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A fair spot for the undefeated player

Posted By: Rich Munitz
Date: Wednesday, 17 June 2009, at 2:38 a.m.

In Response To: A fair spot for the undefeated player (Maik Stiebler)

Yes, of course. Knew I must have goofed somewhere.

Winner of the once-defeated bracket must go 5-1 (6/64) and then either win 2 or 3 straight.

So...

6-0 = 1/64 = 8/512

7-1 = (6/64)/4 = 6/256 = 12/512

8-1 = (6/64)/8 = 6/512

Average once-defeated finalist = 9/512

Ratio Once-Defeated:Undefeated = 9:8 = 9/17:8/17 = 53%:47%

Yes, going 8-1 is indeed a tougher feat than going 6-0. But 7-1 is not. And on average, it is still slightly tougher to reach the finals of the undefeated than the once-defeated bracket.

So based on this correction, it looks like having the two finalists playing just one match 0-0 is pretty close to fair. To be slightly more equitable, the undefeated player should start the finals match ahead 1-0 for any likely match length.

However, since a 1 point spot is then going to mess up match stats, probably best to forget about it altogether. The other way to justify that approach is that half of the time, the undefeated player will have had the tougher job than his opponent, and the other half he will have had it easier. So without knowing which it is ahead of time, call it even.

My conclusion: Double elim format - finalists should play one match from an even score for the title.

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