I think it starts off wrong

The steps and resulting formula do not take into account the initial state of the participant. The way I see it, anybody who is presented with this situation first has the option to choose A, B, or N (for Neither). Obviously, since N does not result in any benefit, it is not likely that a rational participant would choose to decline either A or B. However, it is important for N to be included, because it IS an option. The importance of including N becomes apparent when either A, B, or become a loss of benefit. In other words, if the scenario was...

You have $50. Before you are two envelopes that appear identical to you. One contains $100, while the other contains nothing. You may purchase either envelope for $50, but since that is all you have, you may only purchase one. You may also, if you so desire, choose neither and keep your existing $50. ...everything changes. We can now see how valuable having the option to choose N becomes. Now the "player" has a chance to have either $0, $50, or $100. Unlike the original scenario, we have an additional variable to consider, initial condition. The danger of losing money makes the option to not participate a valuable benefit to be considered. While the scenario I presented does vary from the "officially" stated scenario, I still think that the option to consider the initial condition applies. I'm nobody special, but this is my "math statements" for it.

N is a known amount that the participant currently has. A,B are unknown amounts that the participant does not have. N+A or N+B are comfirmed, however, as being greater-than N. Therefore, both A and B are better options than N alone.

I guess I can't represent this in numbers like other people can, but my understanding is that the value of switching is equal to the value of keeping. There is no further benefit, so except for non-rational reasons (greed, random choice, insanity, preference to choose left over right, etc) there should be no switching.

Maybe someone else can tell me how the math works out? - Nathaniel Mitchell 63.239.183.126 20:30, 9 August 2007 (UTC)

Problematic problem formulation

The reason of the paradox is incorrectly formulated problem: We need to notice that we are dealing with a random experiment.

Lets analyze the problem mathematically. First, two definitios:

- A random variable is a mathematical function that maps outcomes of random experiments to numbers.

- Expected value of a random variable is the sum of the probability of each possible outcome of the experiment multiplied by its payoff ("value").

In order to calculate an expected value we must:

1. Define our experiment (Done in 'the setup' part)

2. Declare / Find-out the set of possible outcomes (constant values), and the probability of each outcome (constant probability)

3. Define the random variable (Define a payoff for each outcome of the experiment)

4. Calculate the expected value of our random variable.

Now, regarding the problem, as formulated:

Argument (1) States ‘I denote by A the amount in my selected envelope’.

This argument can be understood in two manners:

a. We are talking about a specific instance of the experiment. A is the amount in the selected envelope in this specific instance, hence, A is constant.

b. A is a random variable that denotes the amount in the selected envelope in our experiment.

Since arguments (4) and (5) deals with the value of A, and not the expected-value of A, we can rule out option (b).

So we are left with the interpretation (a).

Hence, Arguments 1 through 6 are referring to our specific instance of the experiment, and they are flawless.

The problem is with argument (7): it calculates the expected value of a constant. Expected values can only be calculated for random valiables. This is where the problem is incorrectly formulated.

---

Sorry, but even if we are told that envelope A contains $10, the paradox still holds. To our knowledge (so the paradox would say) we stand a 50% chance of swapping to $20 and a 50% chance of swapping to $5 i.e. gain $10 or loose $5. That is what the formula says. So why is this wrong?

You have to show why it isn't 50%, or why it isn't $20, or why it isn't $5, or why we can't add the terms and divide by 2, or some other logical error.

Let's show a case where the formula works: Say I give you an envelope (A) with some money in it. Say that I also flipped a coin to decide whether to put twice/half that amount in another envelope (B). I give you envelope A, do you want to swap to B?

In this case the formula correctly tells you to swap. However the fixed/variable nature of A is just the same as before. What is different about this new game that means we can use the formula?Fontwell (talk) 18:00, 5 September 2008 (UTC)

solution

The first proposed solution solves the paradox, the rest of the article just muddies the issue in my view. Petersburg 21:11, 12 August 2007 (UTC)

I disagree. The first solution states that step 7 is the cause of the paradox. If one denotes by A the value in a selected envelope then A must remain constant even if the value of A is unknown (the amount of money in the envelope is not changed, nor is the amount of money in the other envelope assigned to A). While the mistake becomes obvious in step 7, the critical mistake is done at step 6. If your envelope contains A then the other envelope contains either 2A or 1/2A, but the chance is 100/0 rather than 50/50. After all, one of the envelopes (2A or 1/2A) never even existed. Oddly enough, step 2 still remains true as a way for one to calculate the propability, even if there really is only one possibility. One must keep in mind that since we chose to denote one envelope by A, the 50/50 chance has already happened and determined the value in the other envelope (2A or 1/2A) as well as the value of A. Thus it would be foolish to use these values linked to the past propability and use them with it as if they were not linked. --NecaEum 91.155.63.118 00:31, 26 August 2007 (UTC)

Solution

The flaw in the switching argument is as follows:

In the calculation we assume two different cases. One in which the envelopes contain A and A/2 and one in which the envelopes contain A and 2A. These two cases give four possible combinations of the content of the envelopes:

1. A in my envelope and A/2 in the other.

2. A/2 in my envelope and A in the other.

3. A in my envelope and 2A in the other.

4. 2A in my envelope and A in the other.

In the erroneous calculation of the expected value we use only two of the four possible values and assume that the probability of each of them is 0.5. The correct calculation would be to sum all the four possible values multiplied by 0.25. As we can see that the possible values of both envelopes are the same, the expected value of them is also the same.

First there are infinite cases, second for the paradox it doesn't matter what we multiply the expected value by as long as its the same factor in both cases.Enemyunknown (talk) 08:47, 18 September 2008 (UTC)

More generally

Expected value is defined as the sum of the multiplication of each possible value of a variable by the probability of its occurrence. In this problem we have no information about possible values. The only thing we know is the proportion between the values in the two envelopes. We cannot calculate expected value based on this proportion only. Yet, we can see that every possible value in one envelope is also possible with the same probability in the other envelope; so the expected value of both envelopes is the same. It is easy to be seen in a case where there is a limited number of possible values with equal probability. For example in one envelope there may be any sum from $1 to $1000 randomly and in the other its double. In the example above we took only two values: A/2 and A. But the same is also true if we have infinite number of possible values at any distribution.

--Rafimoor (talk) 22:22, 2 May 2008 (UTC)

One envelope is opened so there is only one envelope for which we calculate expected value and it can be half or double the amount in the first so I don't see how what you wrote is supposed to be an answer.Enemyunknown (talk) 08:47, 18 September 2008 (UTC)

Simple Solution

I propose this very simple solution:

You entered with $0. You just got an envelope with more than $0 in it. Don't debate about switching indefinitely, it doesn't matter which envelope you take, you just got something for nothing. —Preceding unsigned comment added by 129.3.157.107 (talk) 15:18, 8 May 2008 (UTC)

Very simple solution to the 'Two envelopes paradox'

There is one 'unit' of money in one of the envelopes and two units in the other one. Initially the player selects an envelope at random, and if he decides not to switch the return he would expect would be 1.5 units, i.e. the average of the two which is (1+2)/2 = 1.5 units. If the player decides to switch the expected return would be exactly the same, i.e. (2+1)/2 = 1.5 units.

The 'paradox' arises because the player believes he can double his money by switching when he gets the envelope with 2 units in it. If he initially draws this envelope he has no knowledge whether or not this is the higher amount and if he switches will receive half.

This is the strategy of not switching:

Player draws 1 unit (probability 0.5) and doesn't switch. Payoff: 1 unit.

Player draws 2 units (probability 0.5) and doesn't switch. Payoff: 2 units.

Expectation = (0.5 x 1) + (0.5 x 2) = 1.5 units.

This is the strategy of switching:

Player draws 1 unit (probability 0.5) and switches. Payoff: 2 units.

Player draws 2 units (probability 0.5) and switches. Payoff: 1 unit.

Expectation = (0.5 x 2) + (0.5 x 1) = 1.5 units.

It therefore makes no difference whether the player switches or not. —Preceding unsigned comment added by JandyPunter (talk • contribs) 22:55, 24 June 2008 (UTC)

Comments on Comments

There are an awful lot comments showing why it makes no difference to swap envelopes but they are completely pointless. We know there is no gain in swapping. That is why it is a paradox. What we need is a clear explanation of the logical error(s) in the paradox, which the article gives.

Also, there seems to be a certain amount of confusion in the comments about how there are really only two amounts of money and not three, or how really only one condition is true. Now, I may be missing something but I don't think that this is relevant at all and perhaps it is a misunderstanding of how basic game theory works: All betting revolves around several amounts of money, when in fact there will only ever be one - your horse will win or not but it will only do one of them. The point is that your knowledge is partial and so you have to assign a probability to every possible condition even though only one is "really" true.

While I think the main article is probably correct it still doesn't really satisfy - witness the comments.

My own unhelpful thoughts are that the paradox works well and promotes extended unsatisfying explanations because it carefully mixes several factors that individually would not be too bad but together provide a bit of mess. These are

1) The non specific values for both envelopes.
2) The unspecified distribution of values.
3) The unspecified method of choosing which envelope has more in it.
4) The decision of which envelope has the most in it being prior to the player's choosing of an envelope.

Fontwell (talk) 18:38, 5 September 2008 (UTC)

I agree. The "two envelopes paradox" is resolved in the first section of the article and the other sections are merely cruft. The real value of this paradox is that it illuminates an important point in statistical methods--namely, the difference between "expected value" and statistical probability. 192.91.147.35 (talk) 01:30, 11 September 2008 (UTC)

Problem Supposedly Solved

The problem in the first approach in the article is that you're allowing yourself to double the high value and halving the low value, which you can't.

m = the sum of the money in both envelopes

considering both cases separately T_low = (1/3)*m and T_high = (2/3)*m

A₁=(1/2)*T_low + (1/2)*T_high = (1/2)*m

Now since you can't halve the low value, the probability of doubling it is 1 and vice versa.

A₂=2*(1/2)*T_low + (1/2)*(1/2)*T_high = (1/2)*m = A₁

As opposed to the original approach:

A₂ = (1/2)*2*(1/2)*T_low + (1/2)*(1/2)*(1/2)T_low + (1/2)*2*(1/2)*T_high + (1/2)*(1/2)*(1/2)*T_high = (5/8)*m

Illegal parts and the parts that are affected by them are bolded for your pleasure.

I'm not 100% sure of this, but it adds up and it makes sense. And it shows that there is no paradox, only failing math.

PS. Sorry I don't know latex syntax —Preceding unsigned comment added by 83.254.22.116 (talk) 06:14, 10 September 2008 (UTC)

Possible solution

I have some ideas about this paradox

1. A set of envelopes for which it pays to switch does not exist since the distance between the values is the same no matter from which side it is calculated.
   
2. The reason why it seems to be worth it for both siblings to switch is that the losing one always looses half his money while the winning one gains an equal amount to the one he already has, this is simply due to the way relative differences are calculated. The total sum doesn't change of course.
   
3. I think that if conclusions made for the system when having complete knowledge of the outcomes differ from conclusions made with only partial knowledge the former should be considered correct. This here means that since in case in which we know what is the pair there is no benefit in switching there can be no benefit in switching.
   
4. The hardest problem is with the statement that "one can gain A or lose A/2" which certainly appears to be true (and its not the same numerical value as one comment above states which is easily shown - if you picked 4 the other can be 8 or 2). The problem here is that when we pick C we are comparing the pairs [C/2 C] and [C 2C]. For both pairs gain and loss is equal but in the second pair it is twice as high as in the first. This scaling is the source of paradox. The expected value as defined in the article is also shifted, by calculating gain and loss in relation to expected value we can arrive at the correct result: for first pair the loss is (C/2)/(3C/4)=2/3 and for the second pair the gain is C/(3C/2)=2/3. So both values calculated in respect to expected value (=the middle value of the pair) agree.This can be interpreted that only the results obtained for systems which have the same expected value can be meaningfully compared, in this case that means that gain and loss can only be compared inside pairs and not between pairs. Since the expected value is only multiplied by a constant factor the values can be normalized and compared by taking this factor into account. Enemyunknown (talk) 05:00, 14 September 2008 (UTC)
5. Last point rephrased. This problem deals with infinite sequence of possible pairs of values, the paradox appears cause it appears that each pair is equivalent and while we draw C we can assume both [C/2 C] and [C 2C] this is only partly true we can consider both pairs but the result from one pair cannot be directly compared to the result from another pair. [C/2 C] is a pair shifted to the left in the infinite sequence of pairs, this shifting is crucial cause any gain and loss depends on the position of the pair we want to consider in this sequence: [A/4 A/2] [A/2 A] [A 2A] [2A 4A] it is obvious that shifting to left or right changes potential gain and loss by a factor of 2. It is also obvious that none pair in this sequence have gain different to loss. The paradox arises since each value exists in this sequence twice in two neighboring pairs and considering those two adjacent pairs seems like a very intuitive thing to do but this gives rise to the difference in loss and gain. In order to resolve the paradox either potential gain and loss have to be calculated in the same pair or if we want to compare between pairs the gain and loss have to be calculated in a manner independent of the pairs position in the sequence, this can be done by calculating the gain and loss in respect to the sum of envelope contents, this calculation shows that you can always gain or lose 1/3 of the sum you consider is in the envelopes.
6. This is a great paradox
   Enemyunknown (talk) 18:07, 14 September 2008 (UTC)

Please see if i have a point here:

Answer: The problem lies in the two '1/2' in this equation, 1/2(2A) + 1/2(A/2) =5/4A

First case = {A, 2A} Second case = {A, A/2}

The probability of getting the first case and second case, for an inconsistent variable A, is 0.5. The probability of getting the first case and second case, for a consistent variable A in both the first case and second case, is indeterminate.

Example 1:

A is the value of money in ANY letter. A is 50% the bigger amount and 50% the smaller amount.

Example 2:

A is the value of money in THIS SPECIFIC letter. Let's open the letter and for example, we see $2. Let X define the value in the other letter i.e. X = $1, $4 X is not obtained from a fair toss of the coin, and instead, from an unknown source not given in the question. Thus we do not know the probability of $1 and $4 occuring.

Why is there a difference? In example 1, we are asking for the chances of getting the bigger and smaller value. Since we have the same chance of getting each letter, and there are only two outcomes, the probability is 50%. In example 2, we are asking for two specific outcomes of either {$2, $1} or {$2, $4}. The WAY of obtaining these two specific outcomes cannot be determinated. Thus the probability cannot be determined, and cannot be taken as half. —Preceding unsigned comment added by Kuanwu (talk • contribs) 12:10, 13 March 2009 (UTC)

Comments about the Paradox and the Flaw

First, I have no background in classical paradoxes, so I have no insight into the historical classification or proper statement of this problem or puzzle. Accordingly, my comments are based solely on a logical analysis of the description of the problem as stated in the Wikipedia article as of January 2009. Second, the problem as stated appears to be a paradox because (or in the sense that) it appears that the stated reasoning leads to what can be reasoned to be an incorrect conclusion. Third, if the problem is a paradox, then there must be some flaw in the statements or in the logic of the problem.

About this Discussion

I began writing the following discussion while I was in the process of attempting to understand both the subject Wikipedia article and the paradox itself. Initially, I had no intention of writing a discussion of this length, but I found that as I completed a discussion on each topic, that there was inevitably either an omission, a need for further clarification, or another related topic. Regarding the order of the topics, I initially thought that it would be preferable (for both me and the reader) to have a reasonable understanding of the paradox before I attempted to offer credible comments and discussion about the article. I simply underestimated the amount of discussion for me to do that. When I finally completed all of the related topics, I realized the extent to which my discussion had grown and thought I might be able to summarize the lengthy discussion I had created on the analysis of the paradox. I then added the brief description in the following section. I also did not go back and re-edit what I had previously written, so there may be parts that appear to be redundant or repetitive.

