Yesterday, Brett started us down a dangerous path, but you'll be happy to know I saved us all from potential loss of sleep. I'll explain . . . . Brett was wondering, as I'm sure various of you were, what the results would be from a 64 team single elimination tournament in which there were no upsets and the positions of the teams are clearly defined and well known, but matchups are random. The tournament wouldn't be much good for setting contest up; we all know who the best team is, and they are guaranteed to win. However, there is an interesting question of who gets second place. Certainly, the #2 team has the best chance of getting second, but, if they meet the #1 team before the championship game, they are out of luck.
Brett started from a numerical approach, letting his computer set up 10,000 such tournaments, randomly placing the teams into brackets and seeing how it comes out, and intially supposed that team #2 has a 1/2 chance of 2nd place, #3 a 1/4 chance, #4 a 1/8 chance, and so on and so forth. The result isn't surprising. But quickly, we realized that this can't be correct. First off, it implies that all teams have a chance of winning, when they clearly don't. No team lower than the field size/2 + 1 can make the championship game. Also, that result can't be right because the probability of someone finishing in 2nd place only adds up to 1 if we sum forever, and we don't have infinitely many teams.
So, what is the probability? If you want to figure it out for yourself, quit reading now, because I'm about to tell you. Ok . . . I see that none of you have quit reading. (It's impressive that you've read this far.) After various attempts at finding a formula that fits, I present you with the formula for finding the probability of any team taking second place in such a tournament, generalized for any number of teams and any seed:
where f is the field size and s is the seed and a[s] represents the odds of the nth seed taking 2nd place, where 2 <> (f/2 +1). There, can't you rest easier this weekend now that this problem has been put to rest?
1 comment:
I'll just point out that I also came up with the answer too. Though Clark has a much more neatly written equation for it than I did.
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