# The Two Envelopes Problem – in a nutshell.

The Two Envelopes Problem, also known as the Exchange Paradox, is quite simple to state. You are handed two identical-looking envelopes, one of which, you are informed, contains twice as much money as the other. You are asked to select one of the envelopes. Before opening it, you are given the opportunity, if you wish, to switch it for the other envelope. Once you have decided whether to keep the original envelope or switch to the other envelope, you are allowed to open the envelope and keep the money inside. Should you switch?

Switching does seem like a no-brainer. Note that one of the envelopes (you don’t know which) contains twice as much as the other. So, if one of the envelopes, for example, contains £100, the other envelope will contain either £200 or £50. By switching, it seems, you stand to gain £100 or lose £50, with equal likelihood. So the expected gain from the switch is 1/2 (£100) + 1/2 (-£50) = £50-£25 = £25.

Looked at another way, the expected value of the money in the other envelope = 1/2 (£200) + 1/2 (£50) = £125, compared to £100 from sticking with the original envelope.

More generally, you might reason, if X is the amount of money in the selected envelope, the expected value of the money in the other envelope = 1/2 (2X)+ 1/2 (X/2) = 5/4 X. Since this is greater than X, it seems like a good idea to switch.

Is this right? Should you always switch?

Look at it this way. If the above logic is correct, then after switching envelopes the amount of money contained in the other envelope can be denoted as Y.

So by switching back, the expected value of the money in the original envelope = 1/2 (2Y) + 1/2 (Y/2) = 5/4Y, which is greater than Y, following the same reasoning as before. So you should switch back.

But following the same logic, you should switch back again, and so on, indefinitely.

This would be a perpetual money-making machine. Something is surely wrong here.

One way to consider the question is to note that the total amount in both envelopes is a constant, A = 3X, with X in one envelope and 2X in the other.

If you select the envelope containing X first, you gain 2X-X = X by switching envelopes.

If you select the envelope containing 2X first, you lose 2X-X = X by switching envelopes. So your expected gain from switching = 1/2 (X) + 1/2 (-X) = 1/2 (X-X) = 0.

Looked at another way, the expected value for the originally selected envelope = 1/2 (X) + 1/2 (2X) = 3/2 X. The expected value for the envelope you switch to = 1/2 (2X) + 1/2 X = 3/2 X. These amounts are identical, so there is no expected gain (or loss) from switching.

So which is right? This reasoning or the original reasoning. There does not seem a flaw in either. In fact, there is a flaw in the earlier reasoning, which indicated that switching was the better option. So what is the flaw?

The flaw is in the way that the switching argument is framed, and it is contained in the possible amounts that could be found in the two envelopes. As framed in the original argument for switching, the amount could be £100, £200 or £50. More generally, there could be £X, £2X or £1/2 X in the envelopes. But we know that there are only two envelopes, so there can only be two amounts in these envelopes, not three.

You can frame this as £X and £2X or as £1/2 X and £X, but not legitimately as £X, £2X and £1/2 X. By framing it is as two amounts of money, not three, in the two envelopes, you derive the answer that there is no expected gain (or loss) from switching.

If you frame it as £X and £2X, there is a 0.5 chance you will get the envelope with £X, so by switching there is a 0.5 chance you will get the envelope with £2X, i.e. a gain of £X. Similarly, there is a 0.5 chance you selected the envelope with £2X, in which case switching will lose you £x. So the expected gain from switching is 0.5 (£X) + 0.5 (-£X) = £0.

If you frame it as £X and £1/2 X, there is a 0.5 chance you will get the envelope with £X, so by switching there is a 0.5 chance you will get the envelope with £1/2 X, i.e. a loss of £1/2 X. Similarly, there is a 0.5 chance you selected the envelope with £1/2 X, in which case switching will gain you £1/2 X. So the expected gain from switching is 0.5 (-£1/2 X) + 0.5 (£1/2 X) = £0.

There is demonstrably no expected gain (or loss) from switching envelopes.

In order to resolve the paradox, you must label the envelopes before you make your choice, not after. So envelope 1 is labelled, say, A, and envelope 2 is labelled, say, B. A corresponds in advance to, say, £100 and B corresponds in advance to, say, £200, or to £50, but not both. You don’t know which corresponds to which. If you choose one of these envelopes, the envelope marked in advance with the other letter will contain an equal amount more or less than the one you have selected. So there is no advantage (or disadvantage) in switching in terms of expected value. In summary, the clue to resolving the paradox lies in the fact that there are only two envelopes and these contain two amounts of money, not three.

*Exercise*

You are handed two identical-looking envelopes, one of which, you are informed, contains twice as much money as the other. You are asked to select one of the envelopes. You select an envelope. Before opening it, you are now given the opportunity, if you wish, to switch it for the other envelope. Once you have decided whether to keep the original envelope or switch to the other envelope, you are allowed to open the envelope and keep the money inside. Should you switch? In making your decision, consider an example where the first envelope you chose contained £100. This means that the other envelope contains either £200 or £50.

*References and Links*

ThatsMaths. The Two Envelopes Fallacy. Nov. 29, 2018. https://thatsmaths.com/2018/11/29/the-two-envelopes-fallacy/

Two Envelopes Problem. Wikipedia. https://en.m.wikipedia.org/wiki/Two_envelopes_problem