The Two Envelopes Problem (Solved)

Discussions on solving the paradox of the two envelopes problem have been ongoing for years. However, its solution seems relatively simple.

What is the two envelopes problem?

Imagine you have two envelopes, A and B, each containing an unknown amount of money. You know that one envelope has twice as much money as the other, but you don’t know which one. You randomly choose envelope A.

Now, the question arises: Should you switch to envelope B?

To answer this, we calculate the expected value if you switch:

• If envelope B contains half of A's value, you lose half of A’s amount.
• If B contains twice A’s value, you double the amount.

Our choice is random, therefore both options appear equally likely.

Therefore, if envelope A contains X, then the expected value of money we get after switching to B is:

? (2X) + ? (?X) = 1? X > X

This suggests it’s worth switching!

Assuming this reasoning is correct leads to another paradox because after switching to envelope B, we face the same decision again and might conclude we should switch back to A.

So, what’s really going on here?

When the chosen envelope remains closed

Let’s assume we chose envelope A without opening it. It contains either X or 2X with the same probability. In this case, switching gives a ? chance of gaining X and a ? chance of losing X.

The expected value of such a switch (i.e. the difference of the expected value after and before this switch) is:

?X - ?X = 0

– meaning it doesn’t matter if we switch or not.

When we open the envelope

The situation becomes more complicated when we open the envelope (which is how the two envelopes problem is often described) and find, for example, $10. And say that even after opening the envelope, we still have the option to change our decision.

The reasoning described earlier cannot be applied because the value in the envelope is already known.

The issue is that we’re not choosing between whether the other envelope contains $5 or $20, because that would mean our choice influences the sum placed in the envelopes when the money was put there (i.e., whether there was $15 or $30 in total).

What if we know the total pool

If we know the total sum of money placed in the envelopes (e.g., $30), and we find $10 in the opened envelope, it’s clear that we should switch since the other envelope must contain $20 (though this hardly qualifies as a game).

What if the total pool is unknown

Okay, but how it's possible that the total sum placed in the envelopes is unknown?

Let’s assume that the game organizer has a certain pool of money P, from which a certain sum, less than or equal to P, is randomly selected and then divided into three parts. One part goes into one envelope, and the other two parts go into the other envelope. No one knows how much is in the envelopes. So, the chosen envelope can contain any amount in the range (0; ?P].

Specifically, one envelope will have an amount in the range (0; ?P], and the other will have an amount in the range (0; ?P]. This means that numbers from the range (0; ?P] are twice as likely to appear as numbers from the range (?P; ?P]. In other words, the probability of drawing a number from (0; ?P] is ?, and the probability of drawing a number from (?P; ?P] is ?.

The player draws an envelope with X money inside. If X > ?P, it means that the chosen envelope contains more money than the other (otherwise, the other envelope would contain more than ?P, which would mean the total money exceeds the pool P).

This means that in ? of the cases, switching will always result in losing ? X. In the remaining cases, there’s an equal chance of losing ? X or gaining X.

Therefore, the expected value of switching the envelopes is:

? (-?X) + ?X + ? (-?X) = 0

So, it turns out that for any value of the pool P, switching also doesn’t benefit us.

In an actual decision-making process, knowing or estimating the size of the pool P allows us to sometimes make a switch that increases the expected value of the game (when the open envelope contains less than ? of the possible prize pool, we switch the envelopes).

Why was the original reasoning incorrect?

In the original reasoning, which led to the paradox, it was assumed that the probability that the other envelope contains ?X or 2X is the same, regardless of the value of X in the envelope we opened. This would mean that for any X in the opened envelope, there could be 2X in the other envelope, implying that the pool of money that could be in the envelopes is infinite.

The pool must be finite

If we assume an infinite pool in the two-envelope game, the expected value is also infinite. Switching envelopes in this case would indeed increase the expected value by ?, but it would still be the same infinity. So, it actually does not change anything.

Assuming infinite pools of winnings or stakes in games often leads to absurd conclusions. For example, with an infinite pool of money, we could keep doubling our next bet in a casino until a win eventually covers all previous losses.

In conclusion

The paradox arises from a reasoning error, in which we assign a fifty-fifty chance that the other envelope contains ?X or 2X, regardless of the value X of money found in the envelope already opened. This leads to a hidden assumption that the total pool of money in the envelopes is infinitely large.

Such flawed reasoning led to the paradox, which in fact, doesn’t exist.

What was overlooked is that opening the envelope and learning the amount of money alters the probability.