Talk:Deal or No Deal

UK edition: Am I lising count?

According to Noel in today's episode (on-air as I type) yesterday, the contestant had the 250K in her box. I missed the episode, so I'm not completely sure, but does this make it 3 times now? -- Lardarse 16:51, 8 December 2005 (UTC)

Nope, she was left with a choice between £100,000 and £250,000 (but had dealed earlier on at £31,000) and the £100,000 was in her box. BillyH 20:42, 8 December 2005 (UTC)

She actually had the £100,000 in her box, and the £250,000 was in the other box. The offer "would have been" straight down the middle: £175,000. --Bonalaw 20:45, 8 December 2005 (UTC)

Ok, I changed back the edit I made yesterday, but my cookie expired without me realising it, so it doesn't appear as one of my edits. -- Lardarse 16:41, 9 December 2005 (UTC)

calculating odds

In the US version, I think we can all agree that there is a 3% (1:26) chance of any dollar value appearing in each of the individual cases. Something tells me that this is not accurate with regards to the player.

Is this right: The player is hoping to select a box that contains a specific dollar amount (US$1m). Going back to the old Johnny Carson "someone in this room has my birthday" statistic issue, the odds that the player has selected the $1m box should be much, much less than 3%.

For those of you who don't know the Johnny Carson birthday game, it goes something like this: in any given crowd of people of x size, there is a 50% chance that two people in the group share the same birthday. When this fact was presented on The Tonight Show, Carson inadvertently embarrassed the mathmatician presenting this by asking if anyone in the audience shared a specific birthday (his own). The odds of two people in the room sharing a specific birthday are much lower than the odds of two people in a room sharing any day as their birthday. I believe this also applies to case selection -- any case containing any amount is at 3%, but a specific case containing the big prize is much, much lower. Comments? SpikeJones 02:59, 21 December 2005 (UTC)

The mathematics in Deal or No Deal have nothing to do with the Birthday paradox. The birthday paradox applies to cases where the random numbers are drawn in a way where duplicates are possible. The linked-to article explains it much better than I could.

Deal or No Deal has 26 unique values in 26 unique cases. When the contestant picks his case, there is, of course, a 1 in 26 chance that it was the big money case. Vslashg 03:21, 21 December 2005 (UTC)

Agreed that there is a 1:26 chance that any case contains the big money, but once the game begins with a selected case where the focus is whether that specific case contains a specific amount of money, does that change the odds a la the birthday paradox? SpikeJones 04:18, 21 December 2005 (UTC)

No. The shift in "focus" on Carson's show only "changes" the probability because two completely different things were being measured. Say for the sake of a simple example that there are 50 people in the room. The odds are 97% that at least one pair of people present share a birthday. But if one person speaks out and asks "does anyone share my birthday?", the odds drop to about 12.5%. The change isn't really because we are focusing on one particular birthday, but rather because the first case is looking at all 1225 possible pairs of people in the room, while the second is looking at only 49 pairs.

No such change of focus happens on Deal or No Deal. We agree that each of the 26 cases has a 1-in-26 chance of containing the big money, correct? When the contestant picks his case (let's say #7), absolutely nothing changes. We have no new information, and the contestant's choice changes nothing about the situation. Every case has a 1-in-26 chance of holding the big money, including case #7.

In fact, every time a case that does not contain the big prize is eliminated, the contestant's odds improve. If only five cases remain, and the big money has not yet been uncovered, then each of the five remaining cases shares an equal, one-in-five chance of holding the big money, and this includes the contestant's choice. (This issue is addressed in the third and fourth paragraph of the "Mathematical basis" section of the main article.) Vslashg 06:02, 21 December 2005 (UTC)

There is a major inaccuracy in the article; Expected Value Theory does not include "runs of luck" as that is a superstitious concept that has no bearing on future events. Only reason I don't fix it is because I don't know if the US show honestly incorporates such a silly thing, or if the person who put that in the article was mistaken and I don't really know how to go about finding out which is the case.

There is a very easy way to calculate the contestant's Expected Value at any point in the game, assuming that is the basis behind the contestant's decisions. If the game were to choose one suitcase and then open it, the average amount won would simply be the arithmetic mean of the contents of all suitcases as the formula would be to multiply each suitcase amount by the probability of choosing that suitcase, then add all the numbers together. The probability of opening any individual amount is (1/n) where n is the number of suitcases left, so you would add (1/n)(payout) for each payout which works out to the average.

