Little math problem(s) February 2019

[Albert Chan]

(02-21-2019 10:59 AM)Albert Chan Wrote:  I meant best-of-21 wins, with whatever games needed to achieve it.
Total games are variable, upto max of N = 2n-1 = 41 games.

Expected total games to finish is less than 2n-1
= sum((games played) * (probability of games played produced a winner))
= sum((n+k) * nCr(n-1+k, k) * (p^n (1-p)^k + (1-p)^n p^k), k = 0 to n-1)

-> Best-of-21 wins, p=50%, expected total games ≈ 36.8598 ; maximum expected value
-> Best-of-21 wins, p=60%, expected total games ≈ 34.4171 ; Probability of reaching 41 games = 1 in 18