Author: Bruce Moreland
Date: 19:14:04 08/15/98
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On August 14, 1998 at 21:48:40, fca wrote: >(a) In 1048576 trials (each of 20 throws) the probability of one or more "20 in >a row"s occurring is ________ . About 63% The odds that something bad won't happen in N trials is equal to the odds that it won't happen once, raised to the N. If you play russian roulette with yourself with a six-shooter, and you spin the barrel six times and fire at your head six times, the odds that you will be alive at the end aren't 0%, they are (5/6) ^ 6, which is about 33%. Same basic deal here, only with bigger numbers. Of course, if you just pull the trigger six times, you will be dead, but the spin in between shots is a crucial thing. I saw someone make the mistake of leaving out the spin, same as you did, but I didn't think it was a big deal. >(b) In order for there to be at least a 99.9% probability of getting at least >one "20 in a row", the minimum number of such trials that can be a priori >expected to be needed is _______________ . Whatever value of N you need in order to generate whatever probability you want, in the above formula. I don't want to take the time to turn the formula around, but it's trivial. 7.25 million isn't too far off. >(c) In 1048576 such trials, where exactly 4 "20 in a row"s were observed, there >is a _______ % chance that this result does not contradict the "fair coin" >hypothesis. Good question, and maybe even on topic. I don't know how to do this one without spending more time with the statistics book. Do you? >(d) "20 in a row" has been just been thrown for the first time in the >experiment. The best guess of the trial number on which this happened is >_____________ . One. The odds of getting 20 in a row on the first try are 1/2^20. The odds that you'll miss and get it on the next trial are (1 - 1/2^20) * 1/2^20, which is fractionally less, so by induction, blah blah blah. If two guys who are equally good marskmen take turns firing at each other, bet on the one that gets the first shot. Why did you ask these questions? If you didn't have a reason, I'll ask you one. Describe a method wherein with nothing more than a single ordinary coin you can generate a probability of exactly 1/3. OK. Back on topic for a second. I think it might be useful for the general computer chess population to be able to take a match result and figure out that odds that the weaker program won the match. I bet a lot of the time it is pretty near 50%, and it would be very interesting to know for sure. Since you asked such interesting questions, perhaps you know how we could generate an answer to this one? bruce
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