If you play two games, your chance is $ p^2$. If you play four, your chance is $p^4+4p^3(1-p)$, which becomes greater when $p\gt \frac 13 $. If you play six games, your chance is $p^6+6p^5(1-p)+15p^4(1-p)^2$, which is better than four games when $p \gt \frac 25$ It is tempting to conjecture that you should play $2n$ games (instead of $2n-2$) when $p \gt \frac {n-1}{2n-1}$
| p < | n | P (chances of winning the game) |
|---|---|---|
| $\frac{1}{3}$ | 2 | $p^2$ |
| $\frac{2}{5}$ | 4 | $p^4+4p^3(1-p)$ |
| $\frac{3}{7}$ | 6 | $p^6+6p^5(1-p)+15p^4(1-p)^2$ |
| $\frac{4}{9}?$ | 8 | $p^8 + 8p^7(1-p) + 28p^6(1-p)^2 + 56p^5(1-p)^3$ |
