Saturday, July 2, 2022

279: Improbable Envelopes

 Audio Link

Welcome to Math Mutation, the podcast where we discuss fun, interesting, or weird corners of mathematics that you would not have heard in school.   Recording from our headquarters in the suburbs of Wichita, Kansas, this is Erik Seligman, your host.  And now, on to the math.

Today we’re going to talk about a well-known paradox that a co-worker recently reminded me about, the Two Envelopes Paradox.   It’s similar to some others we have discussed in past episodes, such as the Monty Hall paradox, in that a slightly incorrect use of the laws of probability gives an apparent result that isn’t quite correct.

Here’s how the basic paradox goes.   You are shown two envelopes, and told that one contains twice as much money as the other one, with the choice of envelopes for each amount having been pre-determined by a secret coin flip.  No information is given as to the exact amount of money at stake.   You need to choose which envelope you want.   After you initially point at one, you are told the amount of money in it, and asked, “Would you like to switch to the other one?”   Since you have been presented no new information about which envelope has more money, it should be obvious that switching makes no difference at this point, as either way you have a 50% chance of having guessed the right one.

But let’s calculate the expected value of switching envelopes.   Say the envelope you chose contains D dollars.   If you stick with your current envelope, the expected amount you will get is simply D.  The other one contains either 2D or D/2 dollars, with 1/2 probability for each.   The expected value if you switch is then 1/2*2D + 1/2*D/2, which adds up to (5/4)D.    Thus your expected winnings if you switch are greater than the D you would gain from your first choice, and you should always switch!   But this doesn’t make much sense, if you had no additional information.   Can you spot the flaw in this reasoning?

The key is to recognize that you’re combining two different dollar amounts in your expected value calculation:  the D that exists in the case where you initially chose the smaller envelope is different from the D if you chose the larger one.   The easiest way to see this is if you define another variable, T, the total money in the combination of two envelopes.   In this case, the larger one contains (2/3)T, and the smaller has (1/3)T.   Now your expected winnings become the same as the expected value from switching, (1/2)*(2/3)T + (1/2)*(1/3)T, or simply T/2.  Alternatively, you could have come to the same conclusion by modifying our original reasoning using Bayes’ Theorem, replacing our reuse of D with correct calculations for the conditional values in each envelope.

But weirdly enough, seems to point out an odd strategy that will let you choose whether to switch with a greater than 50% chance of getting the higher envelope.    Here’s how it works.   After you choose your first envelope, choose a random value N using a normal probability distribution.   This is the common “bell curve”, with the important property that any number on the number line has a probability of being chosen, as the ‘tail’ of the bell asymptotically approaches and never reaches 0.   So if the center of the normal distribution is at 100, you have the highest probability of choosing a number near 100, but still a small chance of choosing 1000, and a really tiny chance of choosing 10 billion.   Therefore, if you’ve chosen a number N using this distribution, there is some probability, P, that N is between the dollar values in the two envelopes.   Now assume the envelope you didn’t choose contains that number N, and choose to switch on that basis:  if N is greater than the amount in the envelope you chose, you switch, and otherwise keep your original envelope.

Why does using this random number help?   Well, if your random number N was smaller than both envelopes, then you will always keep your first  choice, and there is no change to the overall 50-50 chance of winning.   If it was larger than both, you’ll always switch, and again no change.   But what if you got lucky, and N was between the two values, which we know can happen with nonzero probability P?   Then you will make the right choice:  you will switch if and only if your initial envelope was the smaller one!    Thus, with probability P, the chance of N being in this lucky interval, you will be guaranteed a win, while the rest of the time you have the original odds.   So the overall chance of winning is (1/2)*(1-P) + 1*P, which is slightly greater than 1/2.   

Something seems fishy about this reasoning, but I can’t spot an obvious error.   I remember also hearing about this solution from a professor back in grad school, and periodically tried searching the web for a refutation, but so far haven’t found one.   Of course, with no information about the actual value of P, this edge can be unpredictably small, so is probably of no real value in practical cases.    There also seems to be a philosophical challenge here:  How meaningful is an unpredictable bonus to your odds of an unknowable amount?   I’ll be interested to hear if any of you out there have some more insight into the problem, or pointers to further discussions of this bizarre probability trick.

And this has been your math mutation for today.