You are given two indistinguishable envelopes, each containing money, one contains twice as much as the other. You may pick one envelope and keep whatever is inside. You pick one envelope at random but before you open it you are given the chance to take the other envelope instead. Is it in your interest to switch envelopes?

The Two Envelope Problem is a classic problem in probability theory and decision-making. The paradox lies in the fact that the expected value of switching appears to be greater than the expected value of not switching, which may prompt one to believe they should always switch. However, the reasoning behind this conclusion breaks down under closer examination.

Let’s first establish some notation. Let X be the amount of money in the first envelope, and let Y be the amount of money in the second envelope. We know that either Y = 2X or X = 2Y. Without loss of generality, let’s assume that we initially picked envelope 1, which contains the amount X. Now, let’s calculate the expected value of switching and not switching.

If we don’t switch envelopes, it’s clear that the expected value is simply X.

Now, consider the case of switching envelopes. Without knowing the value of X, we know that there’s a 50% chance that we’d end up with twice as much money and a 50% chance that we’d end up with half as much. So, the expected value of switching would be the weighted average:

$$E(Switch) = 0.5(2X) + 0.5(\frac{1}{2}X)$$

Calculating the expected value:

E(Switch) = X + 0.25X = 1.25X

It seems that the expected value of switching is greater than not switching envelopes. However, there’s an error in this reasoning.

The problem with the above calculation is that it doesn’t take into account the joint distribution of the amounts X and Y. The premise that Y = 2X and X = 2Y are both equally likely with probability 0.5 is inconsistent – we can’t assign probabilities based on dividing the possibilities by 2 without knowing the actual distribution.

To see this point more clearly, let’s examine a specific example where we know the distribution. Suppose the amounts in the envelopes come from the following discrete distribution:

P(X = 1) = 0.5, P(X = 2) = 0.5

This creates the following joint distribution:

P(X = 1, Y = 2) = 0.25, P(X = 2, Y = 1) = 0.25,

P(X = 2, Y = 4) = 0.25, P(X = 4, Y = 2) = 0.25

Now, let’s compute the expected value of switching envelopes based on this specific distribution:

E(Switch|X = 1) = (0.5)(2) = 1

E(Switch|X = 2) = 0.5(1) + 0.5(4) = 2.5

E(Switch) = E[E(Switch|X)] = 0.5(1) + 0.5(2.5) = 1.75

However, the expected value of not switching simply equals the expected value of X:

E(NotSwitch) = E(X) = 0.5(1) + 0.5(2) = 1.5

In this example, the expected value of switching is higher than not switching, but this result depends on the specific distribution of amounts in envelopes. For different distributions, the result may change – in some cases, indeed, it may be better to not switch.

In summary, the Two Envelopes Problem highlights the counterintuitive nature of some probability problems. The decision of whether to switch or not depends on the underlying distribution of the amounts in the envelopes. In most real-life situations, it is likely that there will not be enough information to accurately determine should you switch or not, hence no definitive answer can be given without specifying the probability distributions in question.