A Brief Description of the Flaw in the Paradox

Basically, the flaw in the paradox is in confusing three related probabilities. First of all it is important to note that the outcome for everything is determined at the time the initial envelope is assigned to the player. However, because the player is unaware of the outcome, the objective of the player is to evaluate the probability based on the available information. If no additional information is provided, as in the version of the paradox where the envelope is unopened, then the reasoning by the player must be based solely on the probability of envelope selection, which is represented to be random or 50:50. This probability is known as a “prior probability.”

There are also two related probabilities associated with the swapping of the envelopes. Because the outcome of a potential envelope swap has effectively been already determined, the actual probability for gain versus loss is either 0:100 or 100:0. The remaining probability is the “prior probability” of the swap as perceived by the player. The tendency of the typical reader and the basis for the paradox is for the reader to transfer the 50:50 probability related to the envelope selection to the win-loss probability of the envelope swap.

Regarding the argument that states that the player can win A and only lose ½A, if A is assumed to be unknown, the statement regarding the player having a probability of winning A and losing ½A is true because A has one value in a winning swap and double that value in a losing swap. Overall, however, there is no net gain or loss. In contrast, if A is assumed to be known (or is known), the statement regarding the player having a probability of winning A and losing ½A is true when there is an equal probability of the other envelope containing an amount equal to either one-half or double the known amount. Otherwise, the statement regarding the player having a probability of winning A and losing ½A is false. If there is no way to assess whether the other envelope contains an amount equal to either one-half or double the known amount, the statement regarding the player having a probability of winning A and losing ½A would have to be considered false because it is not known to be true. Accordingly, there can only be a probability of gain when the statement regarding the player having a probability of winning A and losing ½A is known to be true.

There are also other approaches to analyzing paradox. At least three are fairly popular. One approach relies of what is termed a Bayesian probability analysis that involves creating, proving, and disproving equations. If you are not familiar with the terminology and notation, it is easy to get lost. A second approach relies on an analysis of the distribution from which the amounts are selected thereby showing that the distribution is not uniform and has bias for smaller values thereby offsetting the advantage to win A and lose1/2A. Depending on how the analysis is performed, one can also get lost in equations and notation. A third approach relies on the use of a parallel problem using exaggerated values to illustrate a flaw in the paradox. Although the method works to illustrate a flaw, it does not provide a very comprehensive understanding of what happens in the various variations of the paradox.

Problem Variations and Terminology

In an effort to minimize confusion it may be helpful to note that as of January 2009 the main Wikipedia article discusses four variations of the problem. The first three variations are identified as “The problem,” as “An even harder problem,” and as “A non-probabilistic variant.” For the purpose of my comments I will herein refer to these variations as “the primary version,” “the augmented version,” and “the non-probabilistic version” respectively. The fourth variation is identified in the section “History of the paradox.” For the purpose of my comments I will herein refer to this variation as the “historical version.” Accordingly, it would be helpful in future discussions to be explicit in identifying the respective version or variation.

My Analysis of the Paradox

If you already have a fairly good understanding of the paradox, then you may want to skip ahead to the topics that relate to the subject Wikipedia article. If not, hopefully the following discussion will enhance your understanding.

Key Elements in the Statement of the Paradox

Despite the different variations and the different narratives that are use to present the paradox, several key elements are common and necessary to create the paradoxical condition. Although the Wikipedia article presents these in 12 enumerated statements, it is possible to state the paradox in a more simple form. Accordingly, a basic statement of the paradox is as follows:

An amount of money and double that amount of money are sealed into two envelopes. A player is allowed to randomly select either envelope thereby implying a 50:50 probability in the selection. Thereafter, the player is provided the option to exchange his (or her) envelope for the remaining envelope. If the amount in the player’s initial envelope is designated by A, then it would appear that the player has a 50:50 probability in swapping envelops to gain A or lose one-half A, thus providing an apparent advantage in making the swap.

Although the basic version is sufficient to create the paradox, additional steps or provisions can be added to either emphasize or alter the paradox. One variation is to extend the problem with the argument that if it is advantageous to perform the swap the first time, then by the same reasoning it would be advantageous to repeat the swap again, and therefore again and again, infinitum. This variation helps the reader to discover the paradox in case he (or she) hadn’t already. A second variation is to have the player open the initial envelope and view the amount. This variation complicates the reasoning process by precluding the use of the simplest path of reasoning to solve the paradox.

Three Key Elements that Affect the Probability of the Outcome

First of all, in analyzing any version of the paradox including the four versions of the paradox in the Wikipedia article it is important to note several key elements that affect the probability of the expected outcome. First of all, in all conventional versions of the paradox the outcome is predetermined at the time when the player receives (or is assigned) the initial envelope. By outcome I specifically mean whether or not there is a winning advantage in swapping the two envelopes. Accordingly, it is only because the player is unaware of the predetermined outcome that there exists the perception to the player that there is a probability-related event associated with the outcome. For the player not privy to having any additional clues that reflect on the actual outcome, the probability of the outcome is simply based on the reasoning of player to assess the original theoretical probability of the event. I believe “prior probability” is the term that is normally used to describe this probability. A simple case of this would be in calling the flip of a coin after the coin was flipped but with the outcome not yet known.

A second key element that affects the probability of the expected outcome for the player is having additional information that may provide a truer assessment of the prior probability. For instance, in the coin flip mentioned above, if the player notes that the previous ten coin flips were heads, it may be prudent for the player to consider the possibility of a bias in the coin flip or the possible presence of cheating. Because the probability of ten consecutive heads being a random event is one in 1024, it may be prudent for the player to call heads in an effort to take advantage of the possible bias. And, if cheating is suspected, it may be more prudent for the player to simply not play. Obviously, as the number of consecutive heads increases beyond reasonable expectations, the greater concern the player should have for this apparent improbability. And, although I am mentioning this in my discussion for the purpose of being thorough, I do not see this element as being relevant to any conventional version of the paradox.

A third key element that affects the probability of the expected outcome is a player gaining clues or additional information that alters the prior probability of the event. The classic example of this is the Monty Hall Problem. If you are not familiar with the Monte Hall Problem, it may be beneficial to read about it before attempting to understand this paradox. However, with respect to this paradox, if the player were provided the initial envelope containing an amount prior to the amount being assigned to the second envelope (either one-half or double the amount in initial envelope), the probability of the outcome is not the same as in the normal version of the problem described initially. In this case the order of events has been reversed, whereby the second amount is selected after the assignment of an amount to the initial envelope. In this case, from the perspective of the player, the amount in the initial envelope (designated in the article as A) becomes a fixed, though a-still-unknown amount. Accordingly, swapping the two envelopes under this set of condition does, in fact, always provide the alleged potential advantage of gaining A with only the risk of losing ½A as claimed in the original problem.

In summary, the key element in the above case is that the player receives the initial envelope containing an amount A, and then the second enveloped is assigned an amount equal to either ½A or 2A. Thereafter, when the player is provided a choice to participate in a swap, the swap actually provides a possible gain of A with a possible loss of ½A. Furthermore, if the process were repeated, the administrator would have to reassign new amounts to the remaining envelope after each round. If there were no upper and lower limits restricting the allowable amounts, the process could be repeated indefinitely. Accordingly, the player could never fully lose the initial amount and would have the potential to gain an infinite amount. Calculation of the player’s winnings and losses becomes a simple matter of adding the number of successes and subtracting the number of losses. For example, if the player participated in ten rounds where he (or she) won six times and lost four, the initial amount A will have quadrupled to 4A ( 2^(6 - 4)*A ). Also, to avoid confusion in later discussions, remember that because this case has a different order of events, it is not representative of the other variations discussed in the Wilipedia article. Accordingly, this case was discussed solely for the purpose of providing insight.

Knowing the Amount Contained in the Initial Envelope

Another key element that may have the appearance of affecting the probability of the outcome is having knowledge of the amount contained in the initial envelope. This element is somewhat a paradox in itself. If in the conventional version of the problem the player receives the initial envelope, opens the envelope, and then counts the amount, the claimed potential advantage (of gaining A with only the risk of losing ½A) now becomes more difficult to refute. For example, if the envelope contains $20, the player can easily reason that the second envelope either contains $10 or $40. According, it certainly appears that he (or she) can only gain $20 or lose $10, an apparent advantage (though it will be shown later that this apparent advantage does not always result in an actual advantage).

Regarding the paradox of this case, there are actually two arguments that result in a paradox. Both arguments result from reasoning performed by the player. The first argument is that the player should have reasoned (1) that the amounts in the two envelopes were selected prior to the initial envelope being received by the player, (2) that the event that determined the probability has passed, (3) that the amount in the initial envelope had a 50:50 probability of being the greater or the lesser amount, and (4) therefore there is no advantage to the swap. The second argument is that the player should also have reasoned that the simple action of learning the amount contained in the initial envelope is not an event that can logically be seen as having an effect on the outcome.

Accordingly, the above analysis gets to the heart of the fundamental paradox associated with the two-envelope problem. In brief, how can the above example be so obviously true and yet not actually be true? To answer that, let’s return to the previous example. Accordingly, recall that after the player learns that the initial envelope contains $20, the player easily reasons that the second envelope must contain either $10 or $40, and according, reasons that he (or she) can only gain $20 or lose $10, an apparent advantage. The reason the flaw is so difficult to identify is because the above statement is actually true. The flaw in the reasoning is not in the above statement, but in the fact that there is not a 50:50 probability that the other envelope contains either $40 or $10. This can best be illustrated by using several more examples.

Accordingly, suppose that the two subject envelopes were set up to contain $10 and $20. The player then learns that his (or her) envelope contains $20. In this case the actual probability of the player winning in a swap is zero while the probability of losing $10 is 100 percent. However, because the player only knows the amount in the initial envelope and not the amount in the second envelope, he (or she) is unaware of these probabilities. Worth noting is that the probability of winning and losing is not 50:50 as it may appear to the player, but is actually 0:100. By comparison, suppose that the two subject envelopes were instead set up to contain $20 and $40. As before, the player then learns that his (or her) envelope contains the same $20. In this case, however, the actual probability of the player losing in the swap is now zero while the probability of winning the greater amount of $20 is 100 percent. Again worth noting is that the probability of winning and losing is again not 50:50 as it may appear to the player, but is now 100:0.

Next, if we assemble pairs of amounts to use in setting up our sample problem, we would then discover that when the player swaps amounts, smaller amounts will win at a greater frequency while larger amounts will lose at a greater frequency. For example (and consistent with the above examples), suppose that the pool of amount pairs consists of only pairs of $10 and $20 and pairs of $20 and $40. Again, as before, these amounts are not disclosed to the player. Accordingly, there are then three possible amounts for the initial envelope and four possible outcomes when swapping for the second envelope. As before, let’s represent the amount in the initial envelope as A. Therefore, when A equals $10, the gain is $10. When A equals $20, there are two possibilities; there can be a gain of $20 or a loss of $10. And, when A equals $40, the loss is $20. To assess the overall gain or loss, we need only compare the totals. Accordingly, the total gain over these four possible outcomes is $30 ($10 + $20). Similarly, the total loss over the four possible outcomes is also $30 ($10 + $20). Accordingly, there is no net gain as claimed in swapping for the second envelope.

In brief, depending on the frequency distribution of the amount pairs in the pool, the player will statistically in some form win at a higher rate when the known amount is lower and lose at a higher rate when the known amount is higher. Overall, however, the player will win and lose with a statistical probability of 50:50, and therefore satisfy the original conditions stipulated in the problem. Accordingly, although it is true to state (1) that the probability of A being the smaller amount is one-half and (2) that the probability of A being the larger amount is also one-half, once the value of A is known, however, these statements are no longer true for all specific values of A. Although this appears to violate reason, it does not. Accordingly, recall that the outcome was predetermined before the swap and that the subject probability of winning or losing in the swap is that which is perceived by the player, and this probability can be altered by the available knowledge. Accordingly, if the player learns the amounts contained in both envelopes, few, if any, would question that player’s win-loss probability in the swap hadn’t changed.

In summary, regarding the previously raised question of whether having knowledge of the amount contained in the initial envelope has an effect on the outcome, the answer is that it does not. However, as mentioned above, it does alter the probability as perceived by the player and thereby alters the reasoning used by the player to form his (or her) conclusion. For example, when the player doesn’t know the amount contained in the initial envelope, the simpler path of reason is to represent the initial pair of unknown amounts as simply a single pair, for example C and 2C. However, when the player knows the amount contained in the initial envelope, the path of reasoning using a single pair of amounts represented by a single pair of variables is no longer valid, and it then becomes necessary to represent the possible amounts as two possible pairs having a common member, for example ½A, A, and 2A. And, although both methods will yield the same result (that there is no advantage in swapping), the respective variables represent different amounts and therefore the probability statements that are stated in terms of these variables do not have the same meaning and therefore cannot be interchanged. Accordingly, it is because of our tendency to unknowingly switch between these to different paths of reason that we are easily drawn into the paradox. Furthermore, it is also the reason why the paradox is so difficult resolve.

The Truth about the Argument that a Player Can Win A and only Lose ½A

Regarding the argument that states that the player can win A and only lose ½A, this argument can be either true or false depending on the definition of A. If A is assumed to be unknown and is therefore a variable, the subject argument is true. However, because A has one value in a winning swap and double that value in a losing swap, there is no net gain or loss. In contrast, if A is assumed to be known (or actually is known), then there is no longer an associated 50:50 probability of winning or losing for all specific values of A. And, although A is not assumed to be known in the some versions of the paradox, there is a tendency for the reader to think and reason in terms as if it is known. This is especially true when reasoning in terms of the variable A. Also, the fact that the reader can imagine the player having an envelope containing an amount in the player’s possession creates a tendency for the reader to switch from one line of reasoning (based on the amount being unknown) to the other line of reasoning (based on the amount being known). In either case, however, the player can be observed to win a greater number swaps when A is smaller and to lose a greater number of swaps when A is larger, but overall win and lose a statistically equal number of times. So, although the player can win A and only lose ½A on an individual case, the player does not win A and only lose ½A on an overall basis.

Related to the above, one should be clear in understanding that the act of the player learning or knowing (the amount contained in the initial envelope) does not in any way affect the outcome of the swap. However, it does affect the player’s perception of the swap and thereby the line of reasoning available to the player. For example, if the player knows that the lower and upper limits of the amounts in the envelopes are $1 and $1000 respectively, valid reasoning would indicate the following regarding making a swap: If the initial envelope contains $1, the probability of having a gain is 100 percent. If the initial envelope contains $1000, the probability of having a gain is zero. This should be seen as obvious because one of the possible mating pairs is eliminated by the upper and lower limits. By extending the same reasoning, we can also conclude the following. If the initial envelope contains between $1 and 2$, the probability of having a gain is again 100 percent. Similarly, if the initial envelope contains between $500 and $1000, the probability of having a gain is again zero. And, if the initial envelope contains $2 through $500, the probability of having a gain on any swap is 3:1 or 0.75 (equivalent to a 50:50 probability of winning A and losing ½A).

By the same reasoning, if the amount pairs were restricted to being either $10 and $20 or $20 and $40, in performing the swap the player would always win when having an initial amount of $10, always lose when having an initial amount of $40, and would win and lose 50 percent of the time when having an initial amount of $20. In this example the gains and losses both add to $30, and are thereby equal. However, if we only consider the cases where the initial amount was $20, we would observe the player to have won and lost 50 percent of the time for a net gain of $10 ($20 - $10), an average gain of $5 ($10/2), and a probability of gain on any swap of 3:1 or 0.75 (again the equivalent to a 50:50 probability of winning A and losing ½A).

