5 ms·

You don't need a uniform distribution to get guaranteed better than even odds. And many nonuniform distributions work just fine. There are no hidden conditions

by btilly 2y ago

You don't need a uniform distribution to get guaranteed better than even odds. And many nonuniform distributions work just fine. There are no hidden conditions. It is just a shocking result that we don't expect.

hansvm 2y ago
You need some _known_ distribution though, and it's shocking because the distribution is ommitted from the question, and the presence of the same distribution is snuck into the answer.
- ColinWright 2y ago
  I think you are wrong ... see my answer here: https://news.ycombinator.com/item?id=42371563 https://news.ycombinator.com/item?id=42371563
  - hansvm 2y ago
    Not to be too dismissive, but the title text of this xkcd [0] seems relevant. No matter how nice the explanation is, the fact that the conclusion is wrong suggests that the reasoning has a flaw. I took a stab at what I think that flaw is when I responded to your rebuttal [1]. [0] https://xkcd.com/2217/ https://xkcd.com/2217/ [1] https://news.ycombinator.com/item?id=42372285 https://news.ycombinator.com/item?id=42372285
    - raincole 2y ago
      I dimissed it at first (you can see I made some embarrassing comments above), but the strategy is correct. I'd argue it's even intuitively correct once the problem in question is clarified. It's not saying that after the player see the number in the first envelope, the strategy guarantees a >50% outcome. It's saying that give any distribution, over all possible outcomes, >50% times the strategy will end up pick the larger number. You can say this >50% is the expected winning chance before the player see the number in the first envelope. I'd say this is "intuitve" because, if your strategy can guarantee "when the player see a large number in the first envelope, he's less likely to switch than if he saw a small number", it would be better than blindly switching by coin toss. So intuitively such a strategy exists. The only "trick" here is that since the player doesn't know the initial distribution, they can't tell "how large counts as large?" therefore they needs something that preserves some property over the whole real number line. That's why the strategy involves sampling from a another distribution whose PDF is non-zero everywhere.
    - hansvm 2y ago
      Sorry, I was too dismissive. I'm not crazy about the explanation, but I think you're right.
raincole 2y ago
https://www.alexirpan.com/2015/09/09/the-other-two-envelope-problem.html https://www.alexirpan.com/2015/09/09/the-other-two-envelope-... If your solution is the same as this article's, it's plain wrong. Even the natural number case is plain strong. It's very easy to demostrate as well: consider a trivia case where the distribution is just {P(1)=1/3, P(2)=1/3, P(3)=1/3} and you see 2 in the first envelope. There is no strategy to get a better chance than 50%. Therefore, any strategy that gives a better chance than 50% must implicitly make an assumption over the initial distribution (and therefore excludes a distribution like {P(1)=1/3, P(2)=1/3, P(3)=1/3}) Actually the article is even "wronger" than this, because "started A" and "switched" aren't independent and one can't simply use the product of their probability. The above example is a quick way to demonstrate it's not a general strategy without assumption to get >50% winning chance. Similarily, one can just use {P(1)=1/3, P(2)=1/3, P(3)=1/3} (this is a valid distribution over real numbers!) to demonstrate the real number strategy isn't general. Again, for both natural number and real number case, the discussion over strategies is only meaningful is we know something about the distribution. Interestingly, this article is wrong more or less in the same way as believing switching does give you more expected value in the original "twice money in another envelope" variation. Edit: For people who are interested in the switching strategy, check Randomized Switching in the Two-Envelope Problem (2009). Spoiler: full of discussion over the initial distribution.
- btilly 2y ago
  Sorry, but you're simply wrong. You can read the answer I gave. You can read the answer Colin Wright gave. You can trust that we both have math degrees and know what we are talking about. Or, aw heck, you can try it with an actual program at https://www.perlmonks.org/?node_id=39630 https://www.perlmonks.org/?node_id=39630. (Yes, I wrote that piece of hackery about a decade ago.) I don't actually care how you convince yourself. But the explanation is right. If your random number is outside of the range, you've got even odds. If it is inside of the range, you've got 100% odds. As long as there is a positive probability of being between, you've got strictly better than even, by half of the probability of being between. Many, many distributions guarantee positive odds of being in between. The one I chose for my program was: (log(rand) * (flip_coin() ? 10 : -10 )) Which is the log of a random number between 0 and 1, times 10 times + or - with even odds. The various factors were chosen to fit well with normal human choices that most seek to test it with.

by btilly 2y ago

hansvm 2y ago

You need some _known_ distribution though, and it's shocking because the distribution is ommitted from the question, and the presence of the same distribution is snuck into the answer.

