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BGonline.org Forums
More swiss tournaments, best in three shorter matches and clocks
Posted By: Matt Cohn-Geier In Response To: More swiss tournaments, best in three shorter matches and clocks (Matt Cohn-Geier)
Date: Friday, 10 October 2008, at 11:42 p.m.
So, I have some idea as to how to go about solving this, but I haven't done math in awhile. I'm sure Daniel or Bob or someone will correct me when I get it hopelessly wrong.
Let's take a 9-best-of-17 one-pointer series (a 13-best-of-25 series is much the same, only with longer calculations). With a 100-point ELO difference you are a 52% favorite in a 1-point match.
So your chance of winning the first 9 in a row is (.52)^9. You have one different way to win the first 9 in a row.
Now, suppose you win 8 of the first 9, then win the 9th match. Your probability of doing this is .52^9 * .48. However, you have 9 different ways (or 9 choose 1, hereafter 9c1) to do so. Your opponent could win the first match or the 9th match or any match in between.
Similarly, if you win the first 7 out of 9, then the next two, your probability is .52^9 * .48^2, and you have 10c2 different ways to do so.
If you win the first 6 out of 9 and then the next three, your probability of winning is .52^9 * .48^3, with 11c3 different ways, and so on.
This all adds up to:
1(.52^9) + (9c1)(.52^9 * .48) + (10c2)(.52^9 * .48^2) + (11c3)(.52^9 * .48^3) + (12c4)(.52^9 *.48^4) + (13c5)(.52^9 * .48^5) + (14c6)(.52^9 + .48^6) + (15c7)(.52^9 + .48^7) + (16c8)(.52^9 + .48^8)
The cases where your opponent wins 9 matches aren't counted because then he would win the best-of-17 series.
According to a calculator, this all adds up to:
56.65%
Whereas your ME from -17, -17 is 61.7%. You'd be giving up around 6% ME if you went with the best-of-17 one-pointers.
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