If we assume that the contestant never makes a deal for less than the arithmetic mean of the suitcases, then this means that the worst thing that possibly happens is that the contestant opens all suitcases except for the original randomly selected suitcase (I'm assuming the contestant never switches cases because as others have pointed out, that doesn't do anything). This is strategically the same thing as the contestant just picking a suitcase and then opening it at the outset.

Of course the contestant does get to open a few cases and reevaluate, but by the formula above her EV will still be the arithmetic mean of all the remaining suitcases at any point if she never accepts a deal. And if she does accept a deal that is greater than the arithmetic mean of the remaining suitcases, then her EV actually turned out to be higher than that. This is why the contestant's EV can be more than the average (depending on the Banker's strategy) but never less using this dealing strategy, assuming EV is the basis of the contestant's decisions. It gets a bit more complicated if the contestant decides to accept a -EV deal to avoid risk of ruin (for example, electing to settle for $350K if the remaining cases are .01 and $1M); it may be worth it to accept a "bad" deal on one round if accidentally knocking out one of the big prizes will force you to accept a worse deal on a later round.

BTW, I saw an episode yesterday where the remaining cases were $1M, $500K, and two meaningless little amounts and the deal offered was still under 75% of the contestant's EV ($275K vs. the $375K or so she should expect to make by "gambling"). She then eliminated the $500K case and was offered 81% of her EV which she accepted. So I also have a hard time believing the statment about the banker ever offering above the arithmetic mean as a deal. I haven't seen every episode so I can't say for certain that it's never happened, but it seems doubtful to me as well. People are more likely to settle for a -EV deal as the amounts get larger; http://www.cardplayer.com/poker_magazine/archives/?a_id=15067&m_id=65575 (ignore the poker stuff and read the stuff under subheading "Asymmetric Risk Preferences")Xhad

Moved from article

Please source this if you want to put it in the article:

The most famous international version is arguably the Australian version hosted by Andrew O'Keefe. It currently screens on the Seven Network.

Argued by who? Soo 12:17, 23 December 2005 (UTC)

Clean-Up

I think that this article needs to be cleaned up. Here's what I propose...

The following layout:

• (Introduction)
• Format
• International Versions (heavily cut back)
• Mathematical Basis
• Analyzing decision making under risk
• External Links

This would be achieved through the following steps...

1. A "Format" section be created. This will explain the general format of the show without going into detail about specific international variations.
2. The "US version" section be scrutinised. The subheading "Odds and Probability" and "Strategy / Winning a million dollars" aren't really US-specific and should probably be removed and incorporated into "Mathematical Basis".
3. Pages be created for the Australian, British and American versions of Deal or No Deal (eg, Deal or No Deal (Australia), etc). Fill these pages with the information currently listed for each respective version on the main Deal or No Deal page (variation from general format, briefcase values, other country-specific information such as History for the Australian version). While the content is note-worthy, it distracts from the main article. In addition each version is repetitive in it's explanation of the basic format (which can be solved through the first step)
4. The listings of International Versions be heavily reduced. With the creation of country-specific pages in step 3, the listing on this page could be reduced to something like...

Australia

Main article: Deal or No Deal (Australia)

Deal or No Deal airs in Australia on the Seven Network, and is hosted by Andrew O'Keefe. The program debuted in late 2003 as an hour-long program, airing in prime-time on Sunday nights. In 2004 the show was reduced to a 30 minutes format, airing weekdays at 5:30pm.

Ideally the length of the new Australian, UK and US versions would match the existing versions (Dutch, Italian and Indian).

