Two Envelopes
The Two Envelope Problem and a Variation
“Alice laughed. 'There's no use trying,' she said. 'One can't believe impossible things.' I daresay you haven't had much practice,' said the Queen. 'When I was your age, I always did it for half-an-hour a day. Why, sometimes I've believed as many as six impossible things before breakfast. There goes the shawl again!” ― Lewis Carroll
The Problem
It goes like this: You are given two identical envelopes, A and B, each with money, where one holds twice as much as the other. After picking an envelope but before opening it, you’re offered the chance to switch.
Intuitively, switching seems unnecessary due to the symmetry of the situation. However, a straightforward expected value calculation:
paradoxically suggests it’s always advantageous to switch, as the potential gain of doubling your money appears to outweigh the risk of halving it. Take a second to think about this if you haven’t seen it before.
Classical Resolution
The classical resolution to this paradox points out that the calculation mistakenly treats A as though it represents both a fixed amount and an expected value in different parts of the formula. Specifically, if A is assumed to be the smaller amount, the expectation differs from the case where it is assumed to be the larger amount. A correct calculation would acknowledge that the expected value depends on these conditional assumptions.
To resolve this issue, we redefine our calculation using conditional expectations. Let 𝑥 denote the smaller amount, meaning the two envelopes contain 𝑥 and 2𝑥. Now, the new EV calculation assuming WLOG we’ve picked A is:
Substituting in the right values, we see that
Similarly, we see that the expectation for A is
The Infinite $$$ Version
We note that the mechanism by which the amounts in the two envelopes are determined is crucial for deciding whether to switch. Consider the case where instead of simply filling two envelopes E1 and E2 with fixed amounts and then labeling them as A and B at random (e.g., via a coin toss), we start by placing a fixed amount a in envelope A, and then determine the contents of envelope B based on both a coin toss and the amount in A. If the coin lands heads, B receives 2a; if tails, B receives a/2.
If the player understands this mechanism and knows that they hold envelope A, although they do not know the coin toss outcome or the value a, the switching argument actually holds, and switching is indeed recommended. This is because, under this mechanism, the expected value of the other envelope becomes unbounded. Each time the player switches, the potential gain doubles or halves with a probability of 1/2, resulting in an unbounded expected value. Under these rules, the calculation to "switch" does not converge to a finite recommendation but instead drives the player towards switching indefinitely.
Peeking Inside the Envelope
Assuming that you are playing this game and now the gamemaster allows you to peek inside the first envelope before making a decision to switch or not, is there a strategy that you can employ that will allow you to choose the envelope with more money strictly more than 1/2 of the time?
Hint: Yes, with an infinite number of solutions in mixed strategy.
Solution in Mixed Strategy
The strategy is as follows: You can select ANY probability distribution D which is positive everywhere and select a random splitting number Y from this distribution. Let’s say the amounts in the envelopes are A and B = 2A, and you pick one of the envelopes and observe an amount of money X that you don’t know if is equal to A or B. You use the decision rule that if X < Y then you switch, and if X > Y you keep X; WLOG we can assume that X = Y occurs with probability 0 (in any case you can decide X = Y with a coinflip without affecting the expectancy of the game).
There are six possible situations which we show below:
In the 1st and 3rd row you choose correctly with probability 1/2, in the middle row you are always choosing correctly. As such, your total probability of a good choice is:
We can split the middle term into two halves to get
We note that the 3 probabilities in the first grouping add to 1 and so
Because D was a probability distribution chosen to be positive everywhere
From which we conclude that your probability of making correct prediction is strictly greater than 1/2 under the given decision rule. Intuitively, although we don’t know the specific distribution Player 1 used to select the two numbers, we do have a prior understanding that they must be drawn from some distribution that includes a range of possible values. The random number y we generate is like a benchmark from this prior distribution. So, while we don’t know exactly where x or y falls in relation to the other unknown number, our comparison between x and y leverages weak but useful probabilistic information.
An immediate inference the discerning reader can draw is that if the gamemaster knew what our distribution D was (without knowing what Y we drew from it), they can minimize the probability we guess correctly to a number arbitrarily close to 1/2 by selecting A and B in a part of the distribution where the pdf is very close to 0 (and hence forcing most situations to either y < a or b < y instead of a < y < b).
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