The Two-Envelope Paradox Resolved (Part II)
The Paradox
My previous Substack laid out the two-envelope paradox.You are offered a a choice between two envelopes, Alpha and Beta, one of which contains twice as much money as the other. In the open-envelope variant, you open Alpha and find it contains $100. You are allowed to switch to Beta if you want to. Do you? If you switch to Beta, and are allowed to switch back before opening it, do you? In the closed-envelope variant, you choose Alpha but do not get to open it, but you are still offered the choice to switch.
In both variants you know that one envelope contains twice as much as the other. Your first intuition should be that since you don’t know which is which, you shouldn’t switch. The argument for always switching goes as follows. If amount X is in Alpha, then if the probability of Alpha being the big-money envelope is 50%, the expected value of switching to Beta is .50 (X/2) + .50 (2X) = X/4 + X = 1.25X, which is greater than X, so you should switch.
We will build up a general resolution to the paradox starting from the simplest case and moving to the most general. In all cases, let’s start by specifying that you know the possible values for the amount in the big-money envelope and how probable each of those values is. At the end, we’ll return to the case where you don’t know the probability distribution.
I said in my previous Substack that it looked as if the literature had not discovered this solution. I was wrong. Now, having read a little of the literature and reading the Wikipedia article more carefully, I think others have discovered it too, and that I am part of the dominant faction in the dispute over what is the best solution. Still, I may hope to be the easiest explainer for you to read, especially when it comes down to the “real analysis” and “Bayesian statistics” parts.
Case 1: Probability 1 that big-money = $100
Suppose you know that the only possible amount of money in the big-money envelope is $100. You open envelope Alpha and find it contains $100. You obviously should not switch. If Alpha had contained $50, on the other hand, you should definitely switch.
Case 2: Probability .5 that big-money = $100, and .5 that it equals $200
Suppose you know that the big-money envelope contains either $100 or $200, with equal likelihood. If you open envelope Alpha and find it contains $200, you shouldn’t switch. If Alpha contains $50, you *should* switch. If Alpha contains $100, you need to think a little. The other envelope might contain $50, or it might contain $200. Each is equally likely, so the expected value is .5($50) + .5($200) = $25 + $100 = $125. You should switch. The 50-50 idea that created the paradox fails if the first envelope contains $50 or $200, but works if it contains $100. Switching is a good idea if Alpha = $50 or Alpha = $100, but if Alpha = $200 and you switch, you lose bigtime, so a policy of “Always Switch” is not optimal.
What if we are in the Closed-Envelope Variant? Then you have to choose either “Always Switch” or “Never Switch”; you can’t make your decision depend on the value in the first envelope.
There are four different combinations of envelope values you might have ended up with: (Alpha = $50, Beta = $100), (Alpha = $100, Beta = $50), (Alpha = $200, Beta = $100), and (Alpha = $100, Beta = $200). Each of these combinations is equally likely. The expected payoff from always switching (that is, always getting Beta) is
.25($100) + .25 ($50) + .25 ($100) + .25 ($200) = $25 + $12.50 + $25 + $50 = $112.50
The expected payoff from never switching (that is, always getting Alpha) is
.25($50) + .25 ($100) + .25 ($200) + .25 ($100) = $12.50 + $25 + $50 + $25 = $112.50.
So you should be indifferent about switching, the common-sense answer.
Case 3: Equal probabilities of all values of big-money between $0 and $100
In Case 3 we will use the same method. Suppose Alpha = $100. Then, of course, you’d want to not switch, because you know that there can’t be $200 in Beta and so there has to be $50. What if Alpha = $40? Now you would want to switch. The possibilities for Beta are $20 and $80. If the probabilities were 50-50, the expected value of switching would be .50($20) + .50 ($80) = $10 + $40 = $50, which is better than $40. But in fact there’s even more reason to switch. The values of the big-money envelope are distributed between $0 and $100, so each value has probability 1/100, which is 0.01. The values of the little-money envelope are distributed between $0 and $50, so each value has probability 1/50, which is 0.02. Thus, if you see Alpha = $40, that is twice as likely to be the little-money envelope as the big-money envelope. In turn, we then know that Beta is twice as likely to be the big-money envelope as the little-envelope one. That means our computation should be
So you should definitely switch if Alpha = $40. Repeating this kind of calculation, you should switch envelopes for any values of Alpha between $0 and $50.