So... how does that sound? DynaBlast 19:53, 2 January 2006 (UTC)

Could we perhaps have a 'general problem/format' section before diving into the Australian format? I was trying to understand the basic concept of the show, particularly of the UK version, and it's very confusing the way it is laid out. Could we have a basic summary on the common points of all versions, only later followed by the differences? 12:00, 9 January 2006 (UTC) (Skittle)

Much clearer. Thanks. 57.66.51.165 09:04, 13 January 2006 (UTC)

Clean-Up and Deletions for Mathematical Basis

Okay, I removed almost all of the free-floating US/UK/Australian math & odds & strategy paragraphs and consolidated them under Mathematical Basis. I tried to meld important stuff into the math version that had been written from the point of view of decision theory / game theory / utility theory. I deleted duplications and similar analyses. Here are several orphaned paragraphs that I didn't feel qualified to meld in -- if you think that they add to the article please move them with segueways into appropriate places:

"In the Australian version, where audience members are given the chance to guess the value of their case to win $1,000, it is common practice to leave one's friends or companions until last. So, in this case, the probability of having the top prize from the start is 1 in 26, the probability of having the top prize left in the final two cases from the start is 1 in 13. Having said this, there is no reason to assume that the top prize is any more likely to be in the friend's case than the player's own. However, just as players can rue an early deal if they hold the top prize, so can they if their friend contains the top prize, because they know that their final offer would have been high regardless."

Since I've never seen the Australian version, this passage wasn't entirely clear to me.

"The banker's offers are worked out using a software program which offers a range of values, taking into consideration the probability, the player's value of money, and the current run of luck."

Is this true? I'm not sure how it could be proved one way or the other. I would assume that the banker's offer considers probability and general marginal utility and risk aversion, and then is worked out using marketing execs and focus groups for the purpose of maximizing the excitement-value (i.e. ratings) of the TV show -- I guess this could be loosely interpreted as run-of-luck...

"It has been known for the Banker to offer slightly over the odds (e.g. 53,000) as a form of insurance to avoid a player winning a very high prize."

I doubt Bank offers have much to do with insurance - the show is almost certainly insured independent of "the Banker's" particular offers. If anything, banker's offers are based on what makes the show the coolest, as guessed by whoever projects ratings. Nonetheless it would be neat to include some explicit examples of Deal or No Deal bank offerings that exceeded the arithmetic mean.

--Brokenfixer 02:12, 14 January 2006 (UTC)

Mathematical discussions that are not Monty Hall-related

Offhand, it seems that the likelihood of having a high-value case relative to the likelihood of having any of the remaining cases (including those of low-value) could be calculated using Bayesian inference. This would then allow the player to always make the statistically best decision whether to accept or refuse the banker's buy-out offer. Any truth/value to this? Or if that that, perhaps Combinatorial auctions? (unsigned comment by 151.197.239.92 01:42, 21 December 2005)

The expectation of having a high-value case, and the expected value of your chosen case, can be easily calculated (as the uniform average of all remaining values). So it is easy for the player to make the naive 'statistically best decision' about the banker's offer. However, as noted elsewhere, an offer of certain $500,000 is more valuable than a 50/50 chance at $1million (because of risk aversion as well as the declining marginal utility of money). It might make good economic sense to accept a banker's deal that has lower mathematical value than your briefcase's expected value -- depending upon your particular financial needs. As of Jan 13, the DoND article (math basis section) reflects this analysis, although it doesn't reference Bayesian inference. (... However, references to the Monty Hall problem are confusing and false, except to say that because (among other things) Deal or No Deal briefcases are chosen randomly (without using prior knowledge), Monty Hall does not apply.) --Brokenfixer 22:46, 13 January 2006 (UTC) [edited --Brokenfixer 02:48, 14 January 2006 (UTC)]

Brokenfixer, I think the comment by 151.197.239.92 ("Offhand, ...") was not Monty Hall related. It appears to be about the decision as to accept the buyout offer or not. Note the lack of indenting implying it was not a reply, instead it was meant as a new subject. When it first appeared it was at the top of the Talk page and not under any explicit subject heading. Therefore, I think we should move these three comments (151.197.239.92's, your 13th Jan 22:46, this one by me) out of the Monty Hall threads. If you agree, then you can move the three to another heading. I would have just moved them myself, but you had mad made Monty Hall related comments so I wanted to ask you first.

Ok, moved these comments. I also took out my mad boldfacing and added a sentence to my earlier comment. In response to the Comment, note that all the math behind the decision to accept the buyout offer is currently analyzed in the first Mathematical Basis section, called When to Deal. You and the original commenter (and anyone else reading this) are encouraged to expand or clarify that section, describing exactly why and how (mathematically) one can best play the DoND game.

