Little math problem(s) February 2019

[Albert Chan]

For best of N match puzzle, think binary, 0=lost, 1=win
let probability of better player winning be p, for whole tournament be T

Best of 1: 1 ==> T = p

Best of 2: 011 101 11 ==> T = p^2 *(1 + 2(1-p))

Best of 3: 00111 01011 01101 0111 10011 10101 1011 11001 1101 111
sort by length of games: 1x3 + 3x4 + 6x5, total 1+3+6 = 10 cases
==> T = p^3 * (1 + 3(1-p) + 6(1-p)²)

Best of 4: ... got coefficient of 1,4,10,20. My guess is the trend continues.
==> Best of N: T = p^N * sum(nCr(N-1+k,k) * (1-p)^k, k = 0 to N-1)

Code:

Best-of     Whole Tournament, expected better player winning
1      p =  60%    65%    70%
2         64.8%  71.8%  78.4%
3         68.3%  76.5%  83.7%
4         71.0%  80.0%  87.4%
5         73.3%  82.8%  90.1%  <-- T>90% for p=70%, best-of-5
6         75.3%  85.1%  92.2%
7         77.1%  87.1%  93.8%
8         78.6%  88.7%  95.0%
9         80.1%  90.1%  96.0%  <-- T>90% for p=65%, best-of-9
21        90.3%  97.6%  99.6%  <-- T>90% for p=60%, best-of-21