If, however, Alpha = $51, you should not switch. This can’t be the little-money envelope, because the big-money envelope can’t have more than $100, which is less than twice $51. So for all Alpha between $50 and $100, you should not switch.
We’ve destroyed the reasoning behind the original paradox. In the Open-Envelope Paradox, you’ll end up switching half the time. In the Closed-Envelope Paradox, you’ll be indifferent about switching.
Case 4: Unequal probabilities of values of big-money between $0 and $100
Suppose next that the probabilities of different values between $0 and $100 for the big-money envelope aren’t necessarily equal. We could have a bigger probability of $5 than of $85, for example. What happens?
The reasoning is much the same. If Alpha = $100, you still don’t want to switch, because you know Beta must be half as much, $50, because it can’t be twice as much, $200. That’s equally true for any Alpha between $50 and $100, noninclusive, for the same reason. So the paradox is busted; there are values for which you do not want to switch. It doesn’t matter whehter you want to not switch for small values.
We have also busted the paradox’s reasoning for the Closed-Envelope Variant. The probabilities of .5X and 2X might be 50-50, but not necessarily, and definitely not at X = $0 and X = $100.
We could actually stop here. There is a finite amount of wealth in the universe, so the big-money envelope can’t contain more than a finite upper limit. But it doesn’t matter that we chose $100 for Case 4; any upper limit results in the same reasoning. We’ll go on, though, because it’s interesting to see what happens if the envelope could contain an infinite amount.
Case 5: Unequal probabilities of big money between $0 and infinity, but E(big-money) is finite
What if there is no upper bound to the possible value of the big-money envelope? This is a possible probability distribution in theory, even though values beyond the entire wealth of the universe are impossible. You can’t put equal probabilities on all
the values, of course, because that would make the sum of all the probabilities bigger than 1, which is impossible. But you could have a distribution of values where the probability of a given value gets smaller and smaller as the value get bigger, to where the probability becomes absolutely miniscule but still positive for high values. The bell curve normal distribution is an example of this— the tail heights, which are the probabilities, get smaller and smaller for bigger and bigger values. Imagine using just the right-hand side of the bell curve, the values for 0 and above.
I actually can’t figure out a way to very easily show that sometimes you shouldn’t switch if big-money values are distributed this way. Broome’s 1995 article in Analysis has a simple proof in Appendix B, but it still is beyond the technical level I employ here. (That article and Nalebuff [1989] are the best to read.) He proves that once you see the amount in Alpha, the expected value of Beta cannot always be greater than the amount you see; only for some of the values of Alpha does that occur.
Case 6: Unequal probabilities of values of big-money between $0 and infinity, and E(big-money) might be infinite
Broome’s proof that you won’t always switch envelopes relies on the fact that there exists an expected value of the amount in Beta given the amount seen in Alpha. But for some distributions that expected value does not exist. What it means for it not to exist in those cases is that the probabilities of high values of the big money doesn’t get smaller faster than the size of those high values increases. This is known as “fat tails”, because the right side of the probability density doesn’t tail off fast enough. The bell curve doesn’t have fat tails, but some distributions do, notably the Cauchy distribution and the Pareto distribution with its parameter equal to 1. For us, if the chooser thinks Alpha follows the Cauchy, we need to use the half-Cauchy that includes only positive values but doubles the height so the probabilities still add up to one.
I think it can be shown that when the expected value doesn’t exist because of a fat right tail, the two-envelope paradox cannot be resolved, though I don’t know of a cite. But scholars have pointed out that this is because such a probability distribution introduces all kinds of paradoxes, like dividing by zero does.
Let’s think more about what a fat-tailed distribution means. If the probability distribution has fat tails and only takes positive values, then the expected value could be said to not exist (an integral doesn’t converge) or to equal infinity, depending on taste. “Infinity” means in practice “going on forever”.
The way to think about it is to ask how you react if you were given the simplest possible choice: a first envelope from the distribution or switching to a new one (with no relation of one being double the other in this example). Thus, if the distribution had equal probabilities from 0 to $100, if your first envelope had $49 in it, you’d switch, but if it had $51, you wouldn’t. If you had a half-normal distribution with an expected value of $50, you’d behave the same way, sticking with $51 but switching if the first envelope had $49. If the first had $51, you’d keep it even though there is some probability the second envelope had $200, or $900, or $3 billion, because the probabilities drop off so fast that the expected value of the second envelope is just $50.
