4 ms·

I think the analysis of the strategy's chances of winning is faulty. We can assume that the adversary chooses A and B such that P(A < B) = ½, since otherwise th

I think the analysis of the strategy's chances of winning is faulty. We can assume that the adversary chooses A and B such that P(A < B) = ½, since otherwise the optimal strategy is to always guess A < B or B > A depending on if P(A < B) > ½ or P(A < B) < ½. Now, analyze the game using the standard approach of breaking it down into simple, non-overlapping cases: R < A < B: lose A < R < B: win A < B < R: win R < B < A: win B < R < A: win B < A < R: lose Thus the probability of losing is P(R < A < B) + P(B < A < R) = P(R < A ∧ A < B) + P(B < A ∧ A < R) = P(R < A)*P(A < B) + P(B < A)*P(A < R) = P(R < A)/2 + P(A < R)/2 = 1/2 There is no paradox: no matter how you choose R, you have exactly a ½ chance of winning.

pzone 11y ago
That sounds right to me. The first few sentences of the article, talking about "picking numbers at random" without fixing any sort of probability distributions for those numbers, or even labeling those distributions as P1 P2 and so on, raised enough red flags that I stopped reading. Whether or not the the conclusion of this article is correct (and my intuition still screams "not a chance"), the argument is hand-wavey enough that it doesn't reply to the objection raised here.
- pzone 11y ago
  Update: I read one of the comments below and now I'm convinced by the argument. Here's the intuition for what's happening. (I wish the article had spelled this out sooner) As long as A does not equal B with some positive probability, and as long as you choose a separator R with full support, then there is some probability that R lands between A and B. When that happens, you always win the game. Otherwise, if picking A or B randomly doesn't hurt you, then R landing somewhere else not between A and B doesn't hurt you either. So you've gotten a strategy that, once in a while, helps you separate A and B, and doesn't hurt you otherwise.
  - StefanKarpinski 11y ago
    Yeah, I'm not convinced anymore either. Specifically, I think that the assumption that R < A and A < B are independent events is incorrect, which means that P(R < A < B) = P(R < A)*P(A < B) is not necessarily true. I'll have to think on it more and maybe email Mark about it.
level3 11y ago
One way to think about the game is adversarially. That is, we first have to publicly reveal our distribution for R, and then the opponent arbitrarily chooses A and B. If our distribution is "always choose 0" or "uniformly choose an integer between 1 and 100", then the opponent can always pick A and B so that R will never fall between them. But if we pick with a distribution that is positive everywhere (like the normal distribution), it now becomes impossible for the opponent to choose A and B in a way that reduces our chance to zero. She can make that chance arbitrarily small, but it will always be positive.

pzone 11y ago

That sounds right to me. The first few sentences of the article, talking about "picking numbers at random" without fixing any sort of probability distributions for those numbers, or even labeling those distributions as P1 P2 and so on, raised enough red flags that I stopped reading. Whether or not the the conclusion of this article is correct (and my intuition still screams "not a chance"), the argument is hand-wavey enough that it doesn't reply to the objection raised here.

pzone 11y ago
Update: I read one of the comments below and now I'm convinced by the argument. Here's the intuition for what's happening. (I wish the article had spelled this out sooner) As long as A does not equal B with some positive probability, and as long as you choose a separator R with full support, then there is some probability that R lands between A and B. When that happens, you always win the game. Otherwise, if picking A or B randomly doesn't hurt you, then R landing somewhere else not between A and B doesn't hurt you either. So you've gotten a strategy that, once in a while, helps you separate A and B, and doesn't hurt you otherwise.
- StefanKarpinski 11y ago
  Yeah, I'm not convinced anymore either. Specifically, I think that the assumption that R < A and A < B are independent events is incorrect, which means that P(R < A < B) = P(R < A)*P(A < B) is not necessarily true. I'll have to think on it more and maybe email Mark about it.

level3 11y ago

One way to think about the game is adversarially. That is, we first have to publicly reveal our distribution for R, and then the opponent arbitrarily chooses A and B. If our distribution is "always choose 0" or "uniformly choose an integer between 1 and 100", then the opponent can always pick A and B so that R will never fall between them. But if we pick with a distribution that is positive everywhere (like the normal distribution), it now becomes impossible for the opponent to choose A and B in a way that reduces our chance to zero. She can make that chance arbitrarily small, but it will always be positive.