Question

$A$ and $B$ throw a die alternatively till one of them gets a ${}^{\prime }{6}^{\prime }$ and wins the game. Find their respective probabilities of winning, if $A$ starts first.

Answer 1

Let S denote the success (getting a ‘6’) and F denote the failure (not getting a ‘6’) .

Thus, P(S)=$\frac{1}{6}=p,$ P(F)=$\frac{5}{6}=q$

$P($A wins in first throw)$=P\left(S\right)=p$

$P($A wins in third throw)$=P\left(FFS\right)=qqp$

$P($A wins in fifth throw)$=P\left(FFFFS\right)=qqqqp$

So, $P($A wins$)=p+qqp+qqqqp+\dots$

$=p(1+q^2+q^4+\dots )$

$=\frac{p}{1-q^2}=\frac{\frac{1}{6}}{1-\frac{25}{36}}=\frac{6}{11}$

$P($B wins$)=1–P($A wins$)$

P(B wins) =$1-\frac{6}{11}=\frac{5}{11}$

So, P(A wins)=$\frac{6}{11}$ , P(B wins)=$\frac{5}{11}$