(At each step, one portfolio is to accept the banker's certain offer. The other portfolio is a blend of obtaining the contents of each of the remaining briefcases - first, each briefcase is dimished by the non-linear utility curve of money (degrading the value of very large briefcases) and then those values are averaged. Second, that modified average is reduced by a risk factor which reflects undesirable uncertainty; the final number is compared to the banker's offer. An extremely accurate model might even iteratively take into account predictions of the banker's future offerings, which typically become more favorable as the game progresses. But imho this is Talk Page material and not material for the DoND Article.) --Brokenfixer 02:48, 14 January 2006 (UTC)

I meant to say mad, not made, last night. Sorry about that. That's what I get for commenting so late at night! Aaron McDaid 10:53, 14 January 2006 (UTC)

As Brokenfixer said above, accepting a banker's deal does not only depend on how it compares to the expected value of the suitcase; it also compares to the expected values of future banker's deals! If for instance you have $1, $5, $10, $1,000,000 left, then a somewhat high offer might look good given the chances that the next suitcase you open will be the $1,000,000, decreasing the next banker's offer. This seems extremely key to even a somewhat accurate analysis.

Monty Hall discussions

There are multiple subthreads below regarding the Monty Hall-ness of Deal or No Deal. They originally started off as separate top-level threads, but they are now grouped together here. Hopefully we'll be able to merge or delete these duplicates.

1. Claim that Deal or No Deal (random selection) still has Monty Hall properties, and Responses:

Random selection doesnt I believe destroy the Monty Hall factor. Try imagine 100 boxes and elimate 98 at random. - IF your box survived as one of the final pair ( only 1/99 chance) you must still swap as per MH. If it did not survive - as the £250 k rarely survives - a ' virtual ' new game is started with a new highest prize and the number of boxes left when it was eliminated.- and so forth.

Let H be highest sum on the board Let N be the 2nd highest sum on the board

Let x be the nunber of boxes unopened when the highest number first appeared Let y be the nunber of boxes unopened

)

	''''H/x + N / ( pairs at y = z )''''

H= 250 N= 100

x=    22
y   = 22 	     	£11,797
		
			

H = 100 N= 50 £10,877 x = 11 y= 8

H = 250 N= 100 £14,935 x = 22 y= 8

H = 250 N= 100 £21,364 x = 22 y= 8

AubreyA 21:14, 29 December 2005 (UTC)

Ugh. I'm impressed by the math but Monty Hall doesn't apply here. In your example, there are 100 boxes. Firstly, the chances of your box making it to the final pair with the big prize still up is 1/50, not 1/98. To prove this; pair each box with another box. Pick your favourite box. Eliminate all the boxes except for yours and the one it's paired with. There's a 1/50 chance your pairing contains the big prize. (Within that pair, each box has a 1/2 chance of course).

But if you want iron-clad proof, consider this. Say there are 4 boxes, A, B, C, D, randomly worth $1, 2, 3, or 4. You pick A. Then you eliminate D. It had $4. So now, A, B, and C are randomly worth $1, $2, or $3. Do you swap from A? Answer: doesn't matter. Now you elimate C and it was worth $1. A and B are worth $2 or $3, but one doesn't have more of a probability than the other.

The main thing here is that the cases eliminated are purely random. In Monty Hall, they know door 2 is worthless, and they reveal this. In this game, you go to eliminate case 2, but it could just as likely as your case be the one with the big prize. --Headcase 08:11, 9 January 2006 (UTC)

- You are wrong. The cases eliminated are not "random" because contestant can see what was in them. That is, it doesn't matter do you know what's in the case a priori. It is absolutely doesn't matter who selects the case. The only important thing is that the case which is opened is low-value one. Were you any good at probability at college?

Better than you evidently. Maybe didn't take classes as advanced, but have a better grasp on the concept, unless you're a troll, in which case congratulations, you've made me spend time to disprove you and are good at your task.

Monty Hall never eliminates the right one (right = $100, wrong = $0), so... Chances of picking $100 and then eliminating $0: 1/3 * 1/1 = 1/3 Chances of picking $0 and then eliminating $0: 2/3 * 1/1 = 2/3

Since the eliminated is always the same ($0), it ends there. When $0 is revealed, you have a 1/3 chance of having $100 and a 2/3 chance of having $0.