ColinWright 2y ago
I think you are wrong ... see my answer here: https://news.ycombinator.com/item?id=42371563 https://news.ycombinator.com/item?id=42371563
- hansvm 2y ago
  Not to be too dismissive, but the title text of this xkcd [0] seems relevant. No matter how nice the explanation is, the fact that the conclusion is wrong suggests that the reasoning has a flaw. I took a stab at what I think that flaw is when I responded to your rebuttal [1]. [0] https://xkcd.com/2217/ https://xkcd.com/2217/ [1] https://news.ycombinator.com/item?id=42372285 https://news.ycombinator.com/item?id=42372285
  - raincole 2y ago
    I dimissed it at first (you can see I made some embarrassing comments above), but the strategy is correct. I'd argue it's even intuitively correct once the problem in question is clarified. It's not saying that after the player see the number in the first envelope, the strategy guarantees a >50% outcome. It's saying that give any distribution, over all possible outcomes, >50% times the strategy will end up pick the larger number. You can say this >50% is the expected winning chance before the player see the number in the first envelope. I'd say this is "intuitve" because, if your strategy can guarantee "when the player see a large number in the first envelope, he's less likely to switch than if he saw a small number", it would be better than blindly switching by coin toss. So intuitively such a strategy exists. The only "trick" here is that since the player doesn't know the initial distribution, they can't tell "how large counts as large?" therefore they needs something that preserves some property over the whole real number line. That's why the strategy involves sampling from a another distribution whose PDF is non-zero everywhere.
  - hansvm 2y ago
    Sorry, I was too dismissive. I'm not crazy about the explanation, but I think you're right.

raincole 2y ago

https://www.alexirpan.com/2015/09/09/the-other-two-envelope-problem.html https://www.alexirpan.com/2015/09/09/the-other-two-envelope-... If your solution is the same as this article's, it's plain wrong. Even the natural number case is plain strong. It's very easy to demostrate as well: consider a trivia case where the distribution is just {P(1)=1/3, P(2)=1/3, P(3)=1/3} and you see 2 in the first envelope. There is no strategy to get a better chance than 50%. Therefore, any strategy that gives a better chance than 50% must implicitly make an assumption over the initial distribution (and therefore excludes a distribution like {P(1)=1/3, P(2)=1/3, P(3)=1/3}) Actually the article is even "wronger" than this, because "started A" and "switched" aren't independent and one can't simply use the product of their probability. The above example is a quick way to demonstrate it's not a general strategy without assumption to get >50% winning chance. Similarily, one can just use {P(1)=1/3, P(2)=1/3, P(3)=1/3} (this is a valid distribution over real numbers!) to demonstrate the real number strategy isn't general. Again, for both natural number and real number case, the discussion over strategies is only meaningful is we know something about the distribution. Interestingly, this article is wrong more or less in the same way as believing switching does give you more expected value in the original "twice money in another envelope" variation. Edit: For people who are interested in the switching strategy, check Randomized Switching in the Two-Envelope Problem (2009). Spoiler: full of discussion over the initial distribution.

btilly 2y ago
Sorry, but you're simply wrong. You can read the answer I gave. You can read the answer Colin Wright gave. You can trust that we both have math degrees and know what we are talking about. Or, aw heck, you can try it with an actual program at https://www.perlmonks.org/?node_id=39630 https://www.perlmonks.org/?node_id=39630. (Yes, I wrote that piece of hackery about a decade ago.) I don't actually care how you convince yourself. But the explanation is right. If your random number is outside of the range, you've got even odds. If it is inside of the range, you've got 100% odds. As long as there is a positive probability of being between, you've got strictly better than even, by half of the probability of being between. Many, many distributions guarantee positive odds of being in between. The one I chose for my program was: (log(rand) * (flip_coin() ? 10 : -10 )) Which is the log of a random number between 0 and 1, times 10 times + or - with even odds. The various factors were chosen to fit well with normal human choices that most seek to test it with.