Class proper will resume after Labor Day on 9 September.

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HOMEWORK: Review the old material! Class resumes on 9 September.

Lecture

The so-called two envelope paradox is this. There are two envelopes, both containing money, one with twice as much as the other. You pick one. No peeking. You are then offered to switch. Should you? And once you switch, you are offered again to switch. Should you? Do not read farther. Stop and answer. My prediction about your answer is this: the more formal mathematical education you have, the less likely you’ll answer correctly.

Recall the the badly named “expected value” of an “outcome” is the sum of possible values of the outcomes multiplied by their probability. For a die roll, the possible outcomes are 1, 2, …, 6. Given our usual premises, each has probability 1/6. Thus the expected value is (1 x 1/6) + (2 x 1/6) + … + (6 x 1/6) = 3.5. That number makes sense if you are physicist calculating moments (as in statics) and the like, but of course you can never expect a real die to come up 3.5. Think of the expected value as a probability weighted outcome.

In formal decision analysis, one is supposed to pick outcomes that have higher expected values. For instance, suppose you have the first ordinary die and a second which is marked 2, 3, …, 7. In a new game, you want to pick the die that gives you a better chance of getting a higher score. The expected value of die 2 is (2 x 1/6) + (3 x 1/6) + … + (7 x 1/6) = 4.5. Since 4.5 > 3.5, you would choose the second die.

I now quote from the seemingly endless Wokepedia article on the two envelope paradox.

1. Denote by A the amount in the player’s selected envelope.
2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.
3. The other envelope may contain either 2A or A/2.
4. If A is the smaller amount, then the other envelope contains 2A.
5. If A is the larger amount, then the other envelope contains A/2.
6. Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.
7. So the expected value of the money in the other envelope is
   
8. This is greater than A so, on average, the person reasons that they stand to gain by swapping.
9. After the switch, denote that content by B and reason in exactly the same manner as above.
10. The person concludes that the most rational thing to do is to swap back again.
11. The person will thus end up swapping envelopes indefinitely.
12. As it is more rational to just open an envelope than to swap indefinitely, the player arrives at a contradiction.

The gist is that you should pick an envelope, then you should switch, because the expected value of the envelope you’d switch to is higher than the one you hold. Then, after you have switched, you should switch again, since the expected value of the envelope that you rejected is now higher. And so on ad infinitum. There is no way to know what to do! A paradox!

I wish I could properly convey to you the amount of effort that has gone into “solving” this paradox. Scroll through the Wokepedia article. My favorite is the sci-fi sounding “Nalebuff asymmetric variant”. Maybe you’ll enjoy the peer-reviewed article “Gain from the two-envelope problem via information asymmetry: on the suboptimality of randomized switching”. There at thousands of references and works at scholar.google.com. The discussion goes on and on and on and on some more, just like the switching.

Ask any mathematically untrained person whether he should switch and he will say, and say rightly, “It makes no difference.” That is the right answer. You won’t know one envelope from another. There is no paradox. Why does one seem to arise? Because of the rush to get to the “expected value”. If people stuck with probability, and not swapped in some decision rule, which is not probability, for probability, there never would have been any difficulty. Which I shall now prove to you, if you are mathematically trained and need proof.

Look again at the Wokepedia statement of the problem. There is an equivocation, and it goes by so fast in the heat to get the juicy math, that few see it.

Here’s how you can see it. Take any two amounts, one twice as much as the other. Say, 7 and 14 (dollars, yen, whatever). Call the envelopes F and G. What is the probability F gets 7 and, necessarily, G gets 14 (because it’s a joint event)? Well, 1/2, on these premises. What is the probability F gets 14 and, necessarily, G gets 7? Again, 1/2. You pick up F, say. The chance is has 7 (and G has 14) is 1/2; the chance is has 14 (and G has 7) is 1/2. That’s it. You’re done. It’s 50-50, just as common sense says. Switch or not, it’s the same.

Now let’s look at the Wokepedia expected value explanation. Steps 1 and 2 are correct. These say your envelope has A, which in our example can be 7 or 14. The equivocation, the huge mistake, the great notational blunder, happens in Step 3. Step 3 says “The other envelope may contain either 2A or A/2.”

No it doesn’t. No it can’t. If your envelope has A = 7 (probability = 1/2), then the other envelope can have 2A, all right, but it is impossible it contains A/2. Likewise, if your envelop has A = 14 (probability = 1/2), then the other envelope can have A/2, but it is impossible it contains 2A. All the rest of the argument, and all the hundreds of papers are premised on an impossibility. No wonder there’s a paradox!

But it sounds right, that 2A and A/2, and so it’s off to the expected value races. The equivocation happens by abusing the notation A. I’ve warned us since Day One of this Class that notation can be a tremendous help, or a dire curse. When we forget that notation is only a crutch, a stand in for plain English propositions, then we are apt to make mistakes. We did here.

There is nothing special about 7 and 14. Pick x and 2x. Then Step three would read “The other envelope may contain either 2x or x/2″ if your first envelope had x, or “The other envelope may contain either 2(2x) or (2x)/2″ if your envelope had 2x. Yet there is no x/2, because the lowest amount is x; there is no 2(2x) or 4x, because the highest amount is 2x. The problem comes in forgetting that one either has x or 2x at the start, and that all other values are impossible.

Now if have no training, you will forget about this murdered paradox and move on. But if you have probability training, and especially if you’ve seen it before and marveled at it, you will seek to rescue it. “This stuff today must be wrong! Look at all the other work that has gone into it. The paradox must be real!” This is human nature, but it’s more so in science because intelligent people invest a lot of thought in their work. This is how undead thoughts can still walk.

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