Topics Probability

Question A friend posed a problem that according to him reveals an inconsistency in mathematics. There are two envelopes with money in them, and you are given one envelope. One envelope has twice the amount of money as the other, but you don't know which one is which. The question is, if you are trying to maximize your money, after you are given your envelope, should you switch to the other envelope if given the chance? One analysis is: let a denote the smaller amount. Either you have a or 2a in your envelope, and you would switch to 2a or a, respectively, and since these have the same chance of happening before and after, you don't improve and it doesn't matter if you switch. The other analysis is: let x denote the value in your envelope. The other envelope has either twice what is in yours or it has half of what is in yours. Each of these has probability of .5, so .5(2x) + .5(.5x) = 1.2x, which is greater than the x that you started with, so you do improve and should switch. Is there something wrong with the latter analysis? If so, where does it go wrong? Does this bear on inconsistency in mathematics or in probability?

Accepted:
January 2, 2009

Comments

Wouldn't it be nice if mathematics could be brought down so easily! But, sorry, no cigar this time.

It is indeed true that the probability that you are holding the fat or meager envelope is 50/50. Here are the two cases:

1. If you are holding the meager envelope, then switching gets you from x to 2x for a gain of x.

2. If you are holding the fat envelope, then switching gets you from x to 1/2 x for a loss of 1/2 x.

But note that the "x" in these two cases does not signify the same amount of money. In Case 1, x is the smaller amount. In Case 2, x is the larger amount. In Case 1, your gain from switching is the smaller amount. In Case 2, your loss from switching is 1/2 the larger amount (equal to the smaller amount).

The illusion arises because, at first blush, the situation seems similar to another where someone offers to give you 50/50 odds on either doubling or halving some fixed amount of money you have. There your reasoning goes through and you are well-advised to accept. Your probability-weighted pay-off is 1.25x (a little more than you wrote).

But the problem your friend posed is different. Here you do not have a 50/50 chance of either doubling or halving some fixed amount of money. Rather, you have a 50/50 chance of either doubling a small amount of money or halving a large amount of money.

I'd like to add a little bit to what Thomas has said. Probability problems can be tricky because the answers sometimes depend on small details about exactly what procedure was followed. For example, the problem says that "you are given one envelope." Who gave you the envelope? Did the person who gave you the envelope know which envelope was which? Was he a very stingy person, who might have been more likely to give you the envelope with the smaller amount of money? If so, then the probability that you have the smaller amount might not be 1/2.

But that is clearly not the intent of the problem, so let us assume that the person who gave you the envelope flipped a coin to decide which envelope to give you. Then, as Thomas says, the probability is 1/2 that you have the small amount and 1/2 that you have the large amount. Suppose that you open your envelope and find $100 in it. You now know that the other envelope contains either $50 or $200. Do these two outcomes still have probability 1/2 each? Not necessarily; by opening your envelope you acquired new information, and that information could change the probabilities. The answer depends on what procedure was used to decide how much money to put in the envelopes.

Suppose the person who filled the envelopes used the following procedure: they chose a random integer x from 1 to 100, with each integer being chosen with probability 1/100. Then they put $x in one envelope and $2x in the other. In that case, a short calculation using the laws of conditional probability shows that, yes, the probability is 1/2 that the other envelope contains $50 and 1/2 that it contains $200. The expected value of the amount of money in the other envelope is therefore $125, so you would be well-advised to switch.

But now suppose the envelopes were filled by choosing x between 1 and 50, and putting $x in one and $2x in the other. Now, when you find $100 in your envelope, you know for sure that the other envelope contains $50, and you should not switch.

What if we don't know anything about how the envelopes were filled? Now the question is more difficult, but I would be inclined to say that probability theory cannot tell us the probability of the other envelope containing $50 or $200. Probability theory can only tell us how to compute probabilities if we have a well-defined probability distribution for the possible outcomes of the random event under consideration. The choice of this probability distribution is a matter of interpretation; it is not a purely mathematical issue. If we don't have enough information to determine the distribution, then it is not clear that there is a right answer to the probability question.