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in reply to Spooky math problem

Let x,y be your two numbers. So there are |x-y| possible "correct" guesses (a "correct" guess being one that falls between x and y).

Let z be the set from which i'm randomly picking a number n. This number can be anything (1, pi, 2^.5, e^i*pi, etc). So there's an infinite set of numbers from which i'm choosing one member. (I'm not going to get in to the infiniteness of this infinity, as it's been too many years since i've taken discrete math).

The chances of me picking a "correct" number are |x-y|/|z|, where z is infinite, which is 0.

The assumption of gaining odds based on this imagined third number assumes that there is a last number at some point. Once you can quantify infinity, you can retrieve a decimal approximation of your odds. Until that point, you are dealing with zero.

If you're going to pretend that a number is in between the other two, you might as well pretend you know what the other number is....

Update: Here's the proof in the rec.puzzle FAQ. I see two things wrong with it (i think), but i'm going to check into this before posting:-)

Pick any cumulative probability function P(x) such that a > b ==> P(a) > P(b). Now if the number shown is y, guess "low" with probability P(y) and "high" with probability 1-P(y). This strategy yields a probability of > 1/2 of winning since the probability of being correct is 1/2*( (1-P(a)) + P(b) ) = 1/2 + (P(b)-P(a)), which is > 1/2 by assumption.

Update II, Electric Boogaloo: Here's the problem with the proof. As it is written, it is correct, but the first assumption it makes is false. For this situation (picking two random numbers from an infinite set), it is impossible to create a cumulative probability function P(x) as the solution describes. The cumulative probability function is based on the probability mass function, which is always undefined in this case. The probability mass function for any value in this problem is 1/infinity, which is zero. The cumulative probability function is a sum of probability mass functions, which will be zero no matter how many you add together. So if it were indeed possible to pick a function like the proof describes, the proof would hold. However, since the initial assumption is a contradiction, the proof fails.

Here's a nice discussion on cumulative probability functions, and here's a discussion of the probability mass function. Neat-o. Has anyone seen another proof of this?

BlueLines

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RE (tilly) 2: Spooky math problem
by tilly (Archbishop) on Nov 01, 2000 at 06:09 UTC

What your probability is of being correct depends on what two numbers I have. You only can guarantee that it is better than even, but it could be by a very small amount indeed. There is theoretical discussion on this, and the fact that you don't know the probability turns out to be important. (I don't remember details though.)

However there is a variant of this problem where both you and I pick our numbers independently out of the same distribution. Now you can work out the probability of your being right prior to my picking my numbers. And even though your number has no actual information about mine, you turn out to be right exactly 2/3 of the time. If you disbelieve this it is easy to write a short script to test it.

Again, I got this problem from a probability theorist and it really does work.

A Perl script describing this would be most helpful. All this math talk is giving me a headache. Theoretical physics I can handle. It's conceptual. Math is another matter. Given that I think in Perl, I think that'd be best.
RE (tilly) 2 (no problem): Spooky math problem
by tilly (Archbishop) on Nov 02, 2000 at 02:32 UTC
Questions about how I got my numbers get you into the kind of trouble you describe. But that is outside of the problem.

What is inside is that when you make up a number, you need to use a "good" cumulative distribution. And there are plenty of them indeed to be found in any good probability theory book. In fact I named one. The standard normal, which is the prototypical bell-curve, is a probability function which will work. (Albeit with a tail that falls off very rapidly, so your win is miniscule if my numbers are very far away from zero.)

Trust me. I studied math long before I studied Perl, and I got this problem off of a well-known probability theorist. The solution is good.