Assuming one side is 52.5% vs the other side (no ties so other side is obv 47.5%). What percentage of the time would the 52.5% side win a race to 5,10,15, 20,30, and 50 vs the 47.5% side? I know the 52.5% side would win 55.455% in a race to 4 via monkeyfockers series calculator so I have an idea but too retarded to know how to do the math on it. Thanks in advance.
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Last edited by smoke a bowl; 08-15-12, 03:26 PM. -
i'm going to have a go at this but i may be wrong.
i'd do it (prolly painfully badly) by working out the possible series scores and then how many ways it is possible to get each score...
so 5-0 only 1 way 0.525^5
5-1 can be 6 different ways so 6([0.525^5]*0.475)
5-2 i think can be 21 ways so 21([0.525^5]*[0.475^2]) the 21 is 7 choose 2 which is 7!/(2! * (7-2)!)
5-3 is 8 choose 3 which is 56([0.525^5]*[0.475^3]) the 56 is 8 choose 3 which is 8!/(3! * (8-3)!)
and so on....
there is probably a smart equation somewhere but i'm not smart enough to even suggest what to search for.
BTW please check this out for urself, i only think this is a way to do it.Comment -
Thanks for the try Subs. Hoping there is an easier way/formula though.Comment -
Let r be the number of matches that team T has to win to win the series and p the probability that T will win a single match and let all match results be independent of each other.
Then the probability that T will win the series is equal to
1 - bincdf(r-1, 2*r-1, p),
where bincdf(k, n, p) the cdf of the binomial distribution.Last edited by HUY; 08-17-12, 04:02 PM.Comment