Question

Aman and Bhuvan throw a pair of dice alternately. In order to win, they have to get a sum of 8. Find their respective probabilities of winning if Aman starts the game.

Answer 1

Solution :-

Given A and B throw a dice simultaneously

till one thus gets a 6 and wins the game.

Now, consider winning of A

Let X be obtaining a six by A

Y be obtaining a six by B

A can win the following way :

(i) He obtains a six in the first- trial

(ii) A fails to obtain a six, B fails and A obtain

and so on.

P (A wins) = $P\left(X\right)+P\left(XXX\right)+P\left(YYYYX\right)+....\infty$

$=\frac{1}{6}+\frac{1}{6}\times (\frac{5}{6}{)}^2+\frac{1}{6}\times (\frac{5}{6}{)}^4...\infty$

$=\frac{1}{6}\times \left[\frac{1}{1-(\frac{5}{6}{)}^2}\right]$

$=\frac{1/6}{1-\frac{25}{36}}=\frac{6}{11}$