Comments on: Two Envelopes

By: Michael

Michael — Wed, 03 Dec 2003 01:40:15 +0000

Wow. The first comment and the last comment I read were both spot on. Doug said it right at the beginning: when considering whether you’ll get 2y or y/2, the values of y are different. If x = 200 (why not?), then you are either holding the 200 envelope (and the alternative is 2y = 400), or you’re holding the 400 envelope (and y/2 = 200). Thus, 0.5 * 2y + 0.5 * y/2 = 0.5 * 400 + 0.5 * 200 = 300, which is precisely what your expected value was when choosing originally. So Jonathan is correct: Stick with your original choice and expect 1.5x (300, in my example), or switch and expect 1.5x plus a day in Hell. Easy choice. BTW, two more observations. (1) The way I read the problem is that the choice has to be made before seeing the contents of the selected envelope, so there’s no worries about whether the value is an even or odd number, or whether it’s less than 20. (2) The problem may be easily stated in a way that has nothing to do with distributions with infinite means or how God selected x at all. Just say that x = 20, and require that the choice of whether to switch must be made without opening the envelope. Then, presumably, it’s safe to assume the probability of selecting the 40 envelope is 0.5, but the angel’s “reasoning” still sounds persuasive… or does it?

By: Jonathan

Jonathan — Thu, 06 Nov 2003 21:35:55 +0000

This seems pretty simple to me. You’ve got a choice of two envelopes. Once you’ve picked one, the angel is essentially allowing you to make the identical choice again. In no way does making the choice twice change the odds, since you don’t learn anything about the second choice from having made the first choice. Therefore, the choice the angel is giving you is this: 1) Take a chance between x and 2x in heaven or 2) Take a chance between x and 2x in heaven and also spend a day in hell. Unless you are curious about hell, there’s no reason to take the angel’s deal—it gets you exactly the same thing as the original deal, except you also have to spend time in hell, maybe as punishment for 2nd guessing yourself.

By: Bill

Bill — Thu, 06 Nov 2003 04:30:37 +0000

Let x be the amount in the higher envelope. Assess your prior probabilities on what x is (you can always do this). Let y be the amount you find in the envelope you pick. You now know that either x = y or x = 2y. So, the probabilities of those two possibilities will be proportional to your prior probabilities that x = y and x = 2y before you saw what was in your envelope. So (assuming you want to maximize your expected winnings), if before you saw anything, your probability that x=y was more than twice as high than your probability that x=2y, then you should not switch; otherwise you should switch.

By: Thomas Dent

Thomas Dent — Tue, 04 Nov 2003 14:27:46 +0000

Of course if x>10 and you sneak a look at the paper and it says 16 then you should switch, but this is cheating! (a.k.a. improper use of prior information).

By: Thomas Dent

Thomas Dent — Tue, 04 Nov 2003 14:23:35 +0000

Your expectation is 3x/2 whichever you pick, if the angel has truly given you an envelope at random and doesn’t know which is which. Therefore your expected gain is zero. To invert the angel’s (fallacious) argument and expose it as such, without loss of generality denote the envelope you don’t choose by z. Then the envelope you do choose has an expectation value of (z/2 + 2z)/2 = 5z/4. Therefore you expect to lose z/4 by switching. The fallacy is in comparing the two equally-likely cases by setting y to a fixed value in both. Without any further information, and seeing just a single envelope containing y, we would be justified in considering that y was drawn from the same statistical distribution in both cases, therefore we would take the expectation of y to be the same in both. However, we have the information that God has fixed the contents of both envelopes ahead of time, and since they’re both sitting in front of you they won’t change in value depending on which you pick, so we must be consistent between the two cases and set y = 2x in case A (where the other contains y/2) and y = x in case B (where the other contains 2y). The angel’s argument works only if the contents of the envelopes could magically change to twice (or half) their values when you switched so as to conserve the value of y. Or for example if he tosses a fair coin after you look inside one envelope and then gives you either half or twice the number on the paper. To explain once more, say God has put 20 in one and 40 in the other. If you picked 20, then y = 20 and your expectation is 30 (not 25!). If you picked 40, then y = 40 and your expectation is 30 (not 50!). In both cases the angel’s reasoning is fallacious. This has nothing to do with finite or infinite probability distributions, it’s just about the consistent use of mathematical variables in comparing two cases resulting from the same fixed but unknown initial condition.

By: Keith M Ellis

Keith M Ellis — Tue, 04 Nov 2003 07:41:37 +0000

Note that in Matt Weiner’s two statements of the problem, y is a random integer. In the first problem, 2y is also necessarily an integer. However, in the second problem, y/2 is not necessarily an integer. This should serve as an indication to our intuitions that the two problems are not identical. Alas, I’m not bright enough to comprehend in just these few minutes what is going on in this problem. However, I strongly intuit that we’re unwittingly confused by an assumption that two “terms” (I mean this in the most abstract, conceptual sense) of the problem are equivalent when they are not. Language in some way is leading us astray.

