Killing A Paradox: The Two Envelopes

Class proper will resume after Labor Day on 9 September.

Link to all Classes. Jaynes’s book (first part):

Video

Rumble

Bitchute (often a day or so behind, for whatever reason)

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.

Subscribe or donate to support this site and its wholly independent host using credit card click here. Or use the paid subscription at Substack. Cash App: \$WilliamMBriggs. For Zelle, use my email: matt@wmbriggs.com, and please include yours so I know who to thank. 13 Comments 1. McChuck But how many were going to St. Ives? 2. Hagfish Bagpipe “Do not read farther. Stop and answer.” Okay — why switch? To get more money? What if the picker wants less money? Not enough info given here to affect switcheroo decision. And if I’ve learned anything in this class, it’s don’t assume anything not explicitly stated. Now maybe you’re going to do some fancy Monty Hall footwork and prove, via Stove, Jaynes, Aquinas, Aristotle, and irrefutable algebraic logic that by switching, and then maybe switching back, I could have ended up with a fatter wad of cash in my greedy hands. But I don’t want filthy cash. Not in this little thought experiment, anyway. I want the envelope filled with moar TRVTH. Which one is that? Reading on… 3. 1 – Barry made them all go to St. Ives because that’s where the Margrave Ludwig hung out. 2 – unlike monty hall this one is easy to write a simulation for – meaning ( I think) that$ x 1/n is correct regardless of how many envelopes there are or whether someone chooses to switch or not. (In contrast I do not think it possible to simulate monty without begging the question.)

This is a variant of the Monty Hall Sketch. The training tells us to switch. My training tells me to organize the contestants and taking a cut of the outcomes that are accumulated by always switching.

My extended answer is “Don’t 2nd guess your choice. Make a choice and move forward.”

Zenos Paradox (always halving the distance to the wall and never getting there) ignores my size 15 ft and my distinct desire to get down on my knees and attempt to measure less than a shoe length when it comes to walking…

They are all the conundrum.

You learn the conundrum so that you don’t get trapped by the conundrum.

Then you get married and have children and the conundrums expand exponentially.

5. Hagfish Bagpipe

Okay, I’m back…

“I wish I could properly convey to you the amount of effort that has gone into “solving” this paradox.”

That’s funny, I had no idea, never heard of it, it struck me as a rather silly “paradox”. But then, we are in a Golden Age of Paradoxical Follies.

” 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.”

Thanks. Now that you point it out, even a mathematical blockhead like me can see it. I knew something was stupid about the whole setup, but couldn’t quite put my finger on the precise point the way you did.

“This is how undead thoughts can still walk.”

I know what you mean, but might it be better expressed as; This is how dead thoughts are yet staggering about among us like ridiculous, glassy-eyed, slack-jawed drooling zombies. ?

Dead thoughts are a plague. They plague me, too. I appreciate your guidance in pointing them out. You have sharp eyes, Briggs. You would have made a good mountain man/trapper/Indian fighter, as well as Great Lakes pirate, blockbuster noir detective/thriller/adventure author, Air Force colonel, gunsmith, fly fishing guide, men’s style magnate, archbishop, truck fleet manager, or Olympic fencer. And yet here you are advising retards like me online for peanuts. That’s cool, dude.

.

6. Hagfish Bagpipe

*”Truck Fleet Manager” is no joke, Briggs — that’s a seriously demanding occupation for people who can manage elaborate logic tree complications. You might have enjoyed being a mountain man/trapper/Indian fighter more, but there’s little demand for that nowadays. (But who knows about tomorrow.)

7. Briggs

All,

A reader at SS posed a question.

Simplified, it’s this: Here is $8. In this envelope is either$4 or $16. Would you like to swap the$8 for the envelope?

The EV of the envelope is 4*(1/2) + 16*(1/2) = 10. So if you follow EV, you switch. Should you switch back (without peeking in the envelope)? The EV of the $8 is$8. Which is still less than \$10. So you don’t switch back.

If you follow the EV decision rule. Which is not any universal requirement.

Knowing the first amount changes everything!

8. NLR

Another way to look at it is that the trick is that the envelope game is a single game, but the paradoxical analysis treats it as if it’s a sequence of games, as if it was a more favorable variation of the Saint Petersburg paradox.