If you had a fat-tailed beta distribution though, and the first envelope contained $50, you’d want to switch, because the expected value of the values beyond $50 is infinite. If the first envelope contained $3 billion, you’d still want to switch— because even though you’d very likely see a smaller value in the second envelope, there’s still enough probability that it will have $3.1 billion or $3.5 billion, or $20 billion to make it worthwhile. In fact, if the second envelope had $20 billion in it, and I offered you a third envelope, you’d cheerfully switch again. But this is crazy. So I don’t think you would ever have such beliefs. But if you don’t, then we are back to finite expected value, and the paradox is killed.1
Subjective probabilities
To resolve the paradox, we needed the initial probabilities that the big-money envelope contains the various possible amounts of money from $0 to infinity. For example, there might be equal probabilities that the big-money envelope has $100, $200, $300, and $400. If that is the probability distribution, the small-money envelope will have $50, $100, $150, or $200. We saw that whatever those probabilities are, so long as the expected value in the big-money envelope is finite, the paradox is resolved. But where do those “prior probabilities” come from?
The description in the paradox does not provide us with those probabilities, so we do not know the true probabilities. But you can guess, and you do have some best guess from your personal point of view. We use your best guess, your personal subjective probabilities, since that is all you have. We can then figure out what you should do, and we saw that it was *not* true that you should always switch envelopes in the Open-Envelope Variant, and also that you should be indifferent about switching in the Closed-Envelope Variant.
“But I have no idea of the probabilities!” you may say. That is not true. You do have some idea, even if it is a very weak idea. I will start by illustrating with a simpler example: the population of Elletsville, Indiana. What do you think it is? Most likely you have never heard of Elletsville and you will say, “I have no idea.” But then if I say, “Do you think it’s more likely it has 500 people or 500 million?” think about how you would answer. If you really have no idea, you would say those are equally likely, 50% and 50%. If I asked you to place a bet of $10 on one of those two values, you be willing to choose either one. But you *do* have some idea. You would choose the popultation of 500 for your bet, because you know that a two you’ve never heard of wouldn’t contain 500 million people. I could keep on proposing bets to you, at various odd, and by seeing what bets you accept, we could figure out your personal probabilities on every possible value. You might come out, for example, with equal probabilities of every population from 0 to 20,000 and then declining probabilities all the way to 140,000, after which you’d say the probability is zero that it’s any bigger.2
In the same manner, we could construct your personal, subjective, probability distribution for each possible value of the big-money envelope. If you say you have no idea, I could propose to you an even-money bet that the value is $49.56 exactly, and you would be willing to accept that bet— you would say that a probability of 50% for $49.56 and 50% for everything else from $0 to infinity is as likely a distribution as any other. More likely, you would choose something more similar to equal probabilities over every value for $0 to $50 trillion and zero probability of anything greater. To resolve the Two-Envelope Paradox, it doesn’t matter what you pick so long as you pick some distribution. What you have seen earlier is that whatever the distribution of probabilities, we can tell you what to do once you pick it.
We could figure out your subjective probability distribution by asking what you would bet on, using the information from your choice. I could offer you (a) $1 if the value turns out to be $40, or (b) $1 if it turns out to be $50, and see which offer you’d rather have. That shows which one you think is more probable. Using different kinds of bets, we could work out the shape of your guesses about the big-money distribution. Once we have that, we can tell you how to behave in the Paradox.
Note that I can’t tell you what is the *best* distribution to believe in. In actuality, the big-money envelope has some single amount of money in it, so the *true* distribution is 100% probability on that one value. But can insist that you have *some* idea of the distribution of possibilities, just as I could insist that you really do have some idea of the population of Elletsville even though there is just one true population.
Concluding Remarks
The paradox we started with was that rationality seemed to give us two contradictory answers for whether to switch envelopes. Since we had no idea whether Alpha is the big-money envelope, it seems you should be indifferent about switching; since .5(.5X) + .5(2X) = 1.25X > X, it seems you should want to switch. The resolution to the paradox is that using 50-50 as the probabilities is wrong; you do get some information from seeing the amount in envelope Alpha and there is a definite answer as to whether you switch— though that answer depends on your prior opinions about how much the “big-money” is.
