Question

$A$ and $B$ play for a prize; $A$ is to throw a die first, and is to win if he throws $6$. If he fails $B$ is to throw, and to win if he throws $6$ or $5$. If he fails, $A$ is to throw again and to win with $6$ or $5$ or $4$, and so on : find the chance of each player.

Answer 1

$A$ can win in the first throw or $A$ can win after a looses or $A$ can if $B$ looses twice

$\Rightarrow P\left(A\right)=\frac{1}{6}+\frac{5}{6}.\frac{4}{6}.\frac{3}{6}+\frac{5}{6}.\frac{4}{6}.\frac{3}{6}.\frac{2}{6}.\frac{5}{6}\Rightarrow P\left(A\right)=\frac{1}{6}+\frac{5}{18}+\frac{25}{324}\Rightarrow P\left(A\right)=\frac{54+90+25}{324}=\frac{169}{324}$

$B$ can win if $A$ losses then $B$ win or both losses then $B$ win or when $A$ losses twice

$\Rightarrow P\left(B\right)=\frac{5}{6}.\frac{2}{6}+\frac{5}{6}.\frac{4}{6}.\frac{3}{6}.\frac{4}{6}+\frac{5}{6}.\frac{4}{6}.\frac{3}{6}.\frac{2}{6}.\frac{1}{6}.1\Rightarrow P\left(B\right)=\frac{5}{18}+\frac{1}{27}+\frac{5}{324}\Rightarrow P\left(B\right)=\frac{90+12+5}{324}=\frac{107}{324}$