RE: Spooky math problem

in reply to Spooky math problem

Let's frame this in the simplest terms: You can only 'win' if you know something about the mean of the probability distribution that produced the numbers. As we say in physics, this breaks the symmetry of the problem. You have more information that just the value of one number, and from there you can compare your number with the mean. If your number is higher than the mean, than you are likely to have the higher number. Otherwise it is likely to be the lower. If you have a uniform probability distribution between + and - infinity, one could argue that the mean is zero. Therefore, if the number you viewed is positive, it is likely the higher one and vice versa. However, I don't feel comfortable saying it is the mean - too many infinities are involved. It depends on your definition of zero and infinity. A good definition of zero is that it is the mean of integers, rational and/or irrational numbers. A bell curve, although infinite, has a definite mean and finite integral. The probability is zero at +/- infinity, so there are fewer affronts to nature.

Comment on RE: Spooky math problem

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RE (tilly) 2 (no assumptions): Spooky math problem by tilly (Archbishop) on Nov 02, 2000 at 02:12 UTC
Read the problem again. Were such assumptions needed you can be sure I would have stated them. But they are not. The key is that your random number needs some chance of falling in the interval between the numbers. As long as you can guarantee that, you get better than even odds. But it turns out that is not hard to guarantee. However if the two numbers differ by, say, 1 from each other it turns out that on average you are ahead by all of zero percent (very large tails with probabilities near 50% dominate). So against a mildly malicious opponent you - on average - don't come out ahead. But that is an average across an infinite number of situations, every last one of which you came out ahead in. (Infinity has lots of strange stuff like this.) Again, the probability of your winning depends on the two numbers that I have. But it is always better than even. And you can guarantee that no matter how I tried to produce my numbers.	[reply]

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RE (tilly) 2 (no assumptions): Spooky math problem
by tilly (Archbishop) on Nov 02, 2000 at 02:12 UTC

The key is that your random number needs some chance of falling in the interval between the numbers. As long as you can guarantee that, you get better than even odds. But it turns out that is not hard to guarantee. However if the two numbers differ by, say, 1 from each other it turns out that on average you are ahead by all of zero percent (very large tails with probabilities near 50% dominate). So against a mildly malicious opponent you - on average - don't come out ahead. But that is an average across an infinite number of situations, every last one of which you came out ahead in. (Infinity has lots of strange stuff like this.)

Again, the probability of your winning depends on the two numbers that I have. But it is always better than even. And you can guarantee that no matter how I tried to produce my numbers.

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