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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 $p \gt \frac {n-1}{2n-1}$

$$\begin{array}{l|l|l} p\lt & n & P \;\; \text{(chances of winning the game)}\\ \hline \\ \tfrac{1}{3} & 2 & p^2 \\ \tfrac{2}{5} & 4 & p^4+4p^3(1-p) \\ \tfrac{3}{7} & 6 & p^6+6p^5(1-p)+15p^4(1-p)^2 \\ \tfrac{4}{9}? & 8 & p^8 + 8p^7(1-p) + 28p^6(1-p)^2 + 56p^5(1-p)^3 \\ \end{array}$$

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 $p \gt \frac {n-1}{2n-1}$

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}\$

$$\begin{array}{l|l|l} p\lt & n & P \;\; \text{(chances of winning the game)}\\ \hline \\ \tfrac{1}{3} & 2 & p^2 \\ \tfrac{2}{5} & 4 & p^4+4p^3(1-p) \\ \tfrac{3}{7} & 6 & p^6+6p^5(1-p)+15p^4(1-p)^2 \\ \tfrac{4}{9}? & 8 & p^8 + 8p^7(1-p) + 28p^6(1-p)^2 + 56p^5(1-p)^3 \\ \end{array}$$

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 $p \gt \frac {n-1}{2n-1}$

$$\begin{array}{l|l|l} p\lt & n & P \;\; \text{(chances of winning the game)}\\ \hline \\ \tfrac{1}{3} & 2 & p^2 \\ \tfrac{2}{5} & 4 & p^4+4p^3(1-p) \\ \tfrac{3}{7} & 6 & p^6+6p^5(1-p)+15p^4(1-p)^2 \\ \tfrac{4}{9}? & 8 & p^8 + 8p^7(1-p) + 28p^6(1-p)^2 + 56p^5(1-p)^3 \\ \end{array}$$

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}\$

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}\$

$$\begin{array}{l|l|l} p\lt & n & P \;\; \text{(chances of winning the game)}\\ \hline \\ \tfrac{1}{3} & 2 & p^2 \\ \tfrac{2}{5} & 4 & p^4+4p^3(1-p) \\ \tfrac{3}{7} & 6 & p^6+6p^5(1-p)+15p^4(1-p)^2 \\ \tfrac{4}{9}? & 8 & p^8 + 8p^7(1-p) + 28p^6(1-p)^2 + 56p^5(1-p)^3 \\ \end{array}$$