Another probability “paradox”, the two-envelope problem1, goes like this:
Before you are two envelopes, A and B. One of them contains $X and the other $2X (which is equivalent to $Y/2 and $Y). You pick one envelope and are (1) asked if you would like to keep it or switch, or (2) open it, view its contents, and then asked if you would like to keep it or switch. Which strategy, keeping or switching, is likeliest to win you the big bucks?
No peeking solution
The traditional paradoxical solution to (1) is to argue this. Suppose you pick A, which can be said to have $X (recall you do not peek). There is then a 50% chance that B contains 2X (dropping the dollar signs, but understanding they are still there, lurking) and a 50% chance that B contains X/2. The “expected value“2 of B is said to be
0.5 * 2X + 0.5 * X/2 = 5/4 X.
So clearly you should switch, since the expected value of the envelope you did not pick is larger than X, which is the value of the envelope you hold.
But that’s nuts, because you could have picked up B, in which case the expected value of A is 5/4 X, too. Oh, what to do!
There’s a lot of nervousness about this problem; it has, as I said, been called a “paradox” because it does not conform to our understanding of the theorems of decision analysis.
To solve this, understand that there are unstated, tacit premises here. One is that X > 0; another is that X is real, i.e. X can take infinitely many values, and even approach infinity itself. The first assumption (non-negativity) is unproblematic. The trouble comes from assuming X is “real.”
Now, “real” numbers are one of those creations whose names lie. That is, “real” numbers are unreal, in that most of them cannot be seen nor can they be constructed. Actual-real numbers are integers, or ratios of integers, where the integers can be large but finite. What if instead of assuming X “real”, we assume what is true, that X is an integer and is finite? Does that help us out of the paradox?
Take a close look at that “expected” value. Can we actually ever see “5/4 X”? We cannot. By the premises, we can only ever see X or 2X (or Y/2 or Y). The value “5/4 X” is impossible. Thus, we do not ever “expect” to see it.
Well, this is no answer, is it? It is just fretting over an unfortunate naming of a function of the data. Not quite.
What we have learned, in this paradox, is that expected value, whatever be its name, is not always useful as a criterion for decision making when we have discrete, finite information. We just assumed that it would be.3 Thus, the trivial solution to this paradox is nothing deeper than acknowledging that a one-number summary of a complex situation is, at times, insufficient for a decision criterion.
We can say there is a 50% chance that the envelope you did not pick has 2X and another 50% chance that it has X/2. And that’s it; that’s as simple as we can make it. We cannot do away with either of the probabilities, or bury them inside some function.
The answer, strictly according to probability and not expected value: Switching or keeping makes no difference.
Note carefully that there is not one word here, or anywhere, about “repeated” trials of this game. You only get to play once!
At first blush, seeing $X when you open, say, envelope A is no different than the no-peeking situation. After all, B can still contain $2X or $X/2, right? Well, if X is real, then maybe that’s so (but see here for another argument, which is discussed in Part II).
But if X is actual-real, discrete and finite that is, then the situation is different. Suppose the units of X are dollars: it makes no difference what currency you use; the only point is that X comes to us in discrete, indivisible chunks (we could do pennies if you insist on the most basic unit).
It’s also true that if somebody were to play this game, they would not have an infinite amount of money available. There would be at most N dollars, where N can be as large as you like, just not infinite.
Suppose you pick A and find it contains $1. Do you switch or keep? Switch! Because there is a 0% chance that B contains $1/2, which is an impossible amount because, don’t forget, X comes in discrete units and cannot be found in fractions. In fact, if you see any odd value of $X then you should switch, because we are 100% sure that the other envelope contains $2X! Switching will always double your money if $X is odd.
What about $X even? Well, then we are closer to the no-peeking solution, because we might think there is a 50% chance that the envelope you did not pick has 2X and another 50% chance that it has X/2. But we actually have more information, which says we should switch only up to a point.
The total amount of money in the game is 2X + X = 3X, which must be less than or equal to N. Suppose you open, say, A and find some amount W such that W + 2W > N. Then you are 100% sure that B has W/2! You should keep, and not switch, any envelope where three times its amount is greater than N (because we know that W + W/2 <= N; where "<=" is less than or equal to). We can call these keeper Ws "large." One consequence is that large Ws will always be even.
But we should switch when X is even when we know that 3X < N, because we are gaining information just knowing this fact. Just how that information plays out will be explained in Part II.
Thus, when N is known, and X is actual-real, we have a solution guaranteed to maximize your profit. (Incidentally, expected value calculations sometimes work here, but only when X is odd or “large.”)
In Part II: the solution for unknown or unevenly distributed N.
1I’ve not seen either of these solutions in exactly this way, which means little; that is, the solutions might be out there and are unknown to me, which is the most likely case. Mark D. McDonnell and Derek Abbott offer a repeated-trials “randomly” switching solution for real X when one can peek. They also wisely consider and offer a solution for a bounded N. We examine this in Part II.
Sandy Zabell is credited with bringing this problem to the attention of Bayesian statistics. But he called it the “exchange paradox.”
John Norton also fingered the reliance on expected values to be the sticking point in the no-peek game, but was loath to give up on the device. He invented an “improper probabilities”—i.e. probabilities which are not probabilities—solution to leap from the pit.
2“Expected value” is the sum of everything that can happen multiplied by the probability of everything that can happen. Think of it as a weighted average.
3Expected value can also be used to say that the expected value of the envelope you hold is 0.5 * X + 0.5 * 2X = 3/2 X. But that’s also the expected value of the other envelope. Both are equal—and impossible values—-so switching or keeping are equivalent. Once more, sticking to the probabilities gives the unambiguous answer.