By: Keith M Ellis

Keith M Ellis — Tue, 04 Nov 2003 07:29:46 +0000

Matt Weiner, I retract my claim about your comment. I think you’ve got it right: at least one major error is in thinking that the two scenarios you give are equivalent. They’re not, and your conclusions about each of them are correct, I believe.

By: Keith M Ellis

Keith M Ellis — Tue, 04 Nov 2003 07:16:13 +0000

By “MHP” I mean the “Monty Hall Problem”, which is equivalent to the “Let’s Make a Deal Problem”; also known as the “Car and the Goats Problem” and the “Three Doors Problem”. Matt Weiner, I’m pretty sure your reasoning and conclusion above is false. Dsquared, calling it the “Let’s Make a Deal paradox” is misleading, since it’s not a paradox.

By: dsquared

dsquared — Tue, 04 Nov 2003 07:03:08 +0000

No, it’s not the same as the Let’s Make a Deal paradox. That was basically about a popular confusion over Bayesian reasoning and had a definite solution. This one is about a genuine mathematical point; that there are distributions which don’t have means, while the normal rules of decision theory rather rely on their having them.

By: Keith M Ellis

Keith M Ellis — Tue, 04 Nov 2003 07:02:13 +0000

As something of an amateur expert on the MHP, I have to say that I don’t think this problem is at all similar to the MHP, except in the sense that the words “stay” and “switch” appear in the problem annunciation.

By: Matt Weiner

Matt Weiner — Tue, 04 Nov 2003 01:19:04 +0000

Tom T.—I think so, more or less. Note that the order of writing down the numbers and giving them to you matters. In Brian’s problem, God puts y and 2y in an envelope, and then the angel randomly decides to give you one of them. You shouldn’t switch. Suppose it went like this: God writes a number y in an envelope and has the angel give it to you. He then flips a coin to decide whether the other envelope should contain 2y or y/2. You now have the option to switch. Then you should switch—no matter what y originally was, you’ll gain an expected y/4 by switching. I think this is what matters, not the infinite distribution. Unfortunately I am not capable of formulating exactly what I mean by “this” right now.

By: Tom T.

Tom T. — Tue, 04 Nov 2003 01:06:17 +0000

Isn’t this just a variation on the “Let’s Make A Deal” probability problem that provoked a small firestorm among Marilyn vos Savant’s readers? (Apologies to the non-US readers for the doubly incomprehensible pop-culture reference).

By: cks

cks — Mon, 03 Nov 2003 23:43:15 +0000

There’s another variation on this problem that is interesting. In another case, the angel may tell you that the envelopes contain the values x and 2x in them. You get a chance to choose one of the envelopes and open it. After seeing the value in the envelope, y, you can decide whether to switch. If you randomly choose the first envelope, and then flip a fair coin to decide whether to switch, you’re equally likely to get x or 2x as your final value. The expected payoff for this random strategy is 1.5x. Is there a way to do better than this by using the value for y? It turns out that there is. You can make the decision to switch envelopes random, but dependent on the value of y that you pick. If the probability that you switch envelopes a decreasing function of y, one that goes to zero asymptotically, such as e^{-y}, the conditional probability of switching given that you opened the envelope with the larger value is smaller that the conditional probability of switching given that you opened the envelope with the smaller value. The expected payoff is strictly larger than 1.5x.

By: Brian Weatherson

Brian Weatherson — Mon, 03 Nov 2003 23:39:45 +0000

Mats, you do need to hand-wave a little to get the case that the expected gain from swapping is always y/4 for all y. But you don’t need to do so to get the case that the expected gain from swapping is positive whatever value y takes. (Which is why I presented the Broome case.) In the case I’ve given there, if y = 10 then the (known) gain from swapping is 9, if y > 10 then the expected gain from swapping is (roughly) 0.23y. Now of course the expected value of y is still infinite (and that’s relevant to the puzzle) but the angel doesn’t want to use that. Rather, he argues that if you opened your envelope, you’d take the deal no matter what was there, so you should swap now. Maybe to make the deal sweeter, he says that once you open the envelope, the cost to taking the deal will be 3 days in hell rather than 1 – you know that you’ll take this deal once you open the envelope, and you know it’s worse than the deal he’s offering now, so why not go for it?

By: Mats

Mats — Mon, 03 Nov 2003 23:12:08 +0000

How does the sloppiness then cause the paradox – it’s the same as my stripped pick a number game above. The distribution is infinite. So with any number of days we get in the first envelope, it is infinitely unlikely that we actually should have gotten such a low number, a finite one, from an infinite distribution. So we should actually swap. But this conclusion is only valid for infinitely unlikely cases, and hence irrelevant.