To compare Deal or No Deal, say there are 3 cases, and only one has $100, the rest nothing... Chances of picking $100 and then eliminating $0: 1/3 * 1/1 = 1/3 Chances of picking $0 and then eliminating $100: 2/3 * 1/2 = 1/3 chances of picking $0 and then eliminating $0: 2/3 * 1/2 = 1/3

So if we remove the unfortunate instance where $100 is removed... Chances of picking $100 and then eliminating $0: 1/2 Chances of picking $0 and then eliminating $100: 1/2

Better swap the suitcase, it's going to make a difference.

Hell, I'll even go with an extended version of it: 3 suitcases, $1, $10, $100. Chances of picking $1 then eliminating $10: 1/3 * 1/2 = 1/6 Picking $1, eliminating $10: 1/3 * 1/2 = 1/6 10, 1: 1/6 10, 100: 1/6 100, 1: 1/6 100, 10: 1/6

If we only keep instances where $1 is revealed to be eliminated: Chances of picking $10 then eliminating $1: 1/2 Chances of picking $100 then eliminating $1: 1/2 (no difference)

Copy and paste If we only keep instances where $10 is revealed to be eliminated: Chances of picking $1 then eliminating $10: 1/2 Chances of picking $100 then eliminating $10: 1/2 (no difference)

Copy and paste If we only keep instances where $100 is revealed to be eliminated: Chances of picking $1 then eliminating $100: 1/2 Chances of picking $10 then eliminating $100: 1/2 (no difference)

Hey, let's copy and paste that whole thing and do it with four suitcases: 1, 10, 100, 1000. Chances of picking $1 then eliminating $10: 1/4 * 1/3 = 1/12 Picking $1, eliminating $10: 1/4 * 1/3 = 1/12 1, 1000: 1/12 10, 1: 1/12 10, 100: 1/12 10, 1000: 1/12 100, 1: 1/12 100, 10: 1/12 100, 1000: 1/12 1000, 1: 1/12 1000, 10: 1/12 1000, 100: 1/12

If we only keep instances where $1 is revealed to be eliminated: Chances of picking $10 then eliminating $1: 1/3 Chances of picking $100 then eliminating $1: 1/3 Chances of picking $1000 then eliminating $1: 1/3 (no difference)

If we only keep... no way I'm doing more of this.

In all the Deal or no deal examples, after the elimination the value of the case is still one of the values that remains, completely at random. Of course the value of the case fluctuates, no one denies that (especially the banker). If you picked a case and then eliminated the $10 case, one of the lower valued ones like you say, then now our case as either $1, $100, or $1000. Exactly 1/3 chance of eachof those. So why, may I ask, would I switch? The mean value is more, but switching does nothing (from a probability standpoint. It certainly does something from a "the amount of money that is now in your case" standpoint).

But if the banker offered my even one red cent to change to any other closed case, then if I had no psychological attachment it would be logical to do it. If you majored in stats and passed, please state your college so I can blacklist it. Well, maybe that's unfair to the college. Maybe you just know what the Monty Hall problem is. --Headcase 21:19, 13 January 2006 (UTC)

Headcase's examples and math are all correct. AubreyA's probability computations are wrong for Deal or No Deal. If there are 100 boxes and only one contains $1million, and if 98 are opened at random and the $1million doesn't show (which is unlikely but 2% possible...), then your initial choice has a 50/50 chance of being $1million. (This is directly unlike Monty Hall, where Monty looks behind the 99 remaining doors and forcefully demonstrates to you which of those 99 doors contains the $1million (99% of the time).) --Brokenfixer 23:22, 13 January 2006 (UTC)

Sidenote: However, IF the Banker knows the contents of all of the briefcases, then the Banker could make an offer that communicated information to you. For example, consider a game with only two briefcases, one containing $100 and one containing $0. You choose Briefcase "A" (random choice). I open Briefcase "B", look inside, frown enigmatically, and offer you a penny to switch (or perhaps I offer you a penny to stay with your original choice). To always be bribed by a penny is a bad strategy in my game -- you'll get a penny and open an empty briefcase. (Assuming that the Banker has incentive to reduce your winnings -- for example they get to keep the contents of the other briefcase.) Meanwhile, in Deal or No Deal, unlike my game, you never get any useful information distinguishing unopened briefcases, so there is never mathematical incentive to swap or stick. --Brokenfixer 23:22, 13 January 2006 (UTC)

2. Claim that revealing lower-valued cases randomly still yields Monty Hall, and Responses

In the section "Mathematical basis", it is stated that this game is not a version of Monty-hall problem. This is untrue. It doesn't matter who opens the cases (a person who knows what's in them, or who doesn't know). The only thing that matter for Monty-hall is that the cases which are revealed have average value lower than at start of the game.

