Question

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

Answer 1

Let success (getting a $6$) is denoted by $\text{S}$ and the failure(not getting $6$) is denoted by $\text{F}$.

When a die is thrown then sample space $\left\{1,2,3,4,5,6\right\}$

$$\begin{gathered} \text{P(S)=}\frac{1}{6}\text{=p} \\ \text{P(F)=}\frac{5}{6}\text{=q} \end{gathered}$$

If we throw the die then first $\text{A}$ throw then die $\text{B}$ is thrown, so

$$\begin{gathered} \text{P(A wins in the first throw )=p} \\ \text{P(A wins in the third throw)=qqp} \\ \text{P(A wins in the fifth throw)=qqqqp} \end{gathered}$$

$\text{P(A wins)=p+qqp+qqqqp+ ⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯}$

$$\begin{gathered} ={\text{p+q}}^2{\text{p+q}}^4\text{p+⋯⋯⋯⋯⋯⋯⋯⋯⋯} \\ ={\text{p(1+q}}^2{\text{+q}}^4\text{+⋯⋯⋯⋯⋯⋯)} \\ =\text{p(}\frac{1}{1-{\text{q}}^2}{\text{) (here 1+q}}^2{\text{+q}}^4\text{+⋯⋯⋯⋯⋯ is a G.P.)} \\ =\frac{\text{p}}{1-{\text{q}}^2} \end{gathered}$$

Now we find $1-{\text{q}}^2\text{=1-(}\frac{5}{6}{\text{)}}^2\text{=1-}\frac{25}{36}\text{=}\frac{36-25}{36}\text{=}\frac{11}{36}$

Now we find $\frac{\text{p}}{1-{\text{q}}^2}=\frac{1}{6}÷\frac{11}{36}=\frac{1}{6}\times \frac{36}{11}=\frac{6}{11}$

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