Question

Two players, A and B, alternately keep rolling a fair dice. The person to get a six first wins the game. Given that player A starts the game, the probability that A wins the game is

• A. $\frac{5}{11}$
• B. $\frac{1}{2}$
• C. $\frac{7}{13}$
• D. $\frac{6}{11}$

Answer 1

The correct option is D $\frac{6}{11}$

Let W = {winning event} & L = {loosing even}
$P\left(6\right)=1/6$ & P($\overline{6}$)=$\frac{5}{6}$

$P\left(W\right)=\frac{1}{6}$ & $P\left(L\right)=\frac{5}{6}$

The case can be

W or LLW or LLLLW & so on.

$\therefore P\left(A\right)=\frac{1}{6}+\frac{5}{6}.\frac{5}{6}.\frac{1}{6}+\frac{5}{6}.\frac{5}{6}.\frac{5}{6}.\frac{5}{6}.\frac{1}{6}+...\infty$

$=\frac{1}{6}(1+{\left(\frac{5}{6}\right)}^2+{\left(\frac{5}{6}\right)}^4+----+\infty )$

$=\frac{1/6}{1-{\left(\frac{5}{6}\right)}^2}=\frac{1/6}{1-25/36}=\frac{1/6}{11/36}=\frac{6}{11}$