And this is why we try to hammer into students' head that you can only manipulate series after you've established convergence / that the quantity actually exists in general.
Two envelopes problem
61–70 of 318 posts
Re: Two envelopes problem
#62https://www.youtube.com/watch?v=_NGPncypY68 this video explains the paradox properly, with a very reasonable explanation.
that's an excellent treatment. I'm curious: this video picks a particular example of a distribution of the amounts and shows that the expected profit from switching is not defined. Can one somehow prove that for all possible distributions, either the expected profit from switching is 0 or it is undefined?
the point of the video is that given a distribution, the total expected profit cannot be defined, because the infinite sum of the probabilities of each case don't have an order, and also adds up to positive infinity and negative infinity.
it doesn't really matter what the distribution of the amounts are, as long as there's an infinite number of possibilities in the distribution (ie., it's not a finite amount of possible envelopes).
Re: Two envelopes problem
#63This reminds me of the Monty Hall Problem: https://en.wikipedia.org/wiki/Monty_Hall_problem (except that problem has a more definite solution)
Exactly what I thought too. I find human behavior in such situations fascination. Have penned some thoughts on it - have a read if you like
Re: Two envelopes problem
#64Earlier quoted context omitted.
Well, the simple way to put is that "A" isn't fixed. The "expected value" argument in steps 6-7 is acting like "A" is a single value when "A" will be larger or smaller depending on what envelope you picked.
I think that subtly misses the point. The problem is that you're implicitly using a distribution that... isn't a distribution. And with this particular not-a-distribution, whether you should switch or not does not depend on the value of A . But with any actual distribution (... I think?) it does , at which point... no paradox. It's true that it's not clear (at least to me , but perhaps more generally) what distributi…
No, joe_the_user is exactly correct. The expectation they compute in step 7 requires that A is a constant. The formula seems to just be the regular formula for expectation: Value that B can take times probability that it takes that value, summed over all possible values.
But the "possible values" they put in are A/2 and 2A, which makes no sense. A is a random variable.
Don't forget that random variables are really functions on a probability space with an implicit argument, so (1) when multiple random variables appear in a formula, they're implicitly evaluated at the same argument omega, and the formula must hold for all omega from the probability space, and more importantly, (2) there can't be free-standing omegas in a formula for expectation because expectation is an integral (or sum in this case where the probability space is discrete) over all possible omegas.
Re: Two envelopes problem
#65Earlier quoted context omitted.
Well, the simple way to put is that "A" isn't fixed. The "expected value" argument in steps 6-7 is acting like "A" is a single value when "A" will be larger or smaller depending on what envelope you picked.
Yeah, that's obvious when they say "one envelope contains 2*A and the other A/2". Well no, one doesn't contain one fourth of the other.
Where does it say that?
Re: Two envelopes problem
#66This reminds me of the Monty Hall Problem: https://en.wikipedia.org/wiki/Monty_Hall_problem (except that problem has a more definite solution)
The difference is of course that you do get information in the Monty Hall setup, and here you get none, if I read that correctly.
Re: Two envelopes problem
#67The bug is obvious when they say "one envelope contains 2*A and the other A/2". Well no, one doesn't contain one fourth of the other.
That's not what it's saying. It's saying one envelope contains A and the other contains either 2*A or A/2.
Yes, but doesn't the formula treat it a such?
Re: Two envelopes problem
#68The bug is obvious when they say "one envelope contains 2*A and the other A/2". Well no, one doesn't contain one fourth of the other.
Basically the sleight of hand trick is using a random variable A as though it was a constant. Yes, B = A/2 or B = 2A with probability .5 each, but not for the same value of A.
Why not?
Re: Two envelopes problem
#69https://www.youtube.com/watch?v=_NGPncypY68 this video explains the paradox properly, with a very reasonable explanation.
Yeah the Wikipedia explanation is terrible! Someone rewrite that.
Re: Two envelopes problem
#70There is a related puzzle which I've also heard called the Two Envelopes Problem. The premise is that you are handed two envelopes. Inside each envelope is a piece of paper with a number written on it. I have written down these numbers by sampling from some probability distribution, but you don't know what it is. All you know is that the two numbers are different. You get to pick one envelope and then look at the pap…
I think that's an ill-posed question without knowing (or making assumptions I guess) how the distribution was chosen isn't it?