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Little math problem(s) February 2019
02-22-2019, 02:51 PM (This post was last modified: 02-22-2019 03:01 PM by Albert Chan.)
Post: #16
RE: Little math problem(s) February 2019
(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
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RE: Little math problem(s) February 2019 - Albert Chan - 02-22-2019 02:51 PM



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