In summary, if A is assumed to be unknown, the statement regarding the player having a probability of winning A and losing ½A is true because A has one value in a winning swap and double that value in a losing swap. Overall, however, there is no net gain or loss. In contrast, if A is assumed to be known, the statement regarding the player having a probability of winning A and losing ½A is true when there is an equal probability of the other envelope containing an amount equal to either one-half or double the known amount. Otherwise, the statement regarding the player having a probability of winning A and losing ½A is false. If there is no way to assess whether the other envelope contains an amount equal to either one-half or double the known amount, the statement regarding the player having a probability of winning A and losing ½A would have to be considered to be false because it is not known to be true. Accordingly, there can only be a probability of gain when the statement regarding the player having a probability of winning A and losing ½A is known to be true.

Altering the Frequency Distribution of the Amount Pairs in the Envelopes

In the commonly stated versions of the paradox, there is no consideration given to the frequency distribution of the amount pairs that are selected to be contained in the two envelopes. However, one of the four versions of the paradox that are addressed in the Wikipedia article (titled “An even harder problem”) uses a distribution that is specifically meant to be non-uniform. In this variation, the probability is determined by a mathematical formula that results in the amount pair containing the lower amount to have a frequency probability of 3/5, while the amount pair containing the higher amount to have a frequency probability of 2/5.

An example of two such pairs having a uniform frequency distribution is the case where the frequency of a pair of envelopes containing $10 and $20 is the same as for a pair of envelopes containing $20 and $40. In this case a player learning that the initial envelope contained $20 would expect a 50:50 probability that the other envelope contained either $10 or $40. By comparison, an example of two such pairs having a non-uniform frequency distribution equal to 3/5:2/5 is the case where the frequency of a pair of envelopes containing $10 and $20 is 3/5 while the frequency of a pair of envelopes containing $20 and $40 is only 2/5. In this case a player learning that the initial envelope contained $20 would expect with a 60:40 probability that the other envelope contained either $10 as opposed to $40. Because it appears that the frequency distribution could have an effect on the outcome of the subject problem, we need to understand the effects of frequency distribution well enough to make that determination.

First, let’s consider the case where the player does not know the amount in the initial envelope. In this case the problem is completely independent of the distribution of the amount pairs. The smaller or larger amount of any amount pair, despite the frequency of distribution of the pair, is equally probable to become the amount contained in the initial envelope. Because the amount is not disclosed to the player, the player has no additional information and must therefore base his (or her) decision to swap on reason. Because sound reason indicates that there is no advantage in swapping an unknown amount, swapping becomes a neutral proposition.

Next, let’s consider the case where the player does know the amount in the initial envelope. In this case, if the distribution is uniform and if the player knows the lower and upper limits of the amounts in the envelopes, then the probability of gain can be reasoned over a portion of that range. For example, if the amount pairs were restricted to being either $10 and $20 or $20 and $40, in performing the swap the player would always win when having an initial amount of $10, always lose when having an initial amount of $40, and would win and lose 50 percent of the time when having an initial amount of $20. In this example the gains and losses both add to $30, and are thereby equal. However, if we only consider the cases where the initial amount was $20, we would observe the player to have won and lost 50 percent of the time for a net gain of $10 ($20 - $10), an average gain of $5 ($10/2), and a probability of gain on any swap of 3:1 or 0.75 (the equivalent to a 50:50 probability of winning A and losing ½A).

Next, let’s consider the case where the player knows the amount in the initial envelope and is also aware of a non-uniform distribution of the amount pairs. Again, let’s use the previously used ratio of 3/5 to 2/5. If again the amount pairs were restricted to being either $10 and $20 or $20 and $40, in performing the swap the player would always win when having an initial amount of $10, always lose when having an initial amount of $40, and would win 40 percent of the time and lose 60 percent of the time when having an initial amount of $20. However, because the frequency distribution for the $10 and $20 pair is 3/5, while the frequency distribution for the $20 and $40 pair is 2/5, we have to adjust the gains and losses to account for this. Accordingly, in this example the gains and losses are now $28 ($10 *1.2 + $20 * 0.8) and $28 ($10 * 1.2 + $20 * 0.8) respectively, are slightly less, but are still equal. However, as before, if we only consider the cases where the initial amount was $20, we would observe the player to have won only 40 percent of the time and lost 60 percent of the time for a smaller net gain of only $4 ($20 * 0.8 - $10 * 1.2), an average gain of $2 ($4/2), and a probability of gain on any swap of 3:2 or 0.60 (the equivalent to a 40:60 probability of winning A and losing ½A).

In summary, altering the frequency distribution of the amount pairs only affects the probability of winning or losing in a swap when the amount in the initial envelope is assumed to be known and when there is also a known probability of the other envelope containing an amount equal to either one-half or double the known amount. When this is true, however, the actual probability of a gain or loss is no longer as previously stated whereby a player can win A and only lose ½A. Other than in this specific case, altering the frequency distribution of the amount pairs does not affect the probability of winning or losing in a swap.

The Four Versions of the Paradox in the Wikipedia Article

The following discussion relates to the four mentioned versions of the problem that are addressed in the Wikipedia article. To spare yourself possible confusion, be aware that my previous discussions above were related to cases where either the setup was altered or the player had learned the amount contained in the initial envelope. In contrast, the versions of the problem that are discussed below are all cases where the order in the setup has not been reversed and the player has not learned of the amount contained in the initial envelope. Because of this distinction, the respective analysis and solution may not follow the same path of reason as mentioned in some of my discussions above.

The Primary Version

As mentioned previously, the statement of the paradox in the Wikipedia article is accomplished in 12 enumerated statements. Although the 12 statements may or may not have some basis in the historically correct version of the paradox, this version is not only inconsistent with the popular versions, but appears to be extend beyond the basic version required to for gaining a fundamental understanding of the paradox.

Regarding the discussion of the flaw in the paradox, the flaw in the primary version of the paradox is explained mathematically in the section identified as “Proposed solution.” In brief, the flaw is attributed to a misrepresented and thereby incorrect calculation of the expected outcome stated in step 7 of the problem. However, I also found it possible to analyze the flaw by the path of reason provided below.

First, if the player reasons that there are only two envelopes containing two amounts, and for the purpose of reasoning simply calls these amounts C and 2C, then the player can extend this reasoning as follows. If initially the player randomly chooses the envelope containing the smaller amount C, then A equals C by definition. If the player then chooses to swap that amount, amount A, for what could be perceived as being either ½A or 2A, the actual swap can only be for amount 2A, thereby providing an apparent gain of A. In terms of the C, however, the swap is equivalent to having swapped the initial amount C for amount 2C. Because the player initially selected the envelope containing amount C, the smaller amount, there is no such amount equal to ½A in this case.

By comparison, if initially the player randomly chooses the envelope containing the larger amount 2C, then A equals 2C by definition. If the player then chooses to swap that amount, again amount A, for what again could be perceived as being either ½A or 2A, the actual swap this time can only be for ½A, thereby providing an apparent loss of ½A. In terms of C, again, the swap is equivalent to having swapped the initial amount 2C for amount C. Similarly, because the player initially selected the envelope containing 2C, the larger amount, there is no such amount equal to 2A in this case. Accordingly, despite the truth to the statement that the player can gain A and only lose ½A, we can see that this statement is only true because A has twice the value in a loss than in a win. In terms of C, the player can only win or lose the same amount, amount C.

The source of confusion creating the paradox is in being drawn into false reasoning based on believing that because the amount contained in the initial envelope can be represented by A, that A must be a fixed amount. It is easy to be drawn into this trap because the player can visually see and hold the envelope or, at least, imagine doing so. In this context it is then difficult to revert back to the logic and reason that the outcome was predetermined at the time the initial envelope was assigned.

Also worthwhile noting is that statement 6 in what is called “The switching argument” (stated as, “thus, the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2”) is very open to misinterpretation. The truth of this statement is dependent on the precise definition of variable A. As stated previously, although it is true to state that (1) the probability of A being the smaller amount is one-half and (2) the probability of A being the larger amount is also one-half, once the value of A is known, these statements are no longer true for all specific values of A.

The Augmented Version

In the Wikipedia article the augmented version of the paradox alters the frequency probability of the amount pairs that contain a common amount. For instance, in the primary version of the problem the frequency probability of a pair of envelopes containing $10 and $20 is the same as for a pair of envelopes containing $20 and $40. In this variation, the probability is determined by a mathematical formula that results in the pair containing the lower amount to have a frequency probability of 3/5, while the pair containing the higher amount to have a frequency probability of 2/5. Doing this causes some confusion. Although it might appear that this has an affect on the overall 50:50 win-loss probability of a swap, it only does so in certain conditions. Tracking this distinction appears to add another layer to the confusion already present in the original version of the paradox.

Regarding the mathematical formula ( 3/5 * x/2 + 2/5 * 2x = 11/10 x ) that is used to calculate the expected outcome to be 11/10 x, the formula is incorrect for the same reason that the previous mathematical formula ( ½ * 2A + ½ *A/2 = 5/4 A ) was incorrect in the primary version of the problem. Interestingly, in the primary version of the problem the flaw in the formula was uncovered and was explained in terms of the expected amount contained in the two envelopes. The formula for that calculation was ½ * C + ½ * 2C = 3/2 C. For example, if C is equal to $10 and 2C is equal to $20, then the average or expected value is $15.

If we now apply this same reasoning to the augmented version of the problem, the formula becomes 3/5 * C + 2/5 * 2C = 7/5 C. And, if again we use the same values for C and 2C, the average or expected amount contained in the envelopes is $14. Because the frequency distribution of $10 is stipulated to be 3/5 while the frequency of $20 is stipulated to be the remaining 2/5, the expected amount is correct. Accordingly if this reasoning is applicable and sufficient for the “primary version” of the problem as stated in the Wikipedia article, it should also be applicable and sufficient for the “augmented version.” The article simply does not address this issue and instead discusses reasoning along a completely different line. However, despite all of the discussion, it certainly appears that the solution should be identical to that of the primary version of the problem.

Regarding a solution to uncovering the flaw in this version of the paradox (if that is the objective), I also found it possible to analyze the flaw by using the same path of reason I used in the primary version. Accordingly, I provided that path of reasoning below.

First of all, to minimize confusion over the frequency of the pairs of amounts and the amounts themselves, let’s assign designations for everything. Accordingly, let’s call the pair containing the lower amount X and the pair containing the higher amount Y. In terms of the common member B, it follows that X represents B/2 and B, while Y represents B and 2B.

Next, suppose that the envelopes contain the pair designated as X. And, as before, if the player reasons that there are only two envelopes containing two amounts, and for the purpose of reasoning simply calls these amounts C and 2C, then the player can extend this reasoning as follows. If initially the player randomly chooses the envelope containing the smaller amount C, then A equals C by definition. If the player then chooses to swap that amount, amount A, for what could be perceived as being either ½A or 2A, the actual swap can only be for amount 2A, thereby providing an apparent gain of A. In terms of the C, however, the swap is equivalent to having swapped the initial amount C for amount 2C. Also worth noting is that in terms of B, C equals B/2 and 2C equals B. Therefore the swap was also equivalent to having swapped the initial amount B/2 for amount B.

By comparison, if initially the player randomly chooses the envelope containing the larger amount 2C, then A equals 2C by definition. If the player then chooses to swap that amount, again amount A, for what again could be perceived as being either ½A or 2A, the actual swap this time can only be for ½A, thereby providing an apparent loss of ½A. In terms of C, again, the swap is equivalent to having swapped the initial amount 2C for amount C. Again, worth noting is that in terms of B, C still equals B/2 and 2C still equals B. Therefore the swap was also equivalent to swapping the initial amount B for amount B/2. Accordingly, in terms of either B or C, the gain (B/2 or C) and loss (B/2 or C) is the same.

Next, we simply repeat the same reasoning for the pair of amounts designated as Y. Again we would expect to get the same matching gain and loss. However, now worth noting is that in terms of B, C equals B and 2C equals 2B. Accordingly, in terms of either B or C the gain (now B or C) and loss (B or C) is the same.

In this version of the problem the source of confusion creating the paradox extends one level deeper than in the classic version. Accordingly, not only is there the confusion of being drawn into false reasoning based on believing that because the amount contained in the initial envelope can be represented by A, that A must be a fixed amount, there is also the draw to confuse the probability of the frequency of the amount pairs with the probability of the frequency of the amounts within a pair.

The Non-Probabilistic Version

In the Wikipedia article the non-probabilistic version is stated in two enumerated statements as follows: 1. Let the amount in the envelope chosen by the player be A. By swapping, the player may gain A or lose A/2. So the potential gain is strictly greater than the potential loss. 2. Let the amounts in the envelopes be Y and 2Y. Now by swapping, the player may gain Y or lose Y. So the potential gain is equal to the potential loss.

Although the Wikipedia article states that “so far, the only proposed solution of this variant is due to James Chase,” I find this confusing and a somewhat difficult to believe. This variation does not appear sufficiently different from the primary version for the same reasoning to not apply.

Regarding the truth of the two enumerated statements, statement 1 is false and statement 2 is correct. Accordingly, the part of statement 1 that is incorrect is the stated conclusion (“so the potential gain is strictly greater than the potential loss”). The flaw in the conclusion is the result of faulty reasoning between the previous statement (“by swapping, the player may gain A or lose A/2”) and the stated conclusion. The suggested reasoning is that because the player may gain twice the amount that he (or she) may lose, that there must be a potential gain. This conclusion is, in fact, incorrect, because it does not consider the offset created by the fact that swapping larger values of A will result in a greater number of losses than gains, and that swapping smaller values of A will have a greater number of gains than losses. Although neither of the enumerated statements specifically claims the probability for the second envelope to contain either ½A or 2A to be 50:50, it is implied. Accordingly, the implied 50:50 probability for the second envelope to contain either ½A or 2A, however, does also imply that each and every specific values of A must also have that same 50:50 probability. The details of this were discussed in greater detail previously.

Regarding the solution to the paradox, if the player initially chooses the envelope containing the amount Y, then A equals Y as defined in statement number 1. Accordingly, the player can only gain an amount Y in swapping for the remaining envelope containing the amount 2A. In terms of amount A, the player effectively swaps amount A for amount 2A. Because the player initially selected the envelope containing the amount Y, the smaller amount, there is no such amount equal to ½A. By comparison, if the player initially chooses the envelope containing the amount 2Y, then A equals 2Y as defined in statement 1. Accordingly, the player can only lose an amount Y in swapping for the remaining envelope containing the amount ½A. And, in terms of amount A, the player effectively swaps amount 2A for amount A. Similarly, because the player selected the envelope containing 2Y, the larger amount, there is no such amount equal to 2A. According, the player can only gain or lose the amount Y, and the reasoning is an exact parallel to the reasoning in the primary version of the paradox.

Regarding the paradox of the non-probabilistic version being difficult to resolve as claimed in the article, the above explanation does not indicate that to be true. Perhaps there is some misunderstanding or perhaps this issue requires further research.

The Historical Version

In the Wikipedia article the historical version of the problem that is characterized by the narrative of two strangers wagering over the amount of money in their wallets appears to be another version of the Necktie Paradox rather than a version of the two-envelope problem. However, the historical version differs from the Necktie Paradox in that each man may or may not possess information regarding the contents of his own wallet that may alter his own advantage or disadvantage in the wager. Accordingly, if both men are oblivious to the contents of their wallet, the structure of the wager becomes identical to that of the Necktie Paradox. However, if either man is aware of the contents of his wallet, that knowledge can have a significant effect on the decision to partake or not partake in the wager.

