4 ms·

The two reals are selected via some distribution, and the only way you can do better than chance is if you have some knowledge of that distribution. The questi

by hansvm 2y ago

The two reals are selected via some distribution, and the only way you can do better than chance is if you have some knowledge of that distribution. The question leaves that distribution completely hidden, and your answer smuggles it back in. That feels less like a counter-intuitive math/stats question and more like a badly worded gotcha.

raincole 2y ago
Yes, this two real numbers question suffers from hidden conditions as the original version of Monty Hall problem does, but even more explicitly. We can't have an uniform distribution over all real numbers, so it's quite pointless to discuss if looking into the envelope gives any new information, cause we don't even know the distribution yet.
- 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.
ColinWright 2y ago
This turns out not to be the case. Let's play this game exactly once. You choose two unequal real numbers. I don't know what they are, and I don't know the distribution from which you choose them. You write them down and put them in separate envelopes. I'm allowed to choose one envelope and open it to see the number inside, and my job is then to say which envelope holds the larger number. I claim I have a strategy now which lets me win strictly more than 50% of the time. My strategy is this. I choose a real number R at random from a distribution that has dense support. In other words, for any two reals, L and U with L<U, P(L<R<U) > 0. This is easy to do ... one method is to list the rationals, positive and negative, then roll a die, discarding numbers until you get a 6. Now I flip a coin and thereby choose an envelope at random. I proceed by assuming my chosen number is between your two numbers. There is a non-zero chance this is true ... call it e. So e>0. If I'm wrong then my choice is 50% ... probability is 1-e. If I'm right then my choice is 100%. ... probability is e. Combined, my chance of being right is 0.5(1-e) + e = 0.5+e/2, which is strictly greater than 50%. You can make it as small as you like, and if we play the game repeatedly then you can make it approach 50%. But as it stands, with a one-off game, I can win with a probability that depends on your chosen numbers, but which is strictly bigger than 50%.
- hansvm 2y ago
  Why, exactly, are you allowed to know why some random guess is between the two numbers or not when computing your choice?
  - hansvm 2y ago
    Sorry, too late to edit, I see you're describing how you know your choice was good/bad (I thought you were describing a probabilistic choice).
- ajnin 2y ago
  I'm not a mathematician so please bear with me here, but I think a problem stems from the fact that the set of reals is "infinite". So, whatever interval you choose, there are infinitely more reals outside the interval as inside (by that I mean that you can fit an infinite number of copies of that interval up to infinity). So the probability e is not >0, it is effectively 0. The second problem is, what does it mean to choose a real at random ? There is an implication that you can choose such number, but as a human living in the finite universe there are limitations to your choice. Any number you can write using all the atoms in the universe is infinitely outnumbered by all numbers that you can't. So effectively it is impossible to pick a random real number. You have to pick a real in some interval, implicitly the interval of reals you can write in an envelope. Which is a different problem than stated originally and for which your "e" can be >0.

by hansvm 2y ago

raincole 2y ago

Yes, this two real numbers question suffers from hidden conditions as the original version of Monty Hall problem does, but even more explicitly. We can't have an uniform distribution over all real numbers, so it's quite pointless to discuss if looking into the envelope gives any new information, cause we don't even know the distribution yet.

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.

ColinWright 2y ago

This turns out not to be the case. Let's play this game exactly once. You choose two unequal real numbers. I don't know what they are, and I don't know the distribution from which you choose them. You write them down and put them in separate envelopes. I'm allowed to choose one envelope and open it to see the number inside, and my job is then to say which envelope holds the larger number. I claim I have a strategy now which lets me win strictly more than 50% of the time. My strategy is this. I choose a real number R at random from a distribution that has dense support. In other words, for any two reals, L and U with L<U, P(L<R<U) > 0. This is easy to do ... one method is to list the rationals, positive and negative, then roll a die, discarding numbers until you get a 6. Now I flip a coin and thereby choose an envelope at random. I proceed by assuming my chosen number is between your two numbers. There is a non-zero chance this is true ... call it e. So e>0. If I'm wrong then my choice is 50% ... probability is 1-e. If I'm right then my choice is 100%. ... probability is e. Combined, my chance of being right is 0.5(1-e) + e = 0.5+e/2, which is strictly greater than 50%. You can make it as small as you like, and if we play the game repeatedly then you can make it approach 50%. But as it stands, with a one-off game, I can win with a probability that depends on your chosen numbers, but which is strictly bigger than 50%.

hansvm 2y ago
Why, exactly, are you allowed to know why some random guess is between the two numbers or not when computing your choice?
- hansvm 2y ago
  Sorry, too late to edit, I see you're describing how you know your choice was good/bad (I thought you were describing a probabilistic choice).
ajnin 2y ago
I'm not a mathematician so please bear with me here, but I think a problem stems from the fact that the set of reals is "infinite". So, whatever interval you choose, there are infinitely more reals outside the interval as inside (by that I mean that you can fit an infinite number of copies of that interval up to infinity). So the probability e is not >0, it is effectively 0. The second problem is, what does it mean to choose a real at random ? There is an implication that you can choose such number, but as a human living in the finite universe there are limitations to your choice. Any number you can write using all the atoms in the universe is infinitely outnumbered by all numbers that you can't. So effectively it is impossible to pick a random real number. You have to pick a real in some interval, implicitly the interval of reals you can write in an envelope. Which is a different problem than stated originally and for which your "e" can be >0.