Author: George Tsavdaris
Date: 04:14:19 07/25/03
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On July 24, 2003 at 23:46:50, Tom Kerrigan wrote: >Given a round robin tournament with 13 participants of equal strength, 5 of >which are The King: > >There's a 14% chance The King would take the top 3 spots. I don't know for sure***, but i think you are wrong. Because if the participants have equal strength we can do the following: If N is the number of all possible classifications of the programs then N = 13!/5! as we have five "The King". If A is the number of all possible classifications where "The King" has taken the three first places at least (or four of five) then A = 10!/2! . Then there is a 100·A/N % chance "The King to be in the 3 (at least) top spots. So the chance is ~ 3.5 %. >There's an 11.4% chance of a specific program winning the tournament. (This >problem is apparently more complex than it seems--I would have guessed a >1/13=7.6% chance of any given program winning the tournament.) If B is the number of all possible classifications where a program (not "The King") wins the tournament then B = 12!/5! . So there is a 100·B/N % chance a a program (but not "The King") to win the tournament. And 100·B/N % = ~ 7.69 %. For "The King" the possibility to win is C/N where C = 12!/4!. So 100·C/N % = ~ 38.46 %. -------------------------------------------- *** = I understand that the above may be wrong, because in a chess tournament there are no wins-loses but also draws. But the problem is i don't find a particular reason, this can influence the above procedure. -------------------------------------------- >In other words, you should only be slightly less surprised that The King took >the top 3 spots than if a certain program won a tournament against 12 other >programs of the same strength. In other words again, it's unlikely that The King >is the same strength as the other programs. >
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