Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.2 9/5/84; site sunybcs.UUCP Path: utzoo!watmath!clyde!burl!ulysses!gamma!epsilon!zeta!sabre!petrus!bellcore!decvax!genrad!panda!talcott!harvard!seismo!rochester!rocksanne!sunybcs!colonel From: colonel@sunybcs.UUCP (Col. G. L. Sicherman) Newsgroups: net.puzzle Subject: Re: infinite poker chips Message-ID: <2655@sunybcs.UUCP> Date: Tue, 24-Dec-85 12:36:52 EST Article-I.D.: sunybcs.2655 Posted: Tue Dec 24 12:36:52 1985 Date-Received: Sat, 28-Dec-85 00:44:48 EST References: <33@decwrl.UUCP> <11900003@ada-uts.UUCP> Organization: Travelers' Advisory Lines: 27 > I don't see how you have proved your hypothesis, that it doesn't make > sense to discuss random numbers in an infinite domain. First, the number > of points in the range 0-1 is the same as the number of points in the > range -oo to +oo. Secondly, lets play another little game: > [description of same game, only finite] > Since we both have a 2/3 chance of winning, by your reasoning > we must conclude that it makes no sense to discuss random numbers on a > finite domain. Right. > What this really means is that the reasoning is faulty. The game can > be broken into two random acts: first the selection of the disk, second > the orientation of the selected disk. Since each disk has a winning side > and a losing side, they are all equivalent. Thus only the orientation > matters. Seeing the number imparts no information about this, so the odds > are just fifty/fifty, which is what you would expect. It's not enough to produce a proof of the "correct" result. You must find a flaw in Hayes's original Bayesian proof of the "incorrect" result. And I don't think you can do it.... Of course, the "incorrect" proof doesn't really carry over to the finite case. Each side no longer has a 2/3 chance of winning, because there are so many more 10s than 1s in the game. -- Col. G. L. Sicherman UU: ...{rocksvax|decvax}!sunybcs!colonel CS: colonel@buffalo-cs BI: csdsicher@sunyabva