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Comment by dsukhin | Hacker News Reader

If you have a range (-inf, inf) and you split it at some random C, even though there are still infinite values on either side, the probability you choose another number on the left side of C is (p) and on the right side is (1-p) regardless of how you choose the next number. That's what makes these probabilities finite and makes the math work out.

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But... but... you can't "pick a number uniformly distributed in (-inf, inf)". Like, that's not a thing that is possible [0], so saying anything about the probabilities once you've done so is not reasonable. You can pick a random number in infinite range non-uniformly (e.g. to pick a number in the range [1,inf), pick a number X in (0,1] and use 1/X), but then the probabilities are no longer so neat and tidy.

The game only makes sense if you attach limits to the numbers the other person is allowed to pick, and then the only "split" that works is in the middle of the range. At which point the game is sort-of "obviously true".

[0] https://math.stackexchange.com/a/14169/641751

It is not necessary for the distribution to be uniform, it just has to be nonzero everywhere.

If the distribution of A and B is uniform, the probability that C from this proper distribution is between A and B is zero.

If the distribution of A and B is also a proper distribution the probability that C is between them is positive. But then the distribution of A and B has a middle point such that 50% of the numbers are above and 50% below and the problem is trivial.

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