Killing A Paradox: The Two Envelopes
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- Опубліковано 12 вер 2024
- The so-called two envelopes paradox happens because smart people were too enthusiastic about mathematical notation. Don't let this happen to you!
NEW UPDATE:
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!
Written lecture:
www.wmbriggs.c...
wmbriggs.subst...
Permanent class page: www.wmbriggs.c...
Thank you. Very helpful.
The psychology aspect is the most interesting.
I think the confusion starts earlier and with a more familiar kind of math than expected value. We have two propositions:
P One envelope is going to have a certain amount of money and the other is going to have twice as much
Q If one envelope contains A, the other envelope may contain either 2A or A/2
There's the mistake, thinking Q follows from P, or even thinking it's just a simple unquestionable restatement of P. But is It? We're now talking about three sums of money, A, 2A, A/2 when before we were talking of just two.
If we just stick with P, then as you point out, most people correctly conclude there's no point in switching. It's only when we incorrectly move on to Q that we get the strange result. Because yes, if we switch the envelope with A for one with 2A or A/2 we do indeed stand to gain twice as much as we stand to lose with the same probability. But that's not the condition we started with. Effectively it's been turned into a three envelope game.
Matt, I can't tell you how much I appreciate your putting these lectures up-- FOR FREE-- on YT. But I will try: thank you so much!
A pleasure. Thanks for watching.
Let T be the total in the two envelopes. One has T/3, and the other has 2*T/3. The expectation in your envelope is:
(T/3)*(1/2) + (2*T/3)*(1/2) = (T/6) + (T/3) = T/2.
The expectation in the other envelope is:
(2*T/3)*(1/2) + (T/3)*(1/2) = (T/3) + (T/6) = T/2.
The reason this works, is because the probability for each envelope does not consider its value. And the value is another random variable. The paradoxical solution claims that Pr(MINE=LOW) is 1/2. That is true, but the probability they _use_ should be Pr(MINE=LOW *_AND_* LOW=F), which is unknown.. As Briggs points out, if the values are 7 and 14, then Pr(MINE=LOW *_AND_* LOW=14) is zero.
It is a simple case of needing a joint probability distribution.
Apologies for my comment below Dr Briggs. At first you seemed to be making the same simple mistake as everybody else I've seen on UA-cam, that If one envelope has A then the other has 2A or A/2, which is an invalid inference from saying one envelope has twice the other. So I rushed in with my version of that point before continuing to watch your vid in which you eventually say pretty much the same thing.
But let's suppose we make that inferred proposition our initial condition. We start off not with not that one envelope contains twice the other, but that if one contains A then the other contains 2A or A/2. Then does it make sense to switch? I suggest there's still no point, because you still don't know whether you have the envelope with A before switching, or whether you have the one with 2A or A/2. In the first case you stand to gain twice what you stand to lose with the same probability, but in the second you stand to lose twice what you stand to gain. (I think describing the outcomes like that is much less misleading than the technicalities of "expected value")
have you shown us your book collection?
Ha. Only barely.