For instance, if one of the men were carrying more than an expected amount of money in his wallet, he would be wise to not partake in the wager. Similarly, if one of the men were carrying less than an expected amount of money in his wallet (for example, no money), he would likely have a winning advantage. Also, because either man can have the ability to base the decision of whether to make the wager on his knowledge of the contents of his wallet, the problem provides no rationale for providing credibility to having a 50:50 or random probability for either man winning the wager. However, in the interest of focusing only on the paradox portion of the problem, it would probably be more prudent to play along with the intent of the narrative and not be picky about the unintentional flaws in the narrative.

Despite the history of how the two-envelope problem may have evolved, I disagree that the two-envelope problem is simply another version of the two-wallet problem. First, there is the problem of correlating the parenthetically stated “½” probability to the narrative as I mentioned above. Second, even though neither man knows the amount in the wallet of his rival, either man is not precluded from being fully aware of the amount in his own wallet. And, assuming that this amount is reasonable and thereby does not completely override the presumed 50:50 probability to win or lose, one man knowing the amount does affect the path of reasoning to be distinctly different from the two-envelope problem. Third, the two-envelope problem specifically addresses amounts that differ by a factor of two. Although this does not alter reasoning in the problem, it does simplify the discussion when attempting to represent the amounts as variables. And fourth, because the two-envelope problem involves amounts that are unknown and undisclosed, it allows for the inclusion of the argument that proposed a continuous and endless exchange. That same argument is more difficult to propose in the two-wallet problem because the narrative does not support the exchange of unopened wallets.

Accordingly, it appears out of perspective to relate the two-wallet problem to the two-envelope problem without mentioning the more closely related Necktie Paradox. This is similar to (though perhaps not as important as) misplacing a species in the evolutionary tree.

Bill Wolf (talk) 21:10, 21 February 2009 (UTC)

There is no point in attacking the article for being inconsistent in its explanations. A Wikipedia article only reports the different opinions that can be found in the published literature. And in a case like this where there are many different opinions by many different authors, we can't expect anything else than a set of incoherent ideas. And that is how it should be. Your personal ideas for solutions are unfortunately of no relevance to the Wikipedia article, unless you publish your ideas in a peer reviewed article first. But please feel free to enhance the article regarding structure and wording, if you can avoid the temptation to be guided by your personal opinions while editing the article. iNic (talk) 19:20, 5 October 2009 (UTC)

Solving the real problem

Sorry to be a spoil-sport but I believe that all the above solutions are beside the point. To resolve the paradox it is necessary to debunk the reasoning given in the section entitled "An Even Harder Problem". I believe that my solution does this, but it may be a case of being a legend in my own lunchtime! Please see http://soler7.com/IFAQ/two_envelope_paradox_Q.htm Soler97 (talk) 00:34, 18 July 2009 (UTC)

Tell me if I'm wrong, but...

I'll use the simplest example from the page

${1 \over 2}2A+{1 \over 2}{A \over 2}={5 \over 4}A$

Is the solution to divide the answer by the coefficients (in this case: 2+½)? The end result is 1/2 for whatever values used for the coefficients, which is what logic dictates. Really, please inform me if this has either already been stated and I just missed it, or if I'm just plain wrong. Buddhasmom (talk) 03:32, 3 May 2009 (UTC)

step 6

step six is step four + five, which are mutually exclusive. you cant just add them together. —Preceding unsigned comment added by 76.208.70.63 (talk) 07:50, 17 July 2009 (UTC)

Frequentist vs Bayesian distinction

Does this distinction apply to the 'even harder version' of the paradox? It seems to me that this rigorous statement of the paradox is amenable to frequentist analysis. Can someone comment on this please. Soler97 (talk) 21:37, 22 July 2009 (UTC)

Why would this variant be more rigorous than any of the others? Can you please explain to me how frequentist reasoning can solve the other variants but at the same time not this one? iNic (talk) 01:18, 2 October 2009 (UTC)

The original two-envelope problem is not a paradox

(Here's a brilliant solution to the "paradox" I recently found on the web)

The original two-envelope problem is not a paradox. It simply reflects the fact that each envelope has a positive value, and the fact that the second choice (after having picked an initial envelope) is not a double or nothing bet.

After picking the first envelope, you are guaranteed at least half of the first envelope (not zero), with the prospect of getting twice that. In order for there to be no expected gain from switching, the second envelope would need to have a 50-50 probability of having either zero dollars or double the first envelope.

What is most notable is that the expect value of the entire game does not change after the first and second choices. Before the first choice, there is a 50% chance of getting $c and a 50% chance of getting $2c. Thus, the expected value of the game -- before any selection is made -- is $1.25c. This doesn't change after the initial choice has been made, as noted in the article.

What would be more paradoxical is if the expected value of the game did change after the selection of the first envelope. Since picking an envelope conveys no information about its value relative to the other envelope, there shouldn't be any change in the expected value of the game.

A variation on this game that destroys the so-called paradox would be to have some one pick one of two colors of chips (e.g., red and green). One of the chips is worth $c and the other has a 50-50 probability of being worth either $0 or $2c. In this case, the expected value of the game before the first selection is $c [or .5($c) .5((.5)$0 (.5)(2c)]; and it remains $c regardless of whether you switch chips or not, since each chip is individually has an expected value of $c. —Preceding unsigned comment added by 216.164.204.65 (talk) 17:40, 18 August 2009 (UTC)

If "there is a 50% chance of getting $c and a 50% chance of getting $2c" I would expect the expected value to be $1.5c for that game, and not $1.25c. I can't follow the rest of the reasoning either, despite the brilliance you see here. iNic (talk) 01:38, 2 October 2009 (UTC)

Why is there so much debate over this?

Surely the problem is simply that the expectation calculation is making two false assumptions - firstly that you pick one value consistently, and secondly (as pointed out in the article) that a third value is introduced. So for example to do a monte carlo simulation coming out with an expected switch value of (5/4)A, you'd first have to set it up so that one of the values was A, and the other envelope switched between (1/2)A and 2A at random between trials. Then, you'd make sure that the envelope with A in it is picked every time. Clearly, neither of these things are true in the actual problem. This is also why consistent switching could not work - once you switch you have not picked A any more. —Preceding unsigned comment added by 131.111.213.31 (talk) 12:49, 12 September 2009 (UTC)

I don't know why there is so much debate over this. But I do know that you just added to the amount of text written about this problem. So you might get the answer to your question by asking yourself. iNic (talk) 01:18, 2 October 2009 (UTC)

Here is the correction to the switch argument

The mistake lies in steps 6 and 7, because they combine two different values of the same variable A into a single equation, an inconsistency I'm personally surprised is still in the article. Sorry for the programming-style format, but here is how I found the proper equation for the expected value, and what it results in.

A=small sum

B=large sum

C=first selection

D=other envelope

First, either C=A and D=B, or C=B and D=A

In either case, the equation for expected value is found as follows, simplified:

A/2+B/2=(A+B)/2=(C+D)/2

So, the correct equation for expected value in this situation is (C+D)/2

or ([Item selected]+[Other Item])/[Number of Items]

From this comes

(C+D)/2=((B/2)+B)/2 = (3B/2)/2 = 3B/4

(C+D)/2=(A+2A)/2 = 3A/2

And 3A/2 = 3B/4 by definition

Then,

(3A/2)/3 + (3B/4)/3 = 3A/6 + 3B/12 = A/2 + B/4 = 2A/4 + B/4 = (2A+B)/4 = (4A)/4 = A

(3A/2)/3 + (3B/4)/3 = 3A/6 + 3B/12 = A/2 + B/4 = 2A/4 + B/4 = (2B)/4 = B/2 = A

where C=A, and since A=A, there is equal probability of either envelope having the larger amount. So, there is no difference whether or not one chooses to switch envelopes.

Please correct me if I made any mistakes.

Extremecircuitz^Ubox —Preceding undated comment added 22:27, 25 November 2009 (UTC).

My Solution

It seems obvious to me that arithmetic median is inappropriate to determine expectancy of the amount in the other envelope. Instead use geometrical median:

A: amount in selected envelope B: expected amount in the other envelope

B = SQRT((0.5*A)*(2*A)) = SQRT(0.5*2*A*A) = SQRT(A*A) = A

B = A, or in other words keep your envelope unless you need exercise.

It seems that problem setting question that would fit the arithmetic median expectancy is as follows: "The envelope that you have picked has A money. The other one has C more or C less then your's."

the expectancy math from the original article now makes sense:

B = ((A + C) + (A - C)) / 2 = (A + C + A - C) / 2 = (A + A) / 2 = A

I am very uneducated math-wise. Please respond with arguments.

Dariac —Preceding unsigned comment added by 1.2.3.4 (talk) 05:00, 26 November 2009 (UTC)

Mathematical Fallacy

I just became familiar with the two envelopes problem, while I was searching for another. The solution seem simple and clear to me: Step 8, which states 1.25A > A is mathematically wrong, because A is not a real number. From the description of the problem, expected value of the contents of an envelope does not exist, hence a priori A is undefined. 1.25 times an undefined object is also undefined and is not necessarily greater then the object itself.

As I see, one category of propsed solution is to let a prior distribution, which makes A calculatable, hence removes the paradox. Another category is the observation of A. Suppose we found out that A = $100. Expected value of the second envelope conditional on A = $100, is indeed $125 and you should switch. But again there is no paradox, because you cannot switch back, first envelope is already open. Unless, you regard the interesting fact that you will always take the 'other' envelope as a paradox. I submit that this just intresting, not a paradox.

Solution to original paradox is much simpler

When calculating expected value in:

${\frac {1}{2}}2A+{\frac {1}{2}}{\frac {A}{2}}={\frac {5}{4}}A$

Letter A denotes two different numbers. First A denotes amount of money in the picked envelope in case when it is smaller than in the other envelope. Second A denotes amount of money in he first envelope in case when there is more money than in the other envelope.

So you cannot add two different variables (mistakenly both named A) like that.

In first point A means "amount of money in the picked envelope" but in the above equation first A means "amount of money in the picked envelope provided that we picked envelope with less money" and the second A means "amount of money in the picked envelope provided that we picked envelope with more money". So the symbol A changes meaning in the course of reasoning and that is the cause of the apparent paradox.

That is what an expectation value is.

My concern is that A denotes expectation value of three different random variables across whole reasoning. At the beginning unconditional, and in the calculation two other variables similar to the first one but limited by additional conditions.

What would be you answer to this question? There are three envelopes containing £ 100, £200, £400. You have the £200 envelope. You are offered the option to swap your envelope for one of the others chosen uniformly at random. Do you swap and why? Martin Hogbin (talk) 10:32, 1 June 2010 (UTC)

I swap because expected value of swapping is (£400 + £100)/2 = £250 so it's more then the cost of swapping that is fixed at £200.

I do agree, however, that trying to assign a variable to the value of the unknown envelope that you hold, in the original two enveloped problem, is a bad idea. Martin Hogbin (talk) 10:41, 1 June 2010 (UTC)

It may not necessarily be bad idea. Bad idea is using same letter to denote similar but significantly different concepts along ones reasoning.

But I must agree that original formulation of this paradox by Maurice Kraitchik is much more interesting and it's much harder to spot error there and I think it can't be done without considering distributions. In my opinion problem is that assumption of 0.5 chance of winning is made too hasty because actual probability of winning is dependent on the values of the contents of my wallet and distribution of the value of wallets contents. If I have more then I can win more but I am also less likely to win so it evens out.

Ignorance about conditions you are operating in does not entitle you assume 0.5 probability and expect correct results. It's actually very nice that in this case you can come up with the correct solution without making any assumptions - just by observing clear symmetry of the game.

Do you have any sources for this problem and its solutions? A major problem with the original article is that it was mainly unsourced. Can I also suggest that you sign your posts, just type four tildes ( ~~~~ ) at the end of your text and the system automatically signs and dates it for you. Martin Hogbin (talk) 11:40, 1 June 2010 (UTC)

I'm sorry. I don't have any sources. I just thought this paradox through and these are my conclusions. I'm sorry for not signing but I don't have account here. —Preceding unsigned comment added by 217.113.236.47 (talk) 21:57, 6 August 2010 (UTC)

Not up to Wiki standards

I will confess that I have not read the previous discussion here, but I've got to say this article is not up to Wiki standards. I quote: "There is no solution, which is accepted by all scientists although solutions have been proposed." (Put a "[sic]" after the misused comma.) Well, if there are scientists who cannot see where the misstep is, then there are scientists who are dumb as a box of hammers. My guess is that scientists just have better things to puzzle over. The indefensible mistake is at step seven, although the groundwork is laid earlier, by introducing the symbol A to represent two different quantities in two different contexts. Step six says, "Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2." True enough. We have two correct formulas exhaustively representing possible payouts weighted by probability.

F1: 2A*(1/2)

F2: (A/2)*(1/2)

In F1, "A" unambiguously means the smaller of the two payouts. It is short for "the monetary value of A, given A is the smaller payment." If it means anything else, there is no justification for the formula. Likewise, F2 can only be justified if "A" means the larger of the two payouts. The two formulas cannot be combined as per step 7, because they use the same symbol to mean two different things.

Repair work is easy. Let's call the value of the larger payment "L", and the value of the smaller "S". We know that L = 2*S

In formula F1, it is understood that "A" means the smaller payout. Substitute S for A, giving 2S*(1/2), and in F2, substitute 2*S for A, giving (2*S/2)*(1/2). Now the formulas both use the symbol S to mean the same thing. We may therefore add them up and factor-out S. The result reduces to 3S/2, which is correct.

What is the payout in terms of A? We cannot express it as a product of the form r*A, where r is a real number. We can only express it as a probability distribution, or as an expected value. Jive Dadson (talk) 14:17, 24 December 2010 (UTC)

Edit: The argument uses the symbol A to mean at least three different things. 1) The monetary value of "my selected envelope." 2) The monetary value of envelope, given that it contains the smaller prize, and 3) You know three. Jive Dadson (talk) 14:48, 24 December 2010 (UTC)

Possible Solutions

The switching argument is a form of Ignoratio elenchi in that it does not address the envelope problem as described. Instead, it describes a problem where we are given a fixed amount of money and have the possibility of either walking away with the given amount or of gambling and obtaining either twice or half the initial amount. There are thus three possible outcomes of this game:

1. We leave with the amount in the first envelope (X)
2. We gamble and obtain twice this amount (2X)
3. We gamble and obtain half this amount (X/2)

The probability analysis as given is only valid if all three outcomes can be obtained over repeated trials. Since there are only two envelopes with two (presumably fixed) amounts, it is not possible to have three separate outcomes.

Instead, a more appropriate analysis would label the smaller amount as X and the larger amount as 2X. The expected value of each envelope is thus:

${1 \over 2}X+{1 \over 2}{2X}={3 \over 2}X$

and neither envelope should be preferred over the other. This analysis only permits two possible outcomes over repeated trials: X and 2X, as we would expect from the problem description.

Put another way: if someone were to hand you $100 and then give you the choice of walking away with that money or of flipping a coin and obtaining $200 or $50, you would be well-advised to take the gamble as it has an expected return of $125 compared to the $100 you would obtain without gambling. If you were to repeat this experiment multiple times, each time beginning with $100 and sometimes obtaining $50 after the coin toss and sometimes $200, the probabilities would work out as given in the original analysis. However, in the envelope game this would never happen. If you have $100 in your envelope the other envelope will either always contain $200 or always contain $50 and your expected value will either always be $150 or always be $75. In this case, whether you switch envelopes or not, the expected value (and the value that you will obtain on average over multiple trials) will be constant and unaffected by your decision.

