Bernoulli Trial

Relevant wiki: Uniform Probability (by Outcomes)

$P=$ $_nC_rp^rq^{n-r}$ where: $p$ = probability of success(win), $q$ = probability of failure(loss), $n$ = number of games, $r$ = number of exact wins by Flash, and $_nC_r=\frac{n!}{r!(n-r)!} = \binom{n}{r}$ =binomial coefficient

Because, they are equally strong players, the probability of success is $p=\dfrac{1}{2}$ , thus the probability of failure is $q=1-p=1-\dfrac{1}{2}=\dfrac{1}{2}$

Three games out of four:

Substituting, we obtain

$P=$ $_4C_3(\frac{1}{2})^3(\frac{1}{2})=4(\frac{1}{8})(\frac{1}{2})=\frac{4}{16}$

Five games out of eight:

Substituting, we obtain

$P=$ $_8C_5(\frac{1}{2})^5(\frac{1}{2})^3=56(\frac{1}{32})(\frac{1}{8})=\frac{56}{256}=\frac{7}{32}$

Compare:

$\frac{4}{16}$ $\implies$ $\frac{8}{32}$ (three games out of four)

$\frac{7}{32}$ (five games out of eight)

$\dfrac{8}{32}$ $>$ $\dfrac{7}{32}$ , therefore, the more probable is three out of four games.