Two Envelopes Problem - Randomized Solutions

Randomized Solutions

Suppose as in the previous section that the player is allowed to look in the first envelope before deciding whether to switch or to stay. We'll think of the contents of the two envelopes as being two positive numbers, not necessarily two amounts of money. The player is allowed either to keep the number in Envelope A, or to switch and take the number in Envelope B. We'll drop the assumption that one number is exactly twice the other, we'll just suppose that they are different and positive. On the other hand, instead of trying to maximize expectation values, we'll just try to maximize the chance that we end up with the larger number.

In this section we ask the question, is it possible for the player to make his choice in such a way that he goes home with the larger number with probability strictly greater than half, however the organizer has filled the two envelopes?

We are given no information at all about the two numbers in the two envelopes, except that they are different, and strictly greater than zero. The numbers were written down on slips of paper by the organiser, put into the two envelopes. The envelopes were then shuffled, the player picks one, calls it Envelope A, and opens it.

We are not told any joint probability distribution of the two numbers. We are not asking for a subjectivist solution. We must think of the two numbers in the envelopes as chosen by the arranger of the game according to some possibly random procedure, completely unknown to us, and fixed. Think of each envelope as simply containing a positive number and such that the two numbers are not the same. The job of the player is to end up with the envelope with the larger number. This variant of the problem, as well as its solution, is attributed by McDonnell and Abbott, and by earlier authors, to information theorist Thomas M. Cover.

Counter-intuitive though it might seem, there is a way that the player can decide whether to switch or to stay so that he has a larger chance than 1/2 of finishing with the bigger number, however the two numbers are chosen by the arranger of the game. However, it is only possible with a so-called randomized algorithm, that means to say, the player needs himself to be able to generate random numbers. Suppose he is able to think up a random number, let's call it Z, such that the probability that Z is larger than any particular quantity z is exp(-z). Note that exp(-z) starts off equal to 1 at z=0 and decreases strictly and continuously as z increases, tending to zero as z tends to infinity. So the chance is 0 that Z is exactly equal to any particular number, and there is a positive probability that Z lies between any two particular different numbers. The player compares his Z with the number in Envelope A. If Z is smaller he keeps the envelope. If Z is larger he switches to the other envelope.

Think of the two numbers in the envelopes as fixed (though of course unknown to the player). Think of the player's random Z as a probe with which he decides whether the number in Envelope A is small or large. If it is small compared to Z he will switch, if it is large compared to Z he will stay.

If both numbers are smaller than the player's Z then his strategy does not help him, he ends up with the Envelope B, which is equally likely to be the larger or the smaller of the two. If both numbers are larger than Z his strategy does not help him either, he ends up with the first Envelope A, which again is equally likely to be the larger or the smaller of the two. However if Z happens to be in between the two numbers, then his strategy leads him correctly to keep Envelope A if its contents are larger than those of B, but to switch to Envelope B if A has smaller contents than B. Altogether, this means that he ends up with the envelope with the larger number with probability strictly larger than 1/2. To be precise, the probability that he ends with the "winning envelope" is 1/2 + P(Z falls between the two numbers)/2.

In practice, the number Z we have described could be determined to the necessary degree of accuracy as follows. Toss a fair coin many times, and convert the sequence of heads and tails into the binary representation of a number U between 0 and 1: for instance, HTHHTH... becomes the binary representation of u=0.101101.. . In this way, we generate a random number U, uniformly distributed between 0 and 1. Then define Z = - ln (U) where "ln" stands for natural logarithm, i.e., logarithm to base e. Note that we just need to toss the coin long enough to be able to see for sure whether Z is smaller or larger than the number a in the first envelope, we do not need to go on for ever. We will only need to toss the coin a finite (though random) number of times: at some point we can be sure that the outcomes of further coin tosses is not going to change the outcome of the comparison.

The particular probability law (the so-called standard exponential distribution) used to generate the random number Z in this problem is not crucial. Any probability distribution over the positive real numbers which assigns positive probability to any interval of positive length will do the job.

This problem can be considered from the point of view of game theory, where we make the game a two-person zero-sum game with outcomes win or lose, depending on whether the player ends up with the higher or lower amount of money. The organiser chooses the joint distribution of the amounts of money in both envelopes, and the player chooses the distribution of Z. The game does not have a "solution" (or saddle point) in the sense of game theory. This is an infinite game and von Neumann's minimax theorem does not apply.