Mathematical Analysis

If we denote the possible outcomes for the first envelope as set ${\mathcal {A}}$ and the possible outcomes for the second envelope as set ${\mathcal {B}}$, and use A and B to represent variables sampled from these sets, the proper analysis then calculates:

$E[A|{\mathcal {A}}=\{X,2X\}]={1 \over 2}X+{1 \over 2}{2X}={3 \over 2}X$

$E[B|{\mathcal {B}}=\{X,2X\}]={1 \over 2}X+{1 \over 2}{2X}={3 \over 2}X$

The envelopes thus have equal expectations and neither is to be preferred over the other.

The paradoxical analysis assumes that the value in the first envelope is X (or 2X) and calculates:

$E[A|{\mathcal {A}}=\{X\}]=X$

$E[B|{\mathcal {A}}=\{X\},{\mathcal {B}}=\{X/2,2X\}]={1 \over 2}2X+{1 \over 2}{X \over 2}={5 \over 4}X$

As noted, this analysis involves three possible envelope values: $\{X,\ X/2,\ 2X\}$, only two of which would actually be obtained in repeated trials of the experiment. As the laws of probability only describe expected outcomes over a large number of trials (see Law of large numbers), the probabilities obtained are not applicable to the experiment as described.

This analysis can be repeated with the assumption that ${\mathcal {A}}=\{2X\}$, but this will require that ${\mathcal {B}}=\{X,4X\}$ in order to provide a positive expectation for the swapping argument. This line of reasoning allows us to demonstrate the error in the infinite swapping argument. If we assume that the first envelope has amount X in it, the presumed possible values for the second envelope are ${\mathcal {B}}=\{X/2,2X\}$. After the swap we have two possible cases:

1. $B=X/2$, in which case $A=X$
2. $B=2X$, in which case $A=X$

Because we began our analysis with the assumption that $A=X$, it should be impossible to find any other value or expectation for A. Doing so gives rise to the paradox, and is only possible by assuming that the two cases instead permit the possibilities:

1. $B=X/2$ and ${\mathcal {A}}=\{X/4,X\}$
2. $B=2X$ and ${\mathcal {A}}=\{X,4X\}$

Such possibilities would result in the infinite swapping argument. However, we have now extended our analysis to include 5 possible amounts in the envelopes: $\{X/4,\ X/2,\ X,\ 2X,\ 4X\}$ and it should be clear that there is a problem in our reasoning. Put simply: the two envelopes only allow for two possible outcomes to be present in our problem. If we assume the amounts to be fixed (that is, they do not change between trials of the experiment or after we have swapped envelopes) an analysis involving more than two possible outcomes is not applicable to the envelope problem as described as these outcomes could not all be obtained in repeated trials of the experiment.

Understanding the nuances of this problem can be difficult, and there does not seem to be any solution which is accepted by all scientists. The confusion and arguments surrounding this problem are not similar at all to those of the Monty Hall Problem (although some writers seem to see a similarity). The infinite swapping argument is a falsidical paradox in that it seems to make sense to the casual observer but upon careful examination is easily proved to be flawed. The infinite swapping argument assumes an ever-changing and internally conflicting domain for the set of possible outcomes of each envelope, and it mis-applies the laws of probability which are intended to provide long-term averages over multiple samples from a constant probability space.

Informal solution

A careful analysis shows that the probability assumptions used in the argument for switching are inconsistent with the laws of (subjective) probability. No rational person could ever have the beliefs which are assumed in the outset of the problem statement. To be specific, it could never be the case that the other envelope is equally likely to be double or half the first envelope, whatever the amount in the first envelope. This silly assumption led to a silly conclusion (namely, keep on switching for ever).

Suppose one of the envelopes could contain the amount a. Then apparently, a/2 and 2a are also possible. From this, also a/4 and 4a must also be possible amounts of money, and so on. It turns out that these amounts must not only be considered possible, but by use of the laws of probability it turns out that all these amounts of money must be exactly equally likely.

This certainly leads to a contradiction with a common sense approach to real world problems. The amount of money in the world is bounded so there definitely is an upper limit to the money which could be in the envelopes. Also, we don't allow for arbitrarily small amounts of money in the real world.

However, one need not invoke pragmatic principles to defuse the paradox. Also in mathematics, it is not possible to have uniform probability distributions over infinite, discrete sets. You just can't divide total probability 1 into an infinite number of pieces which are both positive and equal. So also within the abstract world of mathematics, the paradox is resolved by saying that the 2a or a/2 equally likely assumption, whatever a, can never be true. No assignment of probabilities to all possibilities can have this feature.

Formal solution

We now give the same argument in a more rigorous mathematical form.

In the Bayesian paradigm, a person who is reasoning with uncertainty in a self-consistent way is supposed to expresses his or her uncertainty about the possible pair of amounts of money in the two envelopes according to a probability distribution over all pairs. Everything is determined by the probability distribution of the smaller amount of money, since this fixes everything else (the other envelope has twice that amount, and the first envelope given to our subjective Bayesian is equally likely to be either). Given this probability distribution, we can compute the conditional probabilities that the other envelope contains 2x or x/2, given that the first envelope contains x. Suppose that these two conditional probabilities are both 1/2, for any value of x which could occur. It follows that if a random envelope contains an amount of money between 1/2 and 1, the other is equally likely to contain the doubled or halved amount, which is between 1 and 2, or between 1/4 and 1/2 respectively.

Now the smaller amount of money has to be between 1 and 2, or between 2 and 4, or between 4 and 8, ... or between 1/2 and 1, or between 1/4 and 1/2, or between 1/8 and 1/4 ... where in each case let's include the lower end-point of the interval and exclude the upper end-point. To say it a different way, we express the smaller amount of money in binary and look at the position of the leading binary digit; we now may as well round off the rest of the binary fraction to zeros. Binary 101.111 is rounded down to binary 100 or decimal 4 (a number at least as large as 4, and strictly smaller than 8). Binary 0.00101 is rounded down to binary 0.001 or decimal fraction 1/8 (a number at least as large as decimal fraction 1/8 and strictly smaller than 1/4).

After this rounding down, the only amounts of money possible in either (and both) envelopes are ..., 1/8, 1/4, 1/2, 1, 2, 4, 8, ... and all amounts are possible.

Now suppose that the probability that the smaller envelope (rounded down) has an amount $x=2^{n}$ with probability $p_{n}$, where n is any whole number (negative or positive). The probability an arbitrary envelope has x is $(p_{n}+p_{n-1})/2$, the probability that it is the smaller of the two is $p_{n}/2$. The conditional probability that it is the smaller amount can only be 1/2 if $p_{n}/(p_{n}+p_{n-1})=1/2$, and simplifying this equation tells us $p_{n-1}=p_{n}$, for all n=...-2,-1,0,1,2,...

Thus the amount of money in the smaller envelope (rounded down as explained above) is equally likely to be any of an infinite sequence, but there is no probability distribution with this property: we can't divide total probability 1 into an infinite number of both equal and positive probabilities.

The resolution of the paradox is therefore simple: though the assumptions appear reasonable, they are actually inconsistent with one another. No rational person who describes his uncertainty about the world in self-consistent probabilistic terms could ever believe, given that the amount of money in one of the envelopes is x, that the other envelope is equally likely to contain 2x or x/2, whatever x might be.

In the language of Bayesian statistics the uniform distribution over all (rounded down) amounts of money is an improper prior; improper priors are well known to lead (on occasion) to improprietry; in particular to self-contradiction.

Point of view from probability

From the point of view of mathematics, in particular of probability theory, Smulyan's paradox illustrates why not only possibilities but also probabilities must be taken into account when deciding on action in the face of uncertainty. The two points of view in his paradox correspond to focussing on the amount in the smaller envelope, A and the amount in the first envelope, Y. A single probability model, such as described before, incorporates both points of view in one self-consistent mathematical model. Moreover it incorporates prior beliefs about the amounts of money (and if desired also the utilities thereof) into the decision making process.

Two envelopes without probability: summary of solutions

There are quite a number of competing "solutions" to Smullyan's paradox by logicians/philosophers, generally using ideas from the theory of counterfactual reasoning, and all of them very technical. It is very difficult for a non-expert to give succinct summaries.

The situation: you pick one of two closed envelopes, one of which contains twice the amount of money in it as the other, and both amounts are greater than zero. Smullyan claimed that you can say both:

1. By switching, I gain or lose the difference in the amounts in the two envelopes, thus whether I gain or lose, I gain or lose the same amount.
2. By switching, I either gain an amount equal to what's in my envelope or I lose half of what's in it; these amounts are different.

But how can the amount I gain if I gain both be equal, and be not equal, to the amount I lose if I lose?

Actually, to my (scientist's) mind the paradox merely exposes the inadequacy of common language: write out a mathematical description of the situation and the paradox vanishes. Moreover I think that Smullyan's *problem* is not actually probability-free (though his *analysis* - the arguments by which he arrives at these two statements - is probability free). In the beginning of the problem we are told that we pick one of two closed envelopes. The symmetry of this situation and the arbitrariness of our choice is an ingredient which many would formalize using subjectivist probability. This ingredient tells us that we are perfectly indifferent to swapping our envelope (closed) for the other. Neither of Smullyan's two arguments bring this crypto-probability ingredient into play. Their apparent contradiction shows, to my mind, the meaninglessness of the comparison between what you would win if you would win with what you would lose if you would lose. A statement which is meaningless has a meaningless negation and no contradiction results by arriving at both. Is it worth while to pick apart the argument and say where exactly it goes wrong?

Well that is exactly what is done in the philosophy/logic literature on the problem. I recommend especially Chase (1992) and Yi (1999). They disagree (strongly!) on technical details but both agree that there is no logical argument leading from Smullyan's assumptions, without the probability ingredient, to his two conclusions. Yi "fills in" the chain of formal logical steps which must be behind Smullyan's derivations and shows that both are false. Chase adds the probability ingredient and uses it to patch one of the two derivations in order to get to one of the conclusions - the one which is compatible with our indifference to switching. Richard Gill (talk) 14:22, 28 June 2011 (UTC)

Schwitzgebel and Dever

I have been (trying to) read the Schtwitzgebel and Dever paper carefully. It contains nonsense. It is full of logical non-sequiturs and mathematical errors. It does not explain the two (closed) envelopes paradox.

The main point of the paper is that if

E(Y|Z=z)=a(z) E(X)+b(z) and E(X|Z=z)=E(X), both for all z, then E(Y)=E(A).E(X)+E(B)

Here, A=A(Z) and B=B(Z).

OK, it is easy to check that this is a true theorem. The authors give some "exchange"-type examples where this could be thought to be useful, in the sense that a faulty intuitive reasoning also gives that answer, and the conditions of the theorem are satisfied, so the answer is correct despite the wrong intuitive reasoning.

(Non-)Application to the two envelope problem: in the two envelope problem, we take

Y=amount in second envelope (we want to relate E(Y) to E(X))

X=amount in first envelope

Z=amount in first envelope (the condition)

So *if* the expected value of the amount in the first envelope, given the amount in the first envelope, were constant,* then* it would be the case that: if the expectation of the amount in the second envelope given the amount in the first envelope is 5/4 times the amount in the second, then the expectation of the amount in the first is 5/4 times the expectation of the amount in the second. But the very first "if" is not true, so we aren't obliged to draw the "obviously wrong" conclusion that E(Y)=5/4 E(X).

So first of all, they claim that they have shown exactly where the original deduction that you must switch goes wrong. However just because the assumptions of their "new" theorem are not satisfied in this case, does not mean that the assumptions of someone else's theorem are not satisfied! Their new theorem does not explain at all why the argument leading to "keep on switching indefinitely" fails. Their new theorem simply does not apply to TEP so doesn't explain anything about TEP. They didn't show that the conditions of their theorem are not only sufficient but also necessary. They don't know the difference between A implies B and A is equivalent to B. Amazing. But of course, they are philosophers, not logicians, nor mathematicians.

However there is a simple way to see why the original TEP argument is actually in some sense OK.

Let Y be the amount in the second envelope and X be the amount in the first envelope. By symmetry we know that E(Y)=E(X). The argument purports however to show that E(Y)=(5/4) E(X) and hence you must switch. But there is no contradiction between E(Y)=E(X) and E(Y)=(5/4) E(X) when X and Y are two nonnegative random variables. There are in fact *two* solutions: E(X)=E(Y)=0 and E(X)=E(Y)=infinity. So the conclusion of that logical argument is not "you must switch" but "you may switch". And of course, you may switch back and forth (without opening envelopes) as many times as you like. As a friend of mine said today: TEP (closed envelopes) just shows you that if E(X) is infinite, and I give you X, you'll always be dissappointed.

In words: the assumption used at step 6 does correctly imply that the expectation of the amount in the second envelope is (5/4) times the expectation of the amount in the first. Since we know that both amounts are larger than zero, it follows that both expectations must be infinite. So there is no paradox. You may switch. You may switch as often as you like. It doesn't actually make any difference. When you look in your envelope you'll get less than the expected value, so you're in for a dissapoinment anyway.

I will write to Schwitzgebel and Dever pointing out the logical mistakes and the mathematical errors in their paper. I hope they will be prepared to write a correction note. Richard Gill (talk) 15:14, 12 July 2011 (UTC)

They are trying to solve the philosophical problem in the closed envelope case, not any specific mathematical problem. If you don't understand the philosophical problem but see everything as mathematics this paper is very hard to read indeed. iNic (talk) 07:46, 13 July 2011 (UTC)

They are trying to solve a philosophical problem by doing mathematics. And they fail: the same criticism which they give to other earlier such solutions can be leveled at theirs.

The paper is hard to read because it is badly written and contains a lot of nonsense. Are you a philosopher? It has been cited three times by other authors in order to criticise their solution. I do understand what you seem to want to call a philosophical problem, though I would say that it is a problem of logic, or of semantics.

IN fact, most people should find TEP uninteresting! It is a conundrum about logic. It is only interesting if you are interested in logic and possibly also in semantics and mathematics. There is no vast popular literature on it. There is only a vast technical literature.

And interestingly, there is almost no *secondary* or *tertiary* literature on TEP. It is almost all purely research articles, each one promoting the author's more or less original point of view, and each one criticising earlier "solutions". And so it goes on. The three papers which cite those two young US philosophy PhD's do so in order to criticize their solution, and to propose an alternative. According to wikipedia guidelines on reliable sources, the article on TEP should be very very brief and just reproduce the comments in a couple of standard (undergraduate) textbooks on TEP. EG David Cox's remarks in his book on inference. I have no idea if there is a standard philosophy undergraduate text which mentions TEP. iNic hasn't mentioned one. We should take a look at other encyclopedia articles on TEP. I think I will write one for StatProb, and then wikipedia editors can use it. Survey papers are admissable secondary sources for wikipedia provided they do not promote the author's own research. They are a primary source for the latter.

Ordinary people won't be perplexed by TEP. They know by symmetry that switching is OK but a waste of time (if you don't open your envelope). They don't really understand probability calculations anyway, so they know there is something wrong with the argument, but don't care what.

Regarding Smullyan's version, they also know the answer (it doesn't matter whether you switch or not) so they know which argument of Smullyan's is correct. As writer after writer has stated, the problem of both original TEP and of TEP without probability is using the same symbol (original) or the same words (Smullyan) to denote two different things. It's a stupid problem and has a simple resolution.