Too show this, I said “Suppose you think the big-money envelope has probability distribution f(big money),” and then showed that for any reasonable probability distribution, the paradox was resolved. Note that I did *not* try to tell you what probability distribution to use. Usually when people try to address the paradox, they try to find out the “best” probability for big-money and small-money, and whether that’s 50-50 or something else. In my exposition, I don’t care whether you use the best one or any other one— it doesn’t matter. The paradox breaks down as soon as you get specific and choose on particular distribution as the probability of each value of big-money. You don’t have to think about probabilities of probabilities— just use your final result of what probability each value has.
To learn more about the history of the paradox, see Nalebuff (1989), who traces it back to Kraitchik (1942). I’ve
A technical note:
I think Chalmers (2002) has a flaw, one that is most clearly stated in the Wikipedia article as of June 2023 (I have changed Wikipedia’s A and B to my Alpha and Beta.).
If we do not look into the first envelope, then clearly there is no reason to switch, since we would be exchanging one unknown amount of money (Alpha), whose expected value is infinite, for another unknown amount of money (Beta), with the same probability distribution and infinite expected value. However, if we do look into the first envelope, then for all values observed (Alpha = a) we would want to switch because E(Beta|Alpha = a) > a for all a. As noted by David Chalmers, this problem can be described as a failure of dominance reasoning.
Under dominance reasoning, the fact that we strictly prefer Alpha to Beta for all possible observed values a should imply that we strictly prefer Alpha to B without observing a; however, as already shown, that is not true because E(Beta) = E(Alpha) = ∞. To salvage dominance reasoning while allowing E(Beta) = E(Alpha) = ∞, one would have to replace expected value as the decision criterion, thereby employing a more sophisticated argument from mathematical economics.
The defect in the reasoning here is that Beta is not a random draw from the probability density f(Beta) once Alpha = a is observed. Rather, Beta can only take one of two values, .5a or 2a. If f is the uniform density, then f(.5a) = f(.5b). If f is the half-normal or the half-Cauchy or the Pareto with alpha = 1 density, then f(.5a) < f(.5b). But in general, it is quite possible that f(.5a) > f(.5b). Probability densities don’t have to have monotonic tails. It could be that the density is declining everywhere except at 2a, but there it has a probability atom of .5. In that case, you’d definitely want to switch envelopes if Alpha = a, because with probability 1 you’d end up with 2a. On the other hand, if the probability atom was at .5a you wouldn’t want to switch, because with probability 1 you’d end up with .5a. This messes things up technically. It’s still true that for most small Alpha values you would not want to switch but for most large Alpha values you would, but when it’s not true for every small value and every large value, the proof gets hard. And the fact that the expected value is infinite enters in in a much more complicated way.
If you liked this Substack, you’d like "I Am the Very Model of a Modern Major-General," Part I” and “The Two-Envelope Paradox: Stated but Not Resolved” and “Why 0.9999 . . . Equals 1.0000 . . .”.
References:
Broome, John (1995) "The Two-envelope Paradox," Analysis, 55: 6–11, doi:10.1093/analys/55.1.6. Very good. Short and clear.
Chalmers, David J. (2002) “The St. Petersburg two-envelope paradox,” Analysis, 62: 155–157, https://doi.org/10.1093/analys/62.2.155 (April 2002).
Gardner, Martin (1982) Aha, Gotcha. I haven’t seen the book, but Gardner is a good writer.
Gardner, Martin (1989) Penrose Tiles to Trapdoor Ciphers: And the Return of Dr Matrix. I haven’t seen the book, but Gardner is a good writer.
Hoffmann, Christian Hugo (2023) “Rationality applied: resolving the two envelopes problem,” Theory and Decision, 94: 555–573 https://doi.org/10.1007/s11238-022-09906-8(0123456789. Recent and clear.
Kraitchik, M. (1942). Mathematical Recreations. I haven’t seen the book.
Nalebuff, Barry (1989), "Puzzles: The Other Person's Envelope is Always Greener", Journal of Economic Perspectives, 3): 171–181, doi:10.1257/jep.3.1.171.
Samuelson, Paul Anthony (1977) "St. Petersburg Paradoxes: Defanged, Dissected and Historically Described," Journal of Economic Literature, 15: 24-55.
Footnotes
The problem is similar to that in the St. Petersburg Paradox, on which see Samuelson (1977).
Or, you might cheat and google it. Wikipedia says Elletsville had a population of 6,655 in the 2020 census.