Envelopes with Numbers

You are given 2 sealed envelopes with numbers inside. You are told that one of the numbers is twice as much as the other one. You grab one of the envelopes and right before you open it, you make the following calculation:

“If this envelope contains X inside, then the other envelope contains either X/2 or 2X inside. Since the chance that the other envelope contains a larger number is exactly 50%, the expected money I will get after switching is X/4 + X = 1.25X > X. Therefore, I should switch!”

Clearly, this reasoning is wrong, since you can’t possibly deduce which envelope of the two contains a larger number. Where is the mistake?

The trick is that conditionally on the fact that your envelope contains X, it is not true that the other envelope has 50% chance of containing either X/2 or 2X. The reason is that it is impossible that all amounts of dollars appear in the envelopes with the same probabilities (densities). Thus, for example, if it is very unlikely that an envelope contains more than 1000, and you open an envelope with 800 inside, you will not think that the other envelope has 50% chance of containing 1600.

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Responses

1. Thanks Jeff for your example. I also wanna state that the above logic is still true if the prize money can go up to infinity value. But since the game show will set a limit, the contestants who draw an amount near the limit will have 0 chance of doubling it. In order for this logic to work, no matter how high the money was in the first draw, we must still have the same expected chance of doubling it.

2. The problem with the “paradoxical solution” is that X, the value of your envelope, is the realization of a random variable. It has a distribution. In mid’s example, the range of X is {$5,$10} with distribution (0.5, 0.5). In the problem as stated, neither is known.
Here’s the example I prefer: You put one $10 bill in an envelope, and set it aside. Then then pick a number N>1. You divide one$20 bill, and (N-1) $5 bills, between N other envelopes. You select one of these at random, add it to the envelope that was set aside, and give them to the person in this puzzle. When this person selects an envelope, there is a 1/2 chance that it is the higher amount of the two, and a 1/2 chance that it is the lower. If (s)he opens it and sees$10, there is a (N-1)/N chance that this is the higher, and 1/N that it is the lower. So the expected gain by switching depends on N, which the contestant does not know.

3. I agree that the envelopes do not contain a randomly placed prize money in it, because the TV game show normally have a certain preference. Let’s say the TV show held this contest 100 times and each time they use $5 and$10 in the envelopes, then there will be 50 contestants who chose $5 and 50 contestants who chose$10 in the first try. If they use this logic and switch the envelopes, in the end there will still be 50 people who get $5 and 50 people get$10, so the total prize money they’ve won are still the same.

1. Your example with only 2 possible numbers in the envelopes (5 and 10) is great. It shows that if you have X in your envelope, then you will certainly have 15-X if you switch, not 2X with 50% and X/2 with 50%.

1. I would suggest the issues lies with the expectation, stated in the second to last sentence of the second paragraph, of getting money after switching the envelopes. Where does it state there is money in the envelopes! If not in the envelopes are we to assume the individual is being paid to switch envelopes?