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BGonline.org Forums
A fair spot for the undefeated player
Posted By: Rich Munitz In Response To: Format of LA Final? (mfic)
Date: Tuesday, 16 June 2009, at 2:57 p.m.
So looking at the drawsheet, it appears that in a full 64 player bracket, that there are 63 once-defeated players of which 60 will have 8-1 records if they make the finals and 3 will have 7-1 records. That said, due to the flow-downs, assuming all equally skilled players, the chance of a 7-1 player making the final is 50% and the chance of an 8-1 player making the final is 50%. So average record of once-defeated finalist is 7.5-1.
I've done a bit of number crunching to see the likelihood of these outcomes. If we simulate a match with a coin toss, we get the following probabilities:
6-0 (6 heads, 0 tails in 6 tosses) = 1.5625%
7-1 (7 heads, 1 tails in 8 tosses) = 3.1250%
8-1 (8 heads, 1 tails in 9 tosses) = 1.7578%
So the average of the 7-1 and 8-1 outcomes = 2.4414%
Thus, we have two finalists at 1.5625 and 2.4414. And 2.4414/(1.5625+2.4414) = 61%
So it seems to me that the undefeated player is indeed getting shortchanged starting the finals even, regardless of which player he meets in the finals, and that on average, he should be starting the finals with 61% match equity (not 75% and not 50%). Seems in general, the undefeated player should start out with about a 2-0 score advantage to whatever match length.
BTW, for those who did not see the drawsheet (if you care), the flows are modified from my previous post as follows:
8 players left - play down to 4
2 players losing in round of 4 main feed in
6 players left - play down to 3
1 player losing in round of 2 main feed in
4 players left - play down to 2 - play down to 1 = finalist.
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