Well, and if we are allowed to look in our envelope, then everything is different. But no longer very interesting for laypersons. It turns out to be a fact of life that there are "theoretical" situations (but I can simulate them on a computer for you, they are not that theoretical!) where according to conditional expectation value you should switch, whatever. OK, and this is just a fact about a random variable with infinite expectation value: if I give you X you'll always be disappointed compared to its expectation. But to get its expectation I would have to give you the average of millions and millions of copies of X. Eventually the average will be so large that you'll prefer the swap whatever the value of the X I gave you first. Then there are all kinds of nice calculations about whether or not you should switch given a prior distribution of X and there are cute things about what to do if you don't want to trust a prior ... then you should randomize. It's all rather technical, isn't it. Only interesting for specialists. By the way this is *not* theoretical since I can approximate a distribution with infinite expectation with one with very very very large finite expectation. I can create approximately the same paradox without actually using infinite values. Syverson does that. It's very technical and hardly interesting for ordinary folk. Richard Gill (talk) 16:30, 13 July 2011 (UTC)

Open letter to Schwitzgebel and Dever

I have written to Eric Schwitzgebel and Josh Dever, see [1] (posted on my talk page). If they permit it, I will post their reply on my talk page, too. Richard Gill (talk) 18:44, 17 July 2011 (UTC)

Schwitzgebel and Dever's simple solution

The Anna_Karenina_principle: "Happy families are all alike; every unhappy family is unhappy in its own way." Here it's the same: a logically sound reasoning can be logically sound in only one way, the reasoning which leads to a "paradox" can be wrong in many."

SD not only published a paper in the philosophy literature, but also have a little web page with a simple solution to TEP, [2]. Let me summarize their simple solution here. It's quite good. First let me repeat the original TEP paradox, I'll call it TEP-I:

TEP-I

1. I denote by A the amount in my selected envelope.
2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.
3. The other envelope may contain either 2A or A/2.
4. If A is the smaller amount the other envelope contains 2A.
5. If A is the larger amount the other envelope contains A/2.
6. Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.
7. So the expected value of the money in the other envelope is (1/2) 2A + (1/2)(A/2) = 5A/4.
8. This is greater than A, so I gain on average by swapping.
9. After the switch, I can denote that content by B and reason in exactly the same manner as above.
10. I will conclude that the most rational thing to do is to swap back again.
11. To be rational, I will thus end up swapping envelopes indefinitely.
12. As it seems more rational to open just any envelope than to swap indefinitely, we have a contradiction.

For a mathematician it helps to introduce some more notation. I'll refer to the envelopes as A and B, and the amounts in them as A and B. Let me introduce X to stand for the smaller of the two amounts and Y to stand for the larger. I think of all four as being random variables; but this includes the situation that we think of X and Y as being two fixed though unknown amounts of money. It is given that Y=2X>0 and that (A,B)=(X,Y) or (Y,X). The assumption that the envelopes are indistinguishable and closed at the outset translates into the probability theory as the assumption that the event {A=X} has probability 1/2, whatever the amount X; in other words, the random variable X and the event {A=X} are independent.

Schwitzgebel and Dever seem to assume that that steps 6 and 7 together are intended to form a computation of E(B).

In that case, the mathematical rule about expectation values which is being used in step 7 is

E(B)=P(A=X)E(B|A=X)+P(B=X)E(B|B=X).

This means that we can get the average value of B by averaging over the two complementary situations that A is the larger of A and B and that it is the smaller, and then weighing those two so-called conditional averages according to the probabilities of the two situations. Now the two situations have equal probability 1/2, as mentioned in step 6, and those probabilities are substituted, correctly, in step 7. However according to the this interpretation, the two conditional expectations are screwed up. A correct computation of E(B|A=X) is the following: conditional on A=X, B is identical to 2X, so we have to compute E(2X|A=X)=2 E(X|A=X). We are told that whether or not envelope A contains the smaller amount X is independent of the amounts X and 2X, so E(X|A=X)=E(X). Similarly we find E(B|B=X)=E(X|B=X)=E(X).

Thus the expected values of the amount of money in envelope B are 2E(X) and E(X) in the two situations that it contains the larger and the smaller amount. The overall average is (1/2)2E(X)+(1/2)E(X)=(3/2)E(X). Similarly this is the expected amount in envelope A.

Schwitzgebel and Dever's executive summary is "what has gone wrong is that the expected amount in the second envelope given it's the larger of the two is larger than the expected amount in the second envelope given it's the smaller of the two. Indeed. It's rather obviously twice as large!

This is perfectly correct, and very intuitive. But it's not the only thing that goes wrong, in that case. They take no note at all of the fact that the writer finishes with a solution expressed in terms of A, not in terms of E(A). Their explanation of what went wrong is seriously incomplete.

However there is another way to interpret the intention of the writer of steps 6 and 7, and it is also very common in the literature.

Since the answers are expressed in terms of the amount in envelope A, it also seems reasonable to suppose that the writer intended to compute E(B|A). This conditional expectation can be computed just as the ordinary expectation, by averaging over two situations. The mathematical rule which is being used is then

E(B|A)=P(A=X|A)E(B|A=X,A)+P(B=X|A)E(B|B=X,A).

In step 7 the writer correctly sustitutes E(B|A=X,A)=E(2X|A=X,A)=E(2A|A=X,A)=2A and similarly E(B|B=X,A)=A. But he also takes P(A=X|A)=1/2 and P(B=X|A)=1/2, that is to say, the writer assumes that the probability that the first envelope is the smaller or the larger doesn't depend on how much is in it. But it obviously could do! For instance if the amount of money is bounded then sometimes one can tell for sure whether A contains the larger or smaller amount from knowing how much is in it.

Note that some of the literature focusses on the first explanation (the expected value of the amount in the other envelope is different when it is the larger, than when it is the smaller) while some of the literature focusses on the second explanation (the chance that the amount in the first envelope is the larger cannot be assumed to be independent of the amount in that envelope). The original writer was simply mixed up between ordinary expectations and conditional expectations. He is computing an expectation by taking the weighted average of the expectations in two different situations. Either he gets the expectations right but the weights wrong, or the weights right but the expectations wrong. Who is to say? Only one thing is clear: the writer is mixed up and he is using the same symbol to denote different things! Depending on what we imagine he is really trying to do, we can give different analyses of what are the things he is giving the same name, which are actually different. Is he confusing random variables and possible values they can take? Or conditional expectations and unconditional expectations? Who's to say?

Anyway, this suggests to me that original TEP does not deserve to be called a paradox (and certainly not an unresolved paradox): it is merely an example of a screwed-up calculation where the writer is not even clear what he is trying to calculate, hence there are different ways to correct his deriviation, depending on what you think he is trying to do. The mathematics being used appears to be elementary probability theory, but whatever the writer is intending to do, he is breaking the standard, elementary rules. Whether we call "explaining" the "paradox" an exercise in logic or in mathematics is unimportant. Whether it is performed by philosophers, mathematicians or logicians is irrelevant too. It seems to me that original TEP belongs to elementary probability theory.

The idea that there should be a unique solution to this paradox is a wrong idea. The paradox is not a paradox, it's a mess. There are a lot of ways to clean up a mess. Richard Gill (talk) 16:51, 18 July 2011 (UTC)

The job is not to clean up the mess, that is easy (and can be accomplished much faster than what you have done above). Instead, the job is to pinpoint the erroneous step in the presented reasoning, and to be able to say exactly why that step is not correct and under what conditions it's not correct. And of course when it's correct. We want to be absolutely sure we won't make this mistake again in a more complicated situation where the fact that it's wrong isn't this obvious. In other words, a correct solution should present explicit guidelines on how to avoid this trap once and for all, in any imaginable situation. This is the true two envelope problem, and it is still unsolved. S&D have at least understood this--what TEP is all about--while you together with some of the authors of TEP papers haven't. iNic (talk) 01:26, 19 July 2011 (UTC)

The job is to say which if any steps are wrong. There are several wrong steps. They have all been identified in the past. S&D claim no novelty in their results, only novelty in their focus. No mistake could have been made if the author had distinguished random variables from their outcomes, probabilities from conditional probabilities, expectations from conditional expectations. And if he'ld known the standard rules for computing expectations. So there is no danger of making the same mistake in more complex situations. Philosophers do not do actual calculations for hard real problems. Those are left to the professionals, who do their calculations professionally. Philosophy has a big problem as long as it ignores the fact that standard probability calculus and standard notation and standard concepts were introduced precisely in order to avoid such mistakes (Kolmogorov, 1933, solving Hilbert's so-manyth problem, to provide an axiomatic mathematical foundation for probability theory). TEP is not a paradox, it's a problem for the student. A finger-exercise for the beginner in probability theory!

Anyway, I'm having a nice correspondence with S&D at the moment as well as with two other philosophers / logicians and with several mathematicians. I think some nice new results are coming out but obviously they are not for Wikipedia for the time being. Richard Gill (talk) 01:45, 19 July 2011 (UTC)

OK but I don't get it. If "TEP is not a paradox" and only "a problem for the student" and a "finger-exercise for the beginner in probability theory," why on earth are you, four philosophers and several mathematicians writing on some new results about TEP? You are bringing even more contradictions to the table. iNic (talk) 02:07, 19 July 2011 (UTC)

Because it's what I'm paid for, and because it's fun. You should think of TEP, or rather TEP-1, as a kind of joke. After all jokes are built on getting a surprise. Then new people who like that joke and are creative, create new jokes in similar spirit. The whole TEP franchise should be seen as a running gag. Richard Gill (talk) 14:49, 19 July 2011 (UTC)

I have never thought of TEP as a joke, but it never surprised me either. What have surprised me, however, are all the crazy and contradicting ideas people have come up with to try to explain what is surprising to them. They are often quite amusing. I look forward to read your forthcoming paper! iNic (talk) 23:59, 19 July 2011 (UTC)

And I look forward to hearing your comments and criticisms! I'll be posting something it on my home page soon, and I'll put a mention of it on the TEP talk page. Richard Gill (talk) 01:50, 20 July 2011 (UTC)

TEP-2

Just like a great movie, the success of TEP led to several sequels and to a prequel, so nowadays when we talk about TEP we have to make clear whether we mean the original movie TEP-I or the whole franchise.

The first sequel was based on the discovery that there exist distributions of X such that E(B|A)>A. Necessarily, these distributions have to have E(X) infinite. The resolution of TEP-2 is that you will always disappointed when you get X if you were expecting E(X). In practical terms, you don't expect the expected value at all.

Going back to TEP-I, obviously the expected value of what is in the second envelope is bigger when it's the larger of the two, than when it's the smaller of the two. But at the same time it's the same if all these expected values are infinity. One call this decision theory if one likes, but to me this is just more elementary probability.

Actually there is a kind of "return of TEP-1" side to this part of the story. Maybe the original writer of TEP was a Bayesian and was representing his ignorance about X using the improper prior distribution with density (proportional to) 1/x. Argument behind this: we know nothing at all about the smaller amount x except x>0. Therefore we also know nothing about 1 billion times x, or x divided by pi ... in general, we know nothing about y=cx, for any c positive. So the distribution of X which represents our uncertainty must be unaltered when we transform it to the distribution of cX. There is only one such distribution. Anyway, the remarkable feature about this distribution is that it is now the case that given A=a, B is equally likely equal to a/2 or 2a, so it is indeed the case that E(B|A=a)=5a/4, or E(B|A)=5A/4>A. The expectation value does tell us to exchange envelopes. But this is not a paradox: what is in the other envelope has expectation value infinity. Whether we finally get X or 2X, we'll be disappointed when we compare it to its expectation value.

Improper distributions are a bit tricky but all these things are approximately true if we restrict to the amount x to any range say from some tiny epsilon (positive) to some enormous M, and take the density proportional to 1/x on this range. Our knowledge about x is roughly the same as that about cx provided c is not extremely large or small. The expectation value of X is close to M but X itself is almost never as large as its expectation. For most a, given A=a, B is equally likely a/2 or 2a.

Seen this way the paradox is not really a paradox about infinity, since one can see the paradoxical things happening without actually going all the way to infinity.

Also this story shows that the writer of TEP-1 need not actually have been making a mistake in steps 6 and 7 - he was a subjectivist using the completely logical and conventional but improper prior 1/x to represent *complete* ignorance about x>0. The "wrong" step was step 8 - he jumped to the conclusion that whether or not you should exchange can be decided by looking at expectation values. But if your ignorance about x is complete, its expected value according to your beliefs is infinite. And in that case you don't expect the expectation value. Richard Gill (talk) 15:04, 19 July 2011 (UTC)

TEP-3

Next we start analysing the situation when we do look in envelope A before deciding whether to switch or stay. If there is a given probability distribution of X this just becomes an exercise in Bayesian probability calculations. Typically there is a threshhold value above which we do not switch. But all kinds of strange things can happen. If a probability distribution of X is not given we come to the randomized solution of Cover, where we compare A to a random "probe" of our own choosing. More probability. Richard Gill (talk) 17:12, 18 July 2011 (UTC)

TEP-Prequel

This is of course Smullyan's "TEP without probability". The short resolution is simply: the problem is using the same words to describe different things. But different resolutions are possible depending on what one thinks was the intention of the writer. One can try to embed the argument(s) into counterfactual reasoning. Or one also can point out that the key information that envelope A is chosen at random is not being used in Smullyan's arguments. So this is a problem in logic and this time an example of screwed up logic. There are lots of ways to clean up this particular mess. Richard Gill (talk) 17:12, 18 July 2011 (UTC)

Conclusions

The idea that TEP is not solved, that the "paradox" is still not resolved, is wrong. There are a number of variant problems and each problem has a number of solutions. The different solutions to the same problem do not contradict one another. Original TEP is an example of a chain of reasoning which leads to an obviously false conclusion, hence there must be something wrong with the reasoning. The problem is to identify what is wrong. But there are several things wrong, and there are even several levels at which mistakes are being made. As soon as you have identified any of these mistakes you have given "a" solution. Original TEP was solved long ago, and then on the ruins new paradoxes were invented. There is no reason why that process should ever come to an end, either. It does not mean that TEP is *controversial*. It means that it is fruitful, stimulating. Richard Gill (talk) 01:29, 19 July 2011 (UTC)

It would be very interesting if you could show (in a published paper) how all the different ideas presented so far are logically equivalent. But until we have such a result we must confess that the proposed solutions are very different in character indeed. iNic (talk) 11:24, 20 July 2011 (UTC)

The different ideas here are *not* logically equivalent. That is the whole point. We are given what appears a sequence of logical steps. But if it is logic (or mathematics) it is informal logic (or informal mathematics). Definitions are not given. Hence assumptions are only implicit. The writer does not say at each step which theorem of probability theory, or rule of logic, is being used. The context is missing. Hence one can give a different diagnosis of "what went wrong" by construing different intentions (background assumptions...) on the part of the writer. Different contexts. So indeed: the proposed solution are different in character, and they correspond to philosophers, practical statisticians, Bayesian fundamentalists or whoever (educationalists, cognitive scientists, economists, ...) , looking for a context for the argument, each supposing it to be the context most familiar to themselves and their intended readers, and showing "what goes wrong" if one takes TEP as an argument intended to take plcae in "their" context.

