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This is a fantastic and subtle paradox, and not at all taken to quick resolution. A resolution follows, first to spot a problem (if there is one) gets a cookie.

by scythe 13y ago

This is a fantastic and subtle paradox, and not at all taken to quick resolution. A resolution follows, first to spot a problem (if there is one) gets a cookie. Consider the amount in the lower to be f(x), the higher to be x, given f(z) such that f(z) < z for all z. In this way we generalize to all distributions, as the problem clearly applies to all distributions. You open the first envelope, which contains A. The second envelope contains B. The challenge is to calculate the value of B. First we must calculate x. The chance that A = f(x) is 1/2, the chance that A = x is 1/2. The average value of A is (x + f(x))/2, so the average value of x is [f+I]^{-1}(2A), where I is the identity function and [g]^{-1} denotes an inverse function. Now we calculate B. The trick lies in this: both calculations must lie in the same reference frame. So B = x with probability 1/2, and B = f(x) with probability 1/2, giving us B = (x + f(x))/2. The rest is plug-and-chug: B = ([f+I]^{-1}(2A) + f([f+I]^{-1}(2A))/2 --> B = [f+I]([f+I]^{-1}(2A))/2 --> B = 2A/2 thus B = A. Therefore over all probability distributions that can be defined we have the average value of B equal to A in any case. A purely mathematical resolution is satisfying, but I am not in any case an epistemologist, so it may not satisfy people who take a different interpretation of math than me. It works, and I like it.

jamii 13y ago
The problem here is that you are adding together two values for a specific case of x and then adding together the result for all values of x (is that clear? probably not). It's analogous to another age old problem: what is the value of 1 + (-1) + 1 + (-1) + 1 + (-1) ... You could argue that (1 + -1) + (1 + -1) ... = 0 + 0 ... = 0. You could also argue that 1 + (-1 + 1) + (-1 + 1) = 1 + 0 ... = 1. There are actually way to group the numbers in that series to get any integer answer you want :D Since we haven't actually defined a distribution over x the problem is not well-defined anyway. But lets pick, say, P(x=2^i) = 1/(2^i) for all i>1. Then the expected value of A is E(A) = (1/2 * $2) + (1/4 * $4) + (1/8 * $8) ... = +infinity. Similary E(B) = +infinity. Now for E(B-A) E(B-A) = (1/2 * $1) + (1/2 * -$1) + (1/4 * $2) + (1/4 * -$2) ... Like the example above, we can add this up in different ways: E(B-A) = ((1/2 * $1) + (1/2 * -$1)) + ((1/4 * $2) + (1/4 * -$2)) ... = $0 + $0 ... = $0 Or E(B-A) = (1/2 * $1) + ((1/2 * -$1) + (1/4 * $2)) + ((1/4 * -$2) ... = $0.5 + $0 + $0 ... = $0.5 When adding up infinite numbers of things you have to be very careful. I go into more detail in this thread - https://news.ycombinator.com/item?id=6387344 https://news.ycombinator.com/item?id=6387344 Cookie please :D
- scythe 13y ago
  In my original version of the post I had intended to argue for a geometric mean, which was sort of a joke, as it only works where f(x) is a constant k*x. So the statement "A false resolution follows, first to spot the problem..." appears in the post. Upon seeing this an enterprising and intelligent person may have clicked "reply" While writing this I realized that inverting the definition of A gave a different resolution based loosely on operator theory, and posted that instead. A couple minutes later I realized the mistake and changed the beginning to refer to the first to spot "a" problem. The argument you've presented is good, but you have attacked the well-foundedness of the problem itself rather than the logic I've applied. I'm not sure if you get the cookie.
  - jamii 13y ago
    Goddamn, what does it take to get a cookie around here? :D

by scythe 13y ago

jamii 13y ago

The problem here is that you are adding together two values for a specific case of x and then adding together the result for all values of x (is that clear? probably not). It's analogous to another age old problem: what is the value of 1 + (-1) + 1 + (-1) + 1 + (-1) ... You could argue that (1 + -1) + (1 + -1) ... = 0 + 0 ... = 0. You could also argue that 1 + (-1 + 1) + (-1 + 1) = 1 + 0 ... = 1. There are actually way to group the numbers in that series to get any integer answer you want :D Since we haven't actually defined a distribution over x the problem is not well-defined anyway. But lets pick, say, P(x=2^i) = 1/(2^i) for all i>1. Then the expected value of A is E(A) = (1/2 * $2) + (1/4 * $4) + (1/8 * $8) ... = +infinity. Similary E(B) = +infinity. Now for E(B-A) E(B-A) = (1/2 * $1) + (1/2 * -$1) + (1/4 * $2) + (1/4 * -$2) ... Like the example above, we can add this up in different ways: E(B-A) = ((1/2 * $1) + (1/2 * -$1)) + ((1/4 * $2) + (1/4 * -$2)) ... = $0 + $0 ... = $0 Or E(B-A) = (1/2 * $1) + ((1/2 * -$1) + (1/4 * $2)) + ((1/4 * -$2) ... = $0.5 + $0 + $0 ... = $0.5 When adding up infinite numbers of things you have to be very careful. I go into more detail in this thread - https://news.ycombinator.com/item?id=6387344 https://news.ycombinator.com/item?id=6387344 Cookie please :D

scythe 13y ago
In my original version of the post I had intended to argue for a geometric mean, which was sort of a joke, as it only works where f(x) is a constant k*x. So the statement "A false resolution follows, first to spot the problem..." appears in the post. Upon seeing this an enterprising and intelligent person may have clicked "reply" While writing this I realized that inverting the definition of A gave a different resolution based loosely on operator theory, and posted that instead. A couple minutes later I realized the mistake and changed the beginning to refer to the first to spot "a" problem. The argument you've presented is good, but you have attacked the well-foundedness of the problem itself rather than the logic I've applied. I'm not sure if you get the cookie.
- jamii 13y ago
  Goddamn, what does it take to get a cookie around here? :D