I will not change the original article because of my bad english. Can someone do that?

(Unsigned comment by 83.131.134.247 15:31, 13 January 2006)

Your english is good but unfortunately your math is bad. Consider 100 boxes, 99 empty and one with $1 million. You choose one box. 98 remaining boxes are opened at RANDOM, and all happen to show empty. This is unlikely but possible (2% chance). Your chance of having originally chosen $1 million is 50/50, (lucky you), which is totally different from Monty Hall. Try it with a deck of cards (52 "boxes") with the Ace of Spades as a win card, or run a computer simulation. With Random Selection, the last briefcase is worth more than the initially chosen briefcase 50% of the time, REGARDLESS of the values of all of the revealed briefcases. Contrast this with 100-box Monty Hall, where the last briefcase is 99% likely to contain more than the initially chosen briefcase! --Brokenfixer 23:45, 13 January 2006 (UTC)

DoND is not Monty Hall. In Monty Hall, the host is NOT always choosing randomly, sometimes he is choosing deliberately to avoid the car door. In Deal Or No Deal the player IS always choosing randomly up to the choice to swap boxes. It is this difference that is crucial in Monty Hall. This element of non-randomness is what makes the final decision non-random, so to speak. In DoND, everything is random. Aaron McDaid 01:33, 14 January 2006 (UTC)

This is not Monty-Hall. Let's say there were 3 cases left (one with $5, one with $1000, one with $75000).

Option 1: Contestant has $75,000 case. They can

• Open the $5 case, $1000 remaining (would lose money by swapping)
• Open the $1000 case, $5 remaining (would lose money by swapping)

Option 2: Contestant has $1000 case. They can

• Open the $5 case, $75,000 remaining (would gain money by swapping)
• Open the $75,000 case, $5 remaining (would lose money by swapping)

Option 3: Contestant has $5 case. They can

• Open the $1000 case, $75,000 remaining (would gain money by swapping)
• Open the $75,000 case, $1000 remaining (would gain money by swapping)

Now each of the above options, and each choice resulting in those options, are totally random. You've got as much chance as picking the $75,000 case at the start as you do the $5 case (1/26 chance for each in America). So in the end, 50% of the scenarios (3/6) would see you benefit by swapping at the end. The other 50% and you'd lose. This is proof that Monty-Hall does not apply when case selection is random. DynaBlast 16:05, 13 January 2006 (UTC)

In Monty Hall, the player actually has a 66%, not 33%, chance of winning the car if he/she plays the optimum strategy (swapping). In Monty Hall, the player chooses one of three doors, one containing a car, two containing a goat. At this stage the player has a 33% chance of having selected the door with the car. The host now opens a door containing a goat and offers the player the option to change doors, and the player accepts. There are two possibilities:

• The player had originally chosen the door containing the car (33% likely), and then changes to the door containing the other goat.
• The player had originally chosen the door containing a goat (66% likely), and then changes to the door containing the car (all three doors can't contain goats).

This clearly means the player who swaps has a 66% chance of going home with the car. The player who doesn't swap only has a 33% chance of going home with the car.

In the above evaluation of the Monty Hall swapping strategy, I was able to rely on the certainty that the host will definitely open a door containing a goat. To evaluate the swapping strategy for Deal or No Deal, there is no such certainty, so we need a more complicated 'tree' of possibilities, like this: (I'm going to imagine there is $1000, 1c and 2c still in play).

• Player had originally chosen the $1000 box (33% likely). Now a box is opened. It will contain a worthless amount, and the player will go home with the other worthless amount.
• Player had originally chosen a 'worthless' door (66% likely). Now a box is opened. It's 50/50 whether it'll be $1000 or not:

- The box opened contains $1000. This is (66% x 50%) 33% likely. Player goes home with worthless amount.

- The box opened contains the other worthless amount. This is (66% x 50%) 33% likely. Player swaps and goes home with $1000.

Therefore, in Deal or No Deal when you are at three boxes, even if you swap you have only a 33% chance of going home with the $1000.

