given X1,X2,...,Xn (zeta replaced by x for convenience) follow Bernoulli distribution with p = 7/8 with

therefore Sn follows binomial distribution(n,p=7/8).
E(Sn/n) = np/n = p =7/8
Var(Sn) = pq/n =(7/8*1/8)/n
![p[ \left | \frac{S_{n}}{n} -p \right | < k ] \geqslant 1-\frac{Var(S_{n}/n)}{k^{2}}](http://img.homeworklib.com/questions/2e8cbe10-5ec5-11ec-98ef-39924e2539a1.gif?x-oss-process=image/resize,w_560)
choosing k =0.1,
we have

1-(pq/n)/(0.1*0.1) =0.8
n =7/0.128
n = 54.6875
hence n = 55
so he has to play at least n = 55 games to get gain = np = 55*(7/8) =48.125.
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