One should compare TEP to the Aliens movie franchise, where each successive movie had a different director who each brought a very personal touch to their take on the basic story line. Richard Gill (talk) 13:56, 20 July 2011 (UTC)

The problem statement does follow absolutely normal rigor for being a recreational puzzle. Choosing one of two given envelopes from a table is very much within standard ways of formulating puzzle statements. The context is there for sure: choose one of two envelopes containing money. Are you really demanding that every puzzle statement in every book containing recreational puzzles have to be written in a way where they at each step refer to a theorem in mathematics or logic? That is just crazy. Or are you demanding this just for this puzzle and not for anyone else? In that case it begs the question. How should we know, in advance, for which puzzles we need to use a totally formalized language to even be allowed to state the question? iNic (talk) 23:29, 21 July 2011 (UTC)

iNic, are you saying that you believe that the TEP is an open problem, that is to say that the paradox has not been fully resolved. If you want to support this assertion, you must come up with a precise definition of the problem that results in a genuinely paradoxical situation (this comment placed by Marting Hogbin?).

This is not an open problem in probability theory or logic but it is an open problem in the philosophical systems where the problem can be formulated; decision theory and bayesian interpretation of probability theory. An open problem is not necessarily a problem for which no solution has been proposed, it can also be a problem where several different solutions have been proposed and where there still are no consensus which one of the proposed solutions is the correct one. For example, before Newton the problem of gravity was an open problem, but not because of lack of different theories of gravitation. iNic (talk) 23:29, 21 July 2011 (UTC)

Decision theory, and subjectivist probability theory, are mathematical theories, not philosophical systems. People coming from different fields (mathematics, philosophy, logic, economics, foundations of probability, education...) see different issues because that's their jobs, to see the issues which matter to them. All the problems seen by people from different fields are all real problems. There is no such thing as "the correct solution" because all (at least, almost all) the solutions which have been offered to all the issues which can be detected are all simultaneously correct. Nothing much new has been contributed for a long time. Just regurgitation of old issues, translations from one language to another of the same thing. Richard Gill (talk) 17:01, 22 July 2011 (UTC)

If decision theory and subjective interpretation of probability are mathematical theories please tell me where I can find their respective (uncontroversial) axiomatizations.

You think more like an art professor than a math professor. Can you mention some other recreational math puzzle where people from different fields are free to invent their own solutions, as one solution is as good as any other? In modern art one interpretation is indeed as good as any other; you're totally free to invent your own interpretation. No interpretation is wrong because no single interpretation is correct, so the approach is not self-contradictory. You seem to apply the same kind of logic here, right? Someone mentioned postmodernism as a joke at the talk page. But maybe it's not a joke in your case after all? Could this in fact be the first example of true postmodernistic behavior in math and logic? Showcasing that 'truth' itself is not objective but subjective, not only in art but in math and logic too? Can someone please rescue me from this nightmare?

I think I have to invite Alan Sokal into the discussion chasing all the irrational demons away! iNic (talk) 22:50, 22 July 2011 (UTC)

iNic, I did not say that one solution was as good as any other. I did not say that "truth" is not objective in mathematics. You used the word "valid", I hope that the TEP page will only report valid solutions of TEP, and you yourself know there are several different ones. There is not one unique way to translate words about an imaginary situation into a mathematical model or into a logical framework. But anyway, neither your opinions nor my opinions are of any relevance. What do the reliable sources say? There are reliable sources from philosophy who treat TEP like a philosophy problem. There are reliable sources from probability theory who treat TEP like a mathematical puzzle. Plenty of them. Axiomatizations of decision theory, subjective probability: ever heard of de Fineti, Savage? Read some books on mathematical economics, consumer choice... ? Richard Gill (talk) 21:25, 23 July 2011 (UTC)

Namedropping is a nice trick when trying to avoid a question, but it doesn't work on me. There are many candidates for competing theories of bayesian probability/decision theory: Ramsey, de Finetti, Carnap, Savage, Jaynes, Jeffrey, Lindley to just name a few. They have all produced theories of bayesianism that differ a lot. Some of them addresses TEP explicitly and they take care of this problem in very different ways, which, of course, is not surprising. This is why TEP is still an open problem within bayesian decision theory / bayesian probability. When the foundations of a subject is still not settled the subject is commonly referred to as philosophy. This is the case here.

I'm glad you're not a postmodernist and I'm glad that you say that your personal views on the matter is of no relevance here. If you can show that the different proposed solutions (so far) are just different ways of translating the problem into a logical or mathematical framework, that would be very interesting. However, this is not the view any of the authors of the sources have, and no paper so far is promoting this view. Very few even mention other authors and their proposed solutions. Those who do, do it only to criticize their ideas. If you could show that all TEP authors on the contrary should embrace each other in a big hug it would be very interesting new research indeed. Until then, please treat this as your own opinion, and your own opinion only. iNic (talk) 21:08, 26 July 2011 (UTC)

Dubious importance

This is most certainly a paradox. Initially confusing, due to flaw in the construction of the argument, but ultimately obvious. There is no point in switching. It's a fun game to think about. However the article states that it's an unsolved problem in mathematical economics. To confirm that, there's a citation from a philosophy journal (seriously - wtf?). I don't understand why is so much space being given to a minor mathematical puzzle. The Balance puzzle is far more interesting and difficult, but no one is going into the kind of interminable detail this article does. I appreciate that there are one or two confused post-docs writing papers about this, but isn't that true of anything (especially in philosophy)? I would like to see a short article about the simple mathematical puzzle, with a simple explanation of the logical error. Delete the references to the boring and irrelevant papers, delete the feather-light reference to Smullyan's book, include one reference to a pop-math book, and move on. Alternatively, if we are really determined to explore every angle, why don't we bring in the postmodernists - we could even include a reference to "Toward a transformative hermeneutics of quantum gravity" (which gets 20 times more citations than Chalmers's, very nice, paper) --Dilaudid (talk) 15:03, 21 July 2011 (UTC)

What is 'the problem' and 'the logical error' in your opinion? Martin Hogbin (talk) 15:57, 21 July 2011 (UTC)

I mentioned the economic problem in the article - I don't think is a problem. The "Two Envelopes Paradox" is that there is a contrived "switching argument" which suggests it is sensible to switch, but this is in clear disagreement with a much simpler and more obvious argument - symmetry. The switching argument is further contrived to say it is sensible to switch ad-infinitum, which is in conflict with basic reasoning (how can it be advantageous to switch twice??). The logical error in the "switching argument" is the sloppy definition of A - is A a single number (in which case we must assign probabilities which are not 1/2 to it being the larger or smaller number) or is A two different numbers (in which case step 7 is a total mess). These kind of issues are common in basic probability if you don't define notation properly. --Dilaudid (talk) 17:01, 21 July 2011 (UTC)

A represents the (unknown) sum that is in the envelope that you hold. Martin Hogbin (talk) 20:16, 21 July 2011 (UTC)

What do you mean by "represent" it? Is it a number? Dilaudid (talk) 21:33, 21 July 2011 (UTC)

Cutting to the chase, if it's a number, then the probabilities are functions (not 1/2). If it's a function, then what is the thing on the right hand side of the expectation? This is why the (painful) concept of random variables was created, to stop people making basic logical fallacies like this. I don't want to upset you guys (it will only drive you further into your positions), but come on, does anyone with any actual knowledge of probability think this is a real "problem"? It's not exactly Dark Matter :) Dilaudid (talk) 22:16, 21 July 2011 (UTC)

Exactly. No one in probability thinks this is a real problem. It's an example of the mess you get into when you don't distinguish between random variables and their possible values and their expectation values, between expectations and conditional expectations, between probabilities and conditional probabilities. Any attempt to translage the argument into correct, conventional, probability terms, exposes one or two or more errors - it depends on where you think the writer was trying to go. Precisely because of this, any attempt to resolve the paradox in purely philosophical terms will be wordy, convoluted, unappealing - ordinary language is not refined enough that we can easily make the distinctions which are so crucial, our ordinary thinking is clumsy when we deal with subtle probability issues. Because the original argument is not formalized there are different interpretations possible concerning what the writer was trying to do. Therefore it will always remain an open problem to state what is the actual error which the writer was making, since we don't know what the writer was thinking. On the other hand it is closed: the writer gets in a mess because he uses the same symbol (or words) to stand for different things, whatever line of thought he had in mind. It was precisely to avoid this kind of mess that modern probability calculus was invented. Richard Gill (talk) 16:53, 22 July 2011 (UTC)

I understand (I think) but the question that I am trying to answer is, 'Is there a simpler resolution of the TEP that we should add to the article'. I do not think the, undoubtedly correct, statements that you both make can easily be put into a simple and convincing argument for the general reader. What seems to happen is that someone tries to add the argument but ends up by simply saying, 'you should not swap if you have more money in your envelope', or 'the average amount you lose must be equal to the average amount you gain'. Both may be true but they do not resolve the paradox, as doing this requires you to show the fault in the presented line of reasoning not show that there is an alternative one..

On the other hand, I am happy to try to work out a way to explain the problem you have described to the general reader. The thing that started me on this thread was that there seemed to be an assertion by some that there was a really simple resolution that we were missing. I do not think this is the case. There is a quick answer for experts (the question dos not make sense, or at least it is not clear what you are trying to assert) but if this is made too simple it can easily become invalid. Martin Hogbin (talk) 09:20, 25 July 2011 (UTC)

Puzzled

Richard, mentions, 'examples of faulty logical reasoning caused by using the same symbol (or verbal description) to stand for different things at the same time' and Dilaudid seems to be alluding to the same thing yet I do not see that resolution of the paradox anywhere in the article. Can someone tell me if that argument is considered a good resolution of the paradox and, if so, could they present the argument clearly here. Martin Hogbin (talk) 11:48, 22 July 2011 (UTC)

We are not here to judge which proposed solutions are 'good' or 'bad.' Our job is to give a short unbiased account of the more common solutions mentioned in published sources. This proposed solution should definitely be mentioned in the article. I think it is the simplest solution and it should be mentioned as the first proposed solution in the article. iNic (talk) 13:57, 22 July 2011 (UTC)

Falk: "The assertion in no. 6 (based on the notation of no. 1) can be paraphrased ‘whatever the amount in my envelope, the other envelope contains twice that amount with probability 1/2 and half that amount with probability 1/2’. The expected value of the money in the other envelope (no. 7) results from this statement and leads to the paradox. The fault is in the word whatever, which is equivalent to ‘for every A’. This is wrong because the other envelope contains twice my amount only if mine is the smaller amount; conversely, it contains half my amount only if mine is the larger amount. Hence each of the two terms in the formula in no. 7 applies to another value, yet both are denoted by A. In Rawling’s (1994) words, in doing so one commits the ‘cardinal sin of algebraic equivocation’ (p. 100)." Richard Gill (talk) 17:11, 22 July 2011 (UTC)

What is your opinion of this argument? Martin Hogbin (talk) 20:18, 22 July 2011 (UTC)

I do not see how Falk's argument does not apply in this situation, when you should swap, once.

You pick an envelope from a selection containing different amounts of money. A person who knows what is in your envelope then tosses a coin and puts either 1/2 or twice that amount of money in another envelope. You have the option of swapping. Would you? I would. Falk's argument seems to me to show that you should not swap and it is therefore wrong. Martin Hogbin (talk) 22:53, 22 July 2011 (UTC)

In standard TEP, when your envelope is the smaller of the two, it contains on average half what it on average contains when it is the larger. This is also Schwitzgebel and Dever's explanation (not of what is a wrong step in the argument, but why its conclusion is wrong). In your scenario, Martin, your envelope is not different by being smaller or larger. Falk puts into words that in standard TEP, what's in either envelope is statistically different when it's the larger than when it's the smaller. The wrong step in the argument assumes that whether your envelope is the smaller or larger is independent of what's in it. In maths: the wrong step is to assume that the event {A<B} is independent of the random variable A. Falk's explanation: this step is wrong, because A is not independent of {A<B}. It seems that SD are the first philosophers who say this. Falk knows it. The mathematicians knew it long before (they know the symmetry of statistical independence; philosophers don't).

Mathematicians writing for mathematicians can leave their conclusions in technical language and perhaps only semi-explicit: their intended readers understand it. But the philosophers don't understand. And vica versa. Ordinary folk understand neither. Hence the wheel is reinvented so many times. Richard Gill (talk) 17:14, 23 July 2011 (UTC)

If you are pointing out that there is no way to arrange the money in the envelopes such that the probability that B contains half of A and the probability B contains twice A are both equal to 1/2 and the situation is symmetrical then I understand and agree. If this is not what you are saying then could you explain some more please. Martin Hogbin (talk) 18:03, 23 July 2011 (UTC)

I'm not saying that. Put x and 2x in two envelopes, then completely at random write "A" and "B" on the two envelopes. The situation *is* completely symmetric. And the chances A contains twice or half what's in B are both equal to 1/2.

What these folks are saying is that it's asking fir trouble to denote both by "a" what's in A when it's larger and what's in "A" when it's the smaller. Call them "2x" and "x" instead and you don't screw up.

Alternativy, learn about conditional probability and do it with maths. The original TEP argument looks like probability theory but isn't. Philosophers think it's purely a problem of[ logic, not of mathematics. Then what I just explained is the solution. But another solution is to do it properly within probability theory. But in the end it all comes down to the same thing. [User:Gill110951|Richard Gill]] (talk) 18:14, 23 July 2011 (UTC)

How does Falk's argument not apply to my case then? Martin Hogbin (talk) 18:26, 23 July 2011 (UTC)

In your case, what's in your envelope when it's the smaller of the two is the same as what's in it when it's the larger of the two. Your envelope is filled first, say with amount a, only after that is 2a or a/2 put in the other envelope. In TEP two envelopes are both filled with amounts x and 2x. After that, you choose one at random. If it's the smaller amount it's x. If it's the larger amount, it's 2x. This is what Falk is saying, and SD, and a load of other people, though I must say they can find incredibly ugly ways to say it. Richard Gill (talk) 20:40, 23 July 2011 (UTC)

I have just added a bit to my statement above. It now says, ' there is no way to arrange the money in the envelopes such that, for every possible value of A, the probability that B contains half of A and the probability B contains twice A are both equal to 1/2 and the situation is symmetrical. Does this make sense now? Martin Hogbin (talk) 19:13, 23 July 2011 (UTC)

That makes sense to me, though some purists would prefer that you added the word "conditional" in front of probability, though.

However, it is possible to arrange things so that it is almost exactly true. This brings us to TEP-2, "Return of TEP", or "Great Expectations". If you arrange things so that what you just asked is almost exactly true, then for almost every possible value of A, the expectation of B is exactly or approximately 5A/4, so it seems one ought to switch. We seem to have a new paradox. However in that case the amount of money actually in envelope A is almost always far, far, less than its expectation value - expectation values are absolutely useless as a guide to rational choice.

You might object that this corresponds to subjective beliefs about the amount in envelope A, or the smaller of the two amounts X, which are totally unreasonable. However, there are a load of reliable sources which do state that this is not only reasonable, it's even prescribed. If you really know absolutely nothing about X, you also, I think, would be prepared to say that you know absolutely nothing about A, and absolutely nothing about X converted from dollars to Euro's or Renbini or Yen. There's exactly one probability distribution which has this property: it's the probability distribution which makes log X uniformly distributed, so log X is a completely unknown real number somewhere from minus to plus infinity.

The thing is that knowing nothing about a positive number means that however large it is, it could just as well be billions of time larger, and however small it is, it could just as well be billions of time smaller. Because of this your expectation value of what it is, is infinite. So you'll always be disappointed when you get to see the actual value x, which is a finite positive number.

