4 ms·

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-

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 distribution we should assume, but as long as we avoid treating something that isn't a distribution as if it were one we avoid the worst of it.

eru 4y ago
Yes. The big problem is that you can't have a uniform distribution on the natural numbers. (And by extension, you can't have a uniform distribution on the rounded-to-integer version of a distribution on the real numbers.)
- xigoi 4y ago
  Who said there is a uniform distribution on the natural numbers?
  - dllthomas 4y ago
    I think the assumption that "for all x, p(x) = p(2x)" implies a uniform distribution on the natural numbers. Imagine we have a distribution that satisfies that assumption, and then someone tells you they've sampled from that distribution and found that the result is of the form (say) 7*2^k, for some k > 0. That conditional distribution for k would seem to have to be uniform, right?
- dllthomas 4y ago
  We're not dealing with the natural numbers, right? Otherwise, you obviously want to open the envelope and always switch when you see an odd number. Of course, the lack of a uniform distribution on the reals or rationals presents the same problem.
  - eru 4y ago
    > We're not dealing with the natural numbers, right? Well, you can either assume that the envelope's numbers are specified only up to pennies. Or you can just round and say you don't care about sub-pennies. Or you can go with your suggestion, and look at the lack of a uniform distribution over the reals / rationals.
kurthr 4y ago
Well taking the mean of products doesn’t make much sense anyway… but if you take a log and average then you end up with the typical solution. If you take a geometric mean then you end up with 1, which implies there’s no difference either.
mike_hock 4y ago
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.
- jonath_laurent 4y ago
  This comment explains the fundamental reason why the reasoning is incorrect. I offer the same perspective in a different comment (https://news.ycombinator.com/item?id=31569991 https://news.ycombinator.com/item?id=31569991). In my opinion, the Wikipedia article is making a disservice to readers by mentioning unnecessarily complex mathematical arguments involving bayesian reasoning and infinite distributions. I believe all of these are distractions from the more fundamental typing error that invalidates the switching argument.
  - mike_hock 4y ago
    They also fail the article's own premise > in particular, the puzzle is not solved by the very simple task of finding another way to calculate the probabilities that does not lead to a contradiction
    - JoshCole 4y ago
      Honestly, the more I've thought about this problem the more that statement bothers me. It is just such a nonsense goal. The reason the calculations are wrong are fundamentally related to why correcting the calculations leads to the correct probabilities. If you understand why the problem gives you the wrong result, you proceed to correcting it, and you get the right result - that doesn't mean you didn't understand. It means you did. You can tell you did, because you have the correct answer. But if I show the calculations produce the correct answers, well, that is a step too far. We can now dismiss the understanding on the basis that it got the correct answer, apparently? It's no wonder the author of the wiki writes such a bullshit claim - that no one can agree on a definitive conclusion. Their terms of engagement are self-defeating. Correctness is error, because correctness means you didn't /really/ understand. Its a no true scottsman fallacy - get the right answer, and you aren't engaging with the 'real' problem. But its a flawed no true scottsman, because it defines something measurable: it tells us the goal, that it is to avoid this type of mistake in our thinking. And generalized algorithms for solving decisions problems that include this as a case which is successfully solved are many - and they just so happen to be in the imperfect information setting. And that setting has studied the problem of infinite recursion and has solutions for them. Which I can apply. To get the right answer. And, in general, avoid the problems they claim we aim to avoid. More importantly and to the point of this thread - this isn't an intractable debate, because its been solved and used in production settings for literally decades. So what if some people are going to pretend it isn't? This isn't an unsolved problem. The actual game theory math is /well/ beyond this level of complexity. Its contending with things like environments where you have so much complexity you have to reduce to a blueprint abstraction, not stumbling at a decision problem that is quite literally simpler than rock paper scissors.

eru 4y ago

Yes. The big problem is that you can't have a uniform distribution on the natural numbers. (And by extension, you can't have a uniform distribution on the rounded-to-integer version of a distribution on the real numbers.)

xigoi 4y ago
Who said there is a uniform distribution on the natural numbers?
- dllthomas 4y ago
  I think the assumption that "for all x, p(x) = p(2x)" implies a uniform distribution on the natural numbers. Imagine we have a distribution that satisfies that assumption, and then someone tells you they've sampled from that distribution and found that the result is of the form (say) 7*2^k, for some k > 0. That conditional distribution for k would seem to have to be uniform, right?
dllthomas 4y ago
We're not dealing with the natural numbers, right? Otherwise, you obviously want to open the envelope and always switch when you see an odd number. Of course, the lack of a uniform distribution on the reals or rationals presents the same problem.
- eru 4y ago
  > We're not dealing with the natural numbers, right? Well, you can either assume that the envelope's numbers are specified only up to pennies. Or you can just round and say you don't care about sub-pennies. Or you can go with your suggestion, and look at the lack of a uniform distribution over the reals / rationals.

kurthr 4y ago

Well taking the mean of products doesn’t make much sense anyway… but if you take a log and average then you end up with the typical solution. If you take a geometric mean then you end up with 1, which implies there’s no difference either.

mike_hock 4y ago

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.

jonath_laurent 4y ago
This comment explains the fundamental reason why the reasoning is incorrect. I offer the same perspective in a different comment (https://news.ycombinator.com/item?id=31569991 https://news.ycombinator.com/item?id=31569991). In my opinion, the Wikipedia article is making a disservice to readers by mentioning unnecessarily complex mathematical arguments involving bayesian reasoning and infinite distributions. I believe all of these are distractions from the more fundamental typing error that invalidates the switching argument.
- mike_hock 4y ago
  They also fail the article's own premise > in particular, the puzzle is not solved by the very simple task of finding another way to calculate the probabilities that does not lead to a contradiction
  - JoshCole 4y ago
    Honestly, the more I've thought about this problem the more that statement bothers me. It is just such a nonsense goal. The reason the calculations are wrong are fundamentally related to why correcting the calculations leads to the correct probabilities. If you understand why the problem gives you the wrong result, you proceed to correcting it, and you get the right result - that doesn't mean you didn't understand. It means you did. You can tell you did, because you have the correct answer. But if I show the calculations produce the correct answers, well, that is a step too far. We can now dismiss the understanding on the basis that it got the correct answer, apparently? It's no wonder the author of the wiki writes such a bullshit claim - that no one can agree on a definitive conclusion. Their terms of engagement are self-defeating. Correctness is error, because correctness means you didn't /really/ understand. Its a no true scottsman fallacy - get the right answer, and you aren't engaging with the 'real' problem. But its a flawed no true scottsman, because it defines something measurable: it tells us the goal, that it is to avoid this type of mistake in our thinking. And generalized algorithms for solving decisions problems that include this as a case which is successfully solved are many - and they just so happen to be in the imperfect information setting. And that setting has studied the problem of infinite recursion and has solutions for them. Which I can apply. To get the right answer. And, in general, avoid the problems they claim we aim to avoid. More importantly and to the point of this thread - this isn't an intractable debate, because its been solved and used in production settings for literally decades. So what if some people are going to pretend it isn't? This isn't an unsolved problem. The actual game theory math is /well/ beyond this level of complexity. Its contending with things like environments where you have so much complexity you have to reduce to a blueprint abstraction, not stumbling at a decision problem that is quite literally simpler than rock paper scissors.