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This is an old revision of this page, as edited by Jwanders (talk | contribs) at 15:24, 26 August 2014 (→Where is the problem?). The present address (URL) is a permanent link to this revision, which may differ significantly from the current revision.

Where is the problem?

One envelope is $10 another is $20. Pick first envelope... your expected value is now $10*.5+$20*.5=$15 great! Decide to switch to other envelope... your NEW expected value is STILL $20*.5+$10*.5=$15 ????

Where is the problem? Its so obviously devoid of a problem that I am shocked anyone would bother to write this artical... I don't care how important the person is who "discovered" the problem, that person was wrong, and their mistakes should not be allowed to be shown the light of day except in the context of ridicule and disproof. I would sugest this artical be removed from wikipedia. It was a complete waste of my time, I want it back! --THE TERMINATOR 24.13.179.59 (talk) 14:24, 16 August 2014 (UTC)[reply]

I actually bothered to read the "Example" section and THAT is where the problem is. The first thing I notice is how many dollar values are mentioned... lets list them: $10 envelope, $20 envelope, $40 envelope... BOOM that is 3 (yes THREE) envelopes! This violates the original premise of the "problem", namely that there are only TWO envelopes! The problem isn't with the "problem" the problem is with the math used to analyze the "problem"... If your "problem" has THREE envelopes then your in the WRONG wikipedia artical, this artical only has TWO envelopes. I don't know how much more painfully obvious I can make this! PLEASE PROVE ME WRONG! please... --THE TERMINATOR 24.13.179.59 (talk) 14:56, 16 August 2014 (UTC)[reply]
I partially agree. The expectation value of switch may be greater than the amount in the first envelope, but it's equal to the expectation value of the first envelope. This seems like a pretty obvious solution at least to the set-up where the first envelope isn't opened. I'm sure there's some mathematical nuance that I'm missing (seems somewhat more likely than having found a solution that no one else has thought of in the past 100 years), but perhaps that should be addressed in the article?jwandersTalk 15:24, 26 August 2014 (UTC)[reply]

Great work

The article is much more clear and sensible than it was 2 years ago, when I had a crack at it. Congratulations on the great work. Dilaudid (talk) 08:43, 3 October 2012 (UTC)[reply]

I agree with this assessment. I visited the article today after several years and found it vastly improved. I now think I can find some valuable information in it. I want to spend some time going over it carefully. ---Dagme (talk) 04:55, 29 March 2013 (UTC)[reply]

What am I missing?

Surely any actual playable instance of this game has an underlying probability distribution of the amounts in the envelopes? And, because it has to add up to 1 over all amounts, it has to converge to 0 as the amounts get ever higher. So, given this probability distribution, and a value-in-the-first-envelope A, the probability that A is the smaller envelope cannot be exactly 1/2 over all possible values of A.

Consider:

  • You play the game 1,000,000 times. You don't swap, just take the amount in the first envelope. You write down the highest amount you've seen so far.
  • Now, on the 1,000,001th go, you're offered an amount higher than any you've previously seen. Intuitively, would you swap? — Preceding unsigned comment added by 217.150.97.26 (talk) 12:56, 15 October 2013 (UTC)[reply]
I think you've hit the nail on the head, the original argument to support switching assumes that 2X is equally as likely as (1/2)X, and that implies either an upper limit (and we know the maths don't work then, because with a limit of $100 you never swap $100) or constant probability over 0 to infinity with the implication that the expected winnings are infinite, or tend to infinity as we increase the value of the game limit towards infinity. Gomez2002 (talk) 12:39, 6 August 2014 (UTC)[reply]
Yes I agree. Either there is an upper limit, in which case it is easy to show that on average you do not gain by swapping, or there is not, in which case the expected value in both envelopes is infinite. I think this is the conclusion in the literature when you read it. Martin Hogbin (talk) 13:24, 6 August 2014 (UTC)[reply]

Ali Baba Variant

The cited Nalebuff article lists multiple possible answers to the apparent paradox. The only one which implies either Ali or Baba should possibly want to trade is one in which there is a maximum apparent amount of money that might be in the envelope (Nalebuff, The Other Person's Envelope is Always Greener, 1989, pp. 175-177).209.6.139.197 (talk) 04:44, 17 October 2013 (UTC)David Lepzelter[reply]

Nalebuff shows the only scenario where the expected value of (2A+A/2)/2=1.25A can ever apply: If A is the "initial amount" contained in the first envelope (say A), and then the second envelope (say B) is equally likely to be filled – thereafter – with either 2A or A/2. Yes, if Ali got the "first" envelope, then he should switch and can expect to get 1,25*A. Whereas Baba, by switching must expect to get only 0.8*B.