Imagine a game where you are given some amount of money, A to start off with. You have the choice to flip a coin. If it comes up heads, you win double your money, if it comes up tails, you lose half your money. Do you play? In this case, according to the expected value rule, you play as long as you want until you make as much money as you want because you stand to win overall. But as was pointed out in the post, if the second envelope has 2A, then the first just has A, we can’t double again. And if the second had 1/2(A), the the first just has A, we can’t halve again.

9. NLR

For a particularly crazy application of expected value, Sam Bankman-Fried said that he believes in a positive expected value, so he would bet the destruction of the world, if the world would become twice as good, should he win.

Leaving aside the question of what “twice as good” even means, the whole thing with expected value is that you will come out ahead *in the long run*. And maybe the very long run. But if you lose a bet to destroy the world, then you can’t make any more bets, so expected value does not apply here.

10. Riggs

I wonder if you are familiar with the version of the Two Envelope problem formulated by the logician Raymond Smullyan? Smullyan eliminates all kinds of expectation values and probabilities from the paradox, rephrasing it as a purely logical contradicton. It goes like this (if I remember correctly):

First proposition: call the smaller amount x; then the other, larger amount will be 2x. What could you possibly gain by switching? If you have the smaller amount x in your envelope and switch, you gain 2x-x=x. Likewise, if you start with the larger amount 2x and switch, you lose x, the same amount. Therefore, the possible gain equals the possible loss.

Second proposition: call the amount in your envelope x; then the other envelope either contains 2x or x/2 (since we didn’t specify both amounts in advance, as in the text above, this should be a perfectly valid choice). The possible gain is now 2x-x=x again, but the possible loss is suddenly x-x/2=x/2, so the possible gain does not equal the possible loss. Contradiction.

Smullyan, an experienced logician, wrote that he has no idea how to solve this version of the paradox, and that’s saying something. Sometime in the past, when I was still reading the British Journal for the Philosophy of Science (because I thought at the time that philosophers could be helpful in understanding the natural sciences), I found many papers attempting to resolve the problem, but they were using terminology like “in a world where x is the amount of money in the envelope at hand…” etc., and I strongly think that the resolution does’nt lie in philosophical wordplay.

Any thougths?

11. Nate

Riggs, there’s no contradiction in the 2nd item. The second problem simply is not the same problem the same as the first problem.

In the second problem, you have implicitly gained some knowledge of how much you already have. In the first problem you do not. That knowledge makes all the difference in your understanding of your possible gain or loss.

With real numbers:
1st problem, we say the smaller amount = 8. Our possible envelope states are then 8,16 or 16,8.
2nd problem, I call “the amount in my envelope” 8. Our possible envelope states are then 8,16 or 8,4.

These are clearly not the same situation at all, so of course your gain is going to be different.

12. NLR

Another thing is that there are two kinds of uncertainty in this problem. There’s the uncertainty of how much money each envelope contains, but then there’s also the uncertainty of the setup itself, whether the envelopes contain A and 2A or A and 1/2(A). In the original problem, there’s no probability associated with the setup. We don’t know what it is, but it is fixed. That’s why switching doesn’t help.

But suppose the organizers of the game decide that 50% of the time, they will have A, 2A and the other 50% of the time they will have A, 1/2(A). If the first envelope is always A, then it does make sense to switch and you’ll win money overall if you play many times. It doesn’t make sense to switch back, though.

On the other hand, suppose that the envelopes themselves are mixed up. In this case, if we’re in the A, 2A scenario, then either envelope could have A and either could have 2A and likewise in the other scenario. The expected value of this game is 5/8(A) whether you switch or not. 50% of the time, you’ll wind up with A, 25% of the time with 1/2(A) and 25% of the time with 2A.

13. Riggs

Nate, I agree that the situations are subtly different. However, I don’t see how calling the amount in your envelope “x” does add any new information that you didn’t already have. There is money in your envelope, you don’t know how much, but it is a specific amount which you may as well give a name. It seems to be a perfectly innocuous choice, a choice that occurs again and again in countless mathematical problems.

The sense in that the two situations are different is that the two amounts x and 2x (x being the smaller amount) exist “at the same time”, while in the other case, either x and 2x, or x and x/2 exist (x now being the amount in your envelope). That’s what led philosophers to speak about “the world in which…” etc., as I was alluding to above, but comparing either/or scenarios also happens all the time and I still fail to see what is supposed to be illegitimate about this line of thinking that leads to such a drastic change in conclusion.