# Two-Envelope Paradox

The **two-envelope paradox** is a scenario in which a player is presented with two envelopes, each containing an unknown amount of money, and asked to choose one after being given the additional information that one envelope contains twice as much money as the other.

Incorrect, but compelling, reasoning could conclude that it is always optimal for the player to switch his choice if allowed: suppose the envelope the player currently holds contains \(D\) dollars. Then the other envelope will contain either \(2D\) or \(\frac{D}{2}\) dollars, with apparently equal probability. This makes the expected value of switching

\[\frac{1}{2} \cdot (2D) + \frac{1}{2} \left(\frac{D}{2}\right) = \frac{5D}{4},\]

which is greater than the expected value of staying, \(D\). Hence it is always beneficial to switch. But this is intuitively pointless due to the symmetry of the situation, which is why this is sometimes referred to as a paradox.

The apparent paradox is notable because the failure in reasoning illustrates a failure to apply Bayes' theorem and conditional probability, as well as formal logic. Slight modifications to the paradox turn the problem into an important problem in philosophy, especially in detecting implicit reasoning.

## Formal Description

Suppose a player were presented with two envelopes \(A, B\), and told that one contains twice the money of the other. After selecting an envelope, the player is given the opportunity to either keep the envelope (and the money it contains), or to switch to the other envelope.

The *switching argument* concludes that it is always beneficial to switch, by way of the following reasoning: suppose the envelope the player currently holds contains \(D\) dollars. Then the other envelope will contain either \(2D\) or \(\frac{D}{2}\) dollars, so the expected value of switching is

\[\frac{1}{2} \cdot (2D) + \frac{1}{2} \left(\frac{D}{2}\right) = \frac{5D}{4},\]

which is greater than the expected value of staying, \(D\). Obviously, this reasoning must be faulty somewhere (it's absurd to obtain an infinite expected value from two finite envelopes!), but the flaw is not immediately obvious.

## Resolution

The simplest resolution is to closely re-examine the expected value calculation:

\[\frac{1}{2} \cdot (2D) + \frac{1}{2} \left(\frac{D}{2}\right) = \frac{5D}{4}.\]

The error in this calculation lies in a subtle misunderstanding: the two '\(D\)'s in the calculation actually represent different values, that are incorrectly equated. In particular, the '\(2D\)' represents the expected value in the other envelope *given that it is the larger one*, and the '\(\left(\frac{D}{2}\right)\)' represents the expected value in the other envelope *given that it is the smaller one*.

In other words, the additional information "the second envelope contains less money than the first" means that the player should update his belief about the second envelope to be smaller than he would expect without that additional information, and similarly approach the case where the second envelope is the larger one. The correct calculation would be

\[\mathbb{E}[\text{second envelope} = B] = \mathbb{E}(B | A < B)P(A < B) + \mathbb{E}(B | A > B)P(A > B),\]

which, if the envelopes contain \(x\) and \(2x\) money, indeed resolves to \(\frac{3x}{2}\) as intuition would suggest.

Another, perhaps more intuitive, way of understanding the resolution is by considering the total amount of money in the game \(T\), which is fixed regardless of which envelope you choose. But now the amount in your envelope *given that it is the larger one* is well-defined, \(\frac{2T}{3}\). And the amount in your envelope *given that it is the smaller one* is \(\frac{T}{3}\). His expected return from switching is then

\[\mathbb{E}[\text{switch}]=\frac{1}{2} \frac{2T-T}{3} + \frac{1}{2} \frac{T-2T}{3}=0\]

which means there is no advantage or disadvantage to switching, which is exactly as expected.

## Extension to Visible Envelopes

If the situation is slightly modified to allow the player to look inside the envelope prior to making his decision, the apparent paradox continues to hold as this opportunity gives the player no new information (as he would choose to switch, using the naive expected value calculation, regardless of the value of the envelope).

However, it turns out that this opportunity allows the player to choose the larger envelope with probability greater than \(\frac{1}{2}\)! This is an extremely surprising result, possible only by using a **randomized strategy**.

First, the player chooses a continuous random variable \(Z\) that is positive on any interval; the choice of distribution doesn't matter much (for instance, the normal distribution will do fine). The player first selects a random number \(z\) according to this distribution, and compares it to the number in his envelope \(m\).

- If \(z<m\), the player chooses to stay. Intuitively, the player has found that the number in his envelope is "large," so he does not want to switch.
- If \(z>m\), the player chooses to switch. Intuitively, the player has found that the number in his envelope is "small," so he wants to switch.

The analysis of this strategy is relatively straightforward. There are three cases to consider:

- Case 1: \(z\) is less than the numbers in both envelopes.

In this case the player will choose to keep his envelope, no matter if 1.a. the player has the larger amount, or 1.b. they player has the smaller one. The randomization does not help or hurt the player here; the player still has an even chance of having the larger or smaller envelope. - Case 2: \(z\) is greater than the numbers in both envelopes.

In this case the player will choose to switch his envelope, no matter if 2.a. the player has the larger amount, or 2.b. the player has the smaller one. Again, the randomization does not help or hurt the player here; there is an even chance of switching to the larger or smaller envelope. - Case 3: \(z\) is in between the numbers in each envelope.

In this case, if 3.a. the player started with the smaller envelope, the player would choose to switch and thus end up with the larger envelope. If 3.b. the player instead started with the larger envelope, the player would choose to stay. In either case, the player would end up with the larger envelope!

Thus, so long as the third case has a positive probability of occurring, the player will get the larger envelope with probability greater than \(\frac{1}{2}\). But this is indeed the case since \(Z\) is positive on any interval (and thus also the interval between the two envelopes), so this strategy guarantees the player the desired greater than \(\frac{1}{2}\) probability.

## See Also

**Cite as:**Two-Envelope Paradox.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/two-envelope-paradox/