But as, in general, "any" envelope is equally likely to having been picked first, the term (2A+A/2)/2 is nonsensical, because from the outset this implies  "A≠A". So a priori putting up with "2A"=2(A/2) and "A/2"=(2A)/2). Consequently you have to solve this nonsensical term as follows: "The expected value is (A+A)/2". Gerhardvalentin (talk) 08:22, 16 June 2014 (UTC)[reply]

Formatting error

Several times on the page, red text. Can anyone fix this? Shiningroad (talk) 17:11, 8 February 2014 (UTC)[reply]

Odd Solution?

It would seem that with all of those multitudes of words spent on the topic, no-one came up with the obvious: you should always switch envelopes if the amount contained within your envelope is odd and never if it is possible for the amount to be odd, but you find an even quantity.70.90.204.42 (talk) 16:13, 1 May 2014 (UTC)[reply]

Unfortunately, as far as I know, no one has yet published such a solution. So wikipedia can't cite it. Richard Gill (talk) 16:14, 1 June 2014 (UTC)[reply]
That is correct only if we assume integer amounts. The general case is that the amounts can be continuous, so this distinction cannot be made. --Pan-wiki-tsik (talk) 18:21, 4 June 2014 (UTC)[reply]

In a mess again

The body of the article used to start with, 'Here the ways the paradox can be resolved depend to a large degree on the assumptions that are made about the things that are not made clear in the setup and the proposed argument for switching'. This key statement has been moved down the page with attempts at 'resolutions' being placed before it.

It is quite clear that one of the principal problems with resolving this paradox is the uncertainty in what is meant by the question. Without a clear understanding of this fact, all attempts at resolutions are meaningless. Martin Hogbin (talk) 12:51, 16 June 2014 (UTC)[reply]


There are some simple and uncontroversial fundamental principles, that can be applied before we attempt any resolutions.

The setup must be well defined

It must be clear exactly what the rules of the game are. For example, are there any restrictions on the sums that might be in the envelopes and does the player look inside his envelope before making a choice?

The full basis of the argument to switch must be made clear at some point

One problem with this paradox is that intuitively there is no paradox. Most people intuitively think that the correct thing to do is not to bother switching and, unlike the Monty Hall problem for example, their intuition turns out to be correct. As far as I know, there are no sources that say it is sensible to switch for any version of the paradox.

For there to be a paradox at all, a convincing line of reasoning needs to be presented to the reader to support switching. Without this there is no paradox to be resolved. The resolution of the paradox therefore requires the fault (or possibly just ambiguity) in the proposed argument for switching to be found, thus confirming the readers original intuition and the mathematically correct fact that switching is pointless.

A mathematical argument which shows that there is nothing to be gained by switching is not a resolution of the paradox it is merely a confirmation of the obvious.Martin Hogbin (talk) 14:41, 16 June 2014 (UTC)[reply]


In the argument I ignore the Ali Baba variation, where there is still no paradox. In this case it is intuitively obvious that you should swap, a simple argument can be proposed for swapping, and it is easy to show that swapping is advantageous, but it is a different problem. Martin Hogbin (talk) 15:17, 16 June 2014 (UTC)[reply]

Randomized solutions

There's a basically equivalent but simpler formulation than the one in the main article. Select a strictly decreasing function, f(x), defined on the non-negative real numbers, which is 1 at 0 and approaches 0 as x approaches infinity; for instance, exp(-x), 1/(1 + x), 1/log(e + x), etc. If the originally chosen envelope contains an amount A, swap with probability f(A). For any two fixed amounts, A0 < A1, if the smaller amount is chosen, it will be more likely swapped than if the larger amount is chosen. The expected value is (A0 + A1)/2 + (f(A0) - f(A1))(A1 - A0)/2. If the amounts are A and 2A, the expected value is 3A/2 + (f(A) - f(2A))A/2— Preceding unsigned comment added by Thinkatron (talk • contribs) 06:46, 14 August 2014 (UTC)[reply]

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