Little math problem(s) February 2019

[Albert Chan]

If the distribution is normal, to get 90%+ winning, z-score ≥ 1.2816
If n is high enough, binomial distribution approx. well to normal curve:

mean = N*p
std-dev = √(N*p*(1-p))

To simplify, assume game continue to N=2n, even if winner already decided

z-score = (2np - n) / √(2np*(1-p)) ≥ 1.2816
n ≥ ceil( (2p) (1-p) (1.2816/(2p-1))² )

For p=70%, n ≥ ceil(4.3) = 5
For p=65%, n ≥ ceil(8.3) = 9
For p=60%, n ≥ ceil(19.7) = 20

Slightly under-estimated for p=60% (should be best-of-21 wins), but not bad.

update: Tried p = 51%, for 90% chance of winning:
Normal distribution estimate => best-of-2053 wins.
Binomial distribution estimate => best-of-2053 wins.