I was just now talking about subjectivist probability: probability as rational belief. If you want to think of TEP with frequentist probability, then I just want to say that it is easy to arrange that the properties you listed almost exactly hold true. I can write a program on my computer which will print out two amounts x and 2x to go in the two envelopes, and though the event {A<B} won't be exactly independent of the amount A, it will be as close to independent as you like. We can play the game a billion times and never ever meet an occasion when the conditional probabilities that B=a or a/2 aren't each exactly 1/2 (given A=a). So despite what's written in the literature, the paradox of TEP-2 isn't just a paradox of infinity. Unfortunately wikipedia will have to wait 10 years before we can correct that illusion. Richard Gill (talk) 20:53, 23 July 2011 (UTC)

But surely that is still a paradox of infinity. It is only because the expectation is infinite that you do not expect to actually get it.

Also, I think we agree that Falk does not put the argument very well, especially for the layman. Martin Hogbin (talk) 22:04, 23 July 2011 (UTC)

I agree that Falk makes her point rather poorly.

I disagree that the paradox in the case of infinite expectation values is "just" a paradox of infinity. Consider the following completely finite case. Put 1 ct on the first field of a chess board, 2 on the next, 4 on the next.. Pick a field at random and put that amount in one envelope, twice in the other. Pick an envelope at random... The expectation value of what's in both envelopes is finite. Only once in 64 times is the amount in the other envelope *not* equally likely to be half or double that in the first, given the amount in the first. 63 out of 64 times the expected content of B is 5/4 what's in A. (1 in 128 times it's twice, once it's half). The expectation value of A is close to its maximum. Most times, A is far, far smaller than its expectation value. Also the average of many, many independent copies of A is mostly far smaller than the expectation value. So also in this completely finite case, the expectation value is pretty irrelevant. "The long run" doesn't kick in till long after you're dead. This is not an academic example. In finance, meteorology, astronomy, ... one meets this kind of behaviour.Richard Gill (talk) 18:29, 24 July 2011 (UTC)

Three resolutions

Richard, would you agree that three resolutions cover all reasonable formulations of the problem?

This covers all of TEP with closed envelopes except for the Smullyan "no probability" variant. That variant is a problem of the logic of counterfactuals, a topic on which philosophers disagree. But for each popular formal approach to counterfactual reasoning, there is a solution to the paradox. It's all rather technical, and hardly of popular interest.

TEP when you look in A before deciding is interesting for probabilists and for decision theorists, but hardly controversial. Tom Cover's random probe solution is fun. Richard Gill (talk) 18:58, 24 July 2011 (UTC)

The situation not symmetrical

If the situation is not symmetrical then it can be advantageous to swap, but only once.

But the problem description makes it symmetrical! That's why we know there is no point in swapping, and hence that there must be one or more errors in the reasoning. Richard Gill (talk) 18:33, 24 July 2011 (UTC)

Yes but it still needs to be said. It is easy to come up with formulations where the envelopes are not symmetrical. Martin Hogbin (talk) 22:13, 24 July 2011 (UTC)

Agree. It needs to be said very clearly. Richard Gill (talk) 22:51, 24 July 2011 (UTC)

The expectation is infinite

Intuitive arguments involving infinity can easily fail, one way or another.

It's not a problem with infinity, it's a problem of a probability distribution which is so heavy-tailed to the right that its expectation value is in the far end of the distribution and totally unrepresentative of typical values or even of typical averages of less than astronomically many independent values. In the long run we're dead. The mean value is the infinitely long run average. Just not relevant to the real world. Richard Gill (talk) 18:39, 24 July 2011 (UTC)

What about your 'Second resolution' where you say, 'The average amount of money in both envelopes is infinite. Exchanging one for the other simply exchanges an average of infinity with an average of infinity'? Martin Hogbin (talk) 22:37, 24 July 2011 (UTC)

No contradiction. A distribution with infinite expectation value is merely an extreme example of a probability distribution which is so heavy-tailed to the right that its expectation value is in the far end of the distribution and totally unrepresentative of typical values or even of typical averages of less than astronomically many independent values. Richard Gill (talk) 22:59, 24 July 2011 (UTC)

The expectation is not independent of the sum in the original envelope

This covers cases where the possible sums in the envelopes are limited. It is also my first explanation in the 'Introduction section' (which I will now amend slightly).

This same argument can be presented differently in what some have called the 'simplest' argument. However, the x/2x style argument needs to be presented very carefully to be technically correct. The very simple style presented here previously actually misses the point. If presented properly, this argument is only accessible to experts. Martin Hogbin (talk) 09:16, 24 July 2011 (UTC)

Do you mean expectation or probability? The *probability* that the sum of money in Envelop A is the smaller or larger of the two is (in general) not independent of the amount itself. Equivalently:, the amount is (in general) not independent of whether it's the smaller or the larger. (As my chess board example shows, it can be more or less the case, but only in a situation where the expectation value says nothing about what you can realistically expect in a finite lifetime.) Richard Gill (talk) 18:47, 24 July 2011 (UTC)

Yes I did mean probability, the point being that if the probability of doubling your money is not independent of the sum in your original envelope you must take account of this when calculating the expectation. Martin Hogbin (talk) 22:37, 24 July 2011 (UTC)

Right. One could add to this: intuitively, the more money in A, the more likely it's the larger amount. And conversely: given A contains the larger amount, there's typically more in it than when it contains the smaller amount. Richard Gill (talk) 22:57, 24 July 2011 (UTC)

As you can see, I am making this up as I go along but I think what I actually meant was that if the expected gain ratio (expectation/original value) is not independent of the sum in the original envelope. I think this applies to every finite distribution, even your example and, in my opinion, it is sufficient to resolve the paradox. If every step does not always hold good, the simple but compelling argument fails. Of course, in your example, there are other reasons why you might not choose to swap but what I am trying to do is to find the simplest way, with the fewest different resolutions, to cover every reasonable formulation of the problem. There will always be a continuing cat and mouse game between formulators and solvers for those interested. Martin Hogbin (talk) 08:56, 25 July 2011 (UTC)

The expected gain ratio depends on the amount in the original envelope if and only if the probability the original envelope contains the smaller quantity depends on the amount in it. That's because it's a two point distribution on a/2 and 2a hence the mean is a times (1-P(A<B|A=a))/2 +2P(A<B|A=a) which is the same as a times (3P(A<B|A=a)-1)/2. The expected gain ratio is therefore ( 3P(A<B|A=a) - 1 ) / 2. It depends on a if and only if P(A<B|A=a) depends on a. So if A is bounded this quantity has to depend on a. So yes, also my chessboard example has P(A<B|A=a) depends on a, as I said. It is constant and equal to 1/2 for all possible values of a except for the smallest and the largest. It's almost constant. According to conditional expectation, switching is almost always justified. But we agreed that switching is a waste of time, by symmetry. The chessboard example is an example for which you will amost always be disappointed when comparing the mean value of A or B with their actual values. Richard Gill (talk) 14:56, 25 July 2011 (UTC)

I understand that. It is very similar to playing a martingale. With a very large bankroll you will win nearly every time, with an infinite bank roll you will win every time but you bankroll never gets any larger.

My point was that even with your chessboard version, the first resolution works in that it is 'simply impossible that whatever the amount A=a in the first envelope, it is equally likely that the second contains a/2 or 2a', even though this is almost always true. To resolve the paradox you only need to show that any step is not completely correct. The fact that you almost certainly will not get your expectation is another reason not to swap. Martin Hogbin (talk) 15:40, 25 July 2011 (UTC)

On the other hand, I see that your chessboard version could still be used as an argument to swap. Oddly enough there is very nearly a real game using the chessboard version, and this is 'Deal or No Deal'. If a contestant is given the option to swap their box at the start of the game (which they do not usually get, in the UK at least) the expectation value in their new box will, most likely, be greater than the value in the box they hold. In most cases, however, if they swap they will be disappointed Get less than the expectation). Martin Hogbin (talk)

Question

Am I right in saying that the abbreviated expectation calculation given in step 7 in the article is something of a fraud? It only applies if step 6 holds for every possible vale of A. If this is not the case then the expectation must be calculated properly by taking the average of all possible sums, weighted by their probabilities. Martin Hogbin (talk) 08:30, 26 July 2011 (UTC)

You are right, under one common interpretation of what the writer is tring to do. But it's not clear whether the writer is computing the conditional (given A) or the unconditional expectation of B. Both can be computed (in principle) by separating out the two cases, the two complementary events, $A<B$ and $B<A$.

Expectation of B, conditional on A : the writer is presumably using $E(B|A)=E(B|A<B,A)P(A<B|A)+E(B|A>B,A)P(A>B|A)$. Now, given A and given $A<B$, we know B=2A; similarly, given A and given $A>B$, we know B=A/2. So the writer could then procede to $E(B|A)=2AP(A<B|A)+AP(A>B|A)/2$. If this was his intention, he procedes to substitute $P(A<B|A)=P(A>B|A)=1/2$. So either (within this reconstruction) he is mistakenly assuming that the event $A<B$ is independent of the random variable A, or he is knowingly assuming that this is the case. In the first case, he should have realised that for any (proper) probability distribution of X=min(A,B), it is not possible that $A<B$ is independent of A. In the second case, as would be conventional in some Bayesian schools, he's expressing our complete ignorance of X (beyond the fact $X>0$) through the improper distribution with probability density proportional to 1/x. In that case, it's step 8 which is problematic - his prior distribution has infinite expectation, and whatever value of A or B you would actually see, you will be disappointed when you compare them to their expected value.

Expectation of B, not conditional on A : the writer is presumably using $E(B)=E(B|A<B)P(A<B)+E(B|A>B)P(A>B)$. He correctly substitutes $P(A<B)=P(A>B)=1/2$ but mysteriously replaces $E(B|A<B)$ by 2A and $E(B|A>B)$ by A/2. How can two numbers equal two random variables? Or does he not distinguish between A and E(A)? It would be correct to write $E(B|A<B)=E(2X|A<B)=2E(X)$ since the event $A<B$ is independent of the amounts in the envelope X, 2X. Similarly $E(B|A>B)=E(X|A>B)=E(X)$. This would finally lead to the (true) statement $E(B)=3E(X)/2=E(A)$. This seems to be the interpretation of the young philosophers Schwitzgebel and Dever, and leads them to their diagnosis "what goes wrong is that A is smaller when $B>A$ than when $B<A$". Indeed, it's equal to X when $B>A$ but it's equal to 2X when $A>B$. On average, two times smaller in the first case.

Most people in probability think that the writer is trying to compute the conditional expectation of B given A, but philosophers who don't know probability theory do not make these subtle distinctions between conditional and unconditional expectations and actual values. (That's what makes S&D's paper so hard for me to read, and why it's so weird they try to explain what is going wrong, in part, by doing some new probability theory of their own). Still it is true that the event $\{A<B\}$ is independent of the random variable A if and only if the random variable A is independent of the event $\{A<B\}$. So whether the writer was going for a conditional or an unconditional expectation, his reasoning breaks down for the same very intuitive reason: "A is (typically) smaller when $B>A$ than when $B<A$". I say typically because under the improper prior with density 1/x, whether $A>B$ or $B>$A does not tell us anything at all about the value of A. Conversely, the value of A tells us nothing about whether $A>B$ or $B>A$.

Who's to say what the writer was intending? What he knew, what he didn't know? Was he a complete amateur who got more or less everything mixed up (he got more or less mixed up, depending on which way you imagine he was trying to go), or was he a very sophisticated Bayesian using an improper prior? Some people think that improper distributions have no place at all in mathematics, others think they are a vital part of the Bayesian's toolbox. Richard Gill (talk) 11:57, 26 July 2011 (UTC)

TEP Synthesis

The following discussion creates a synthesis between the solution of Schwitzgebel and Dever, representing the philosophers, and the solution of myself, representing the probabilists. We need to be aware of the symmetry of (statistical) dependence and independence. In particular, random variable A is independent of event {A<B} if and only if event {A<B} is independent of random variable A.

S&D's diagnosis of what goes wrong was that the amount A in the first envelope is different, namely smaller, when it is the smaller of the two ("equivocation"). I strengthened this to the observation that on average, it's twice as small as what it is, on average, when it is the larger.

This observation proves that the amount A is statistically dependent on the event {A<B}.

My diagnosis was that step 6 assumes that the fact whether or not A is smaller, is independent of the actual amount A. In other words: step 6 assumes that the event {A<B} is is independent of A.

By symmetry of statistical independence, this is the same thing as independence of A on {A<B}.

There's one tiny speck of dirt in these arguments. If E(X) is infinite then twice infinity is infinity, we can't necessarily conclude that what's in the first envelope is smaller, on average, when it's the smaller of the two. And there's an extreme case where our uncertainty about X is so large that whether or not it's twice as large makes no difference. Knowing we have the larger of the two amounts doesn't change our information about it, at all. Consider the uniform distribution on all integer powers of 2: ..., 1/8, 1/4, 1/2, 1, 2, 4, 8, ... Yes I know this probability distribution doesn't exist in conventional probability theory, but arbitrarily good approximations to it do exist: you might like to think of it as a "real number" addition to the set of rational numbers.

Working with this distribution (or with close approximations, and then going to the limit), it *is* the case that the probability distribution of 2X is the same as that of X, both have infinite expectation values, E(B|A=a)=5a/4 for all a (integer powers of 2), and P(B=2a|A=a)=0.5=P(B=a/2|A=a) for all a. So at the boundary of reality, steps 6 and 7 of the TEP argument do go through. The reason now why switching only gives an illusory gain is that when expectation values are infinite, you will always be disappointed by what you get. They are no guide to decision making.

Richard Gill (talk) 06:44, 1 August 2011 (UTC)

My simple resolution - has this been covered in the literature?

First let me define what I am talking about - in the current article there is a "Switching argument" with 12 steps, and a 4 sentence "basic setup". I will try to put the "basic setup" into mathematical language:

S, L are positive real numbers. A, B are random variables, distributed as $P(A,B=L,S)=0.5$ and $P(A,B=S,L)=0.5$.

Note that this world is perfectly symmetrical - the labels A and B can be swapped without making any difference. Note also that I actually used A in the setup, whereas in the paradox it does not appear until step 1 - this should make no difference, but makes things easier to explain.

I think there are actually two sets of statements here (call them TEP-1a and TEP-1b). TEP-1a consists of both the set-up and the switching argument, and deduces steps 1-6 from the setup. TEP-1b uses steps 1-6 as definitions of a random variable B' and ignores the set-up. I assert the confusion generated by the paradox results from switching between the two variants. This is because the expectation of B is actually different under the two variants.

Under variant TEP-1a, the set up can be used to show statements 2, 3, 4, 5 are true (e.g. 4 states if A = S, then A,B = S,L, therefore B = L). Statement 6 it is a stretch, as A is taking two different values, one a quarter of the other, in different parts of the statement. This makes statement 7 nonsensical. Under variant TEP-1a, from the setup it is clear that $E(B)=E(A)$.

Under variant TEP-1b, we take statements 1-6 as a definition of the world, so in summary (based on steps 1 and 6) we can say:

A is a positive real number, define B as a random variable distributed as $P(B={\frac {1}{2}}A)=0.5$ and $P(B=2A)=0.5$.

Here the expectation of B, $E(B)={\frac {5}{4}}A$, but the problem is not symmetrical. So under TEP-1b it makes sense to switch once, and only once.

I'm satisfied with this resolution up to a point. What I find tricky is that there is a difference in the way that A is defined between TEP-1a and TEP-1b, and I'm not sure exactly how to express it.

Note that the paradox is always explained in terms of the set up, because the set up is a far clearer way to explain the problem than items 1-6 of the switching argument. However the set up is abandoned before we attempt to calculate the expectation, this is because without abandoning the set-up, it is too simple to calculate the expectation in a non-paradoxical way. Dilaudid (talk) 20:10, 8 August 2011 (UTC)