--Aaron McDaid 18:40, 15 January 2006 (UTC)

3. (Nov 2005) Notice that UK section calls DoND a Monty Hall problem

I have just archived the thread that was here, as it contained little that is not in the other Monty-Hall subthreads. --Aaron McDaid 17:58, 15 January 2006 (UTC)

Other comments

Excellant show. I think this shoukld be brought back on a weekly basis and it's a good show to watch until Survivor returns. -- Eddie

The US explanation...sucks. The Australian version makes much more sense. The US version fails to mention that a person decides more based on what the next bank offer will be rather than what their case contains. 68.7.151.162 04:12, 21 December 2005 (UTC)

"see the banker's offer after the deal"? what does it mean

What does this recent addition to Deal_or_No_Deal#UK_version mean? I don't understand it as it's written:

The main difference is that once a person has made a deal with the banker, we are able to see the banker's offer after the deal with the deal that the contestant has made. This will give a clearer picture to the TV audience on whether the contestant has made a good deal.

--Aaron McDaid 00:32, 19 January 2006 (UTC)

It's badly worded, and I'm not sure of the details of the UK version of DOND, but it sounds like the following features of the Australian version: After making a deal, the contestant continues to open briefcases. After opening a number of cases the Bank sometimes shows what it would have offered had the contestant chosen to keep playing. This usually occurs if the contestant knocks out large value after making their deal (the bank revealing the lower offer that would have resulted from this) or if the contestant manages to avoid large values (showing that had they kept playing they may have recieved a larger bank offer). This makes it clear whether or not the contestant has optimised their winnings through correct timing of taking a deal. - DynaBlast 04:17, 19 January 2006 (UTC)

Q4 just made an edit that looks good. I'll probably be watching the UK DOND on Saturday, so I can check it out. The UK version always (from very early January 2006 at least) followed through and got the other offers from the Banker after a player dealed. I think this new feature is simply to display the number on screen in case the viewer forgot. i.e. They're not displaying any new information, just helping the viewer to remember. --Aaron McDaid 10:39, 19 January 2006 (UTC)

US page moved, mathematical analysis

I would think that the mathematical analysis that was moved to Deal or No Deal (USA) applies (at least most of it) to any version, and perhaps belongs on this page rather than being on the US page? Elpaw 10:38, 23 January 2006 (UTC)

I agree that it can go on this page, but that section is currently very much based on the American version (a million dollars, for example) and can do with a little globalising. Bluejam 15:58, 26 January 2006 (UTC)

add Average vs. Expected Payout?

It'd be nice to have a up-to-date table of the average payout on each version of the show and the expected payout per episode assuming the player turns down every offer. It seems to me that if the Banker makes offers that are below the expected value of the game most of the time, he should expect to pay more over the course of the show. On the other hand, when players do accept a low offer, he makes up for the losses. It'd be interesting to see which factor is stronger.

Do odds change?

I was puzzled over this question: at the beginning the contestant obviously has a 1 in 26 chance of getting the briefcase with the $3M. Suppose the contestant then opens ten cases which are small prizes and worthless, so there are now 16 cases remaining. Are the odds that the contestant has the $3m briefcase now 1 in 26 or 1 in 16? -Abscissa 04:46, 4 March 2006 (UTC)

1 in 16. Because you only had a 16 in 26 chance of reaching this stage in the first place. It would only be 1 in 26 if the chance of the $3M already being knocked out was zero (as in the Monty Hall paradox). --Bonalaw 10:30, 4 March 2006 (UTC)

Deal or No deal is dodgy!

Anyone aware of this? : http://forum.digitalspy.co.uk/board/showthread.php?t=349545&page=7&pp=25

Endemol have a problem with their random generator apparently, you can predict whats in the boxes

1 10,000 2 35,000 3 100 4 1p 5 50 6 1 7 750 8 50p 9 20,000 10 250,000 11 500 12 1,000 13 75,000 14 50,000 15 10 16 3,000 17 100,000 18 250 19 5 20 15,000 21 5,000 22 10p

she has 15 000 in her box (Sat 4th march)

Already covered in the Deal or No Deal (UK) article. --Bonalaw 11:17, 5 March 2006 (UTC)

you wreak havoc on my emotions. let's DANCE!!!