Question

A and B alternately throw a pair of symmetrical dice. Who ever throws a sum of 9 points first will be declared as winner. If A starts the game, the probability of his winning is

• A. $\frac{7}{17}$
• B. $\frac{8}{17}$
• C. $\frac{9}{17}$
• D. $\frac{6}{17}$

Answer 1

Answer: (C)

$\frac{9}{17}$

Let's denote event S for sum to be 9.
Then S will happen when the dice show the combination of {3,6 and 4,5}
$P\left(S\right)=2!\times \{P\left(3\right)P\left(6\right)+P\left(4\right)P\left(5\right)\}\Rightarrow P\left(S\right)=2\times \{\frac{1}{6^2}+\frac{1}{6^2}\}=\frac{1}{9}$
If A starts the game then:
$P\left(Awins\right)=P\left(S\right)+P\left(\overline{S}\right)P\left(\overline{S}\right)P\left(S\right)+P\left(\overline{S}\right)P\left(\overline{S}\right)P\left(\overline{S}\right)P\left(\overline{S}\right)P\left(S\right)+........................\infty$
This is an infinite G.P sum with first term=$P\left(S\right)$ and common ratio=$P\left(\overline{S}\right)P\left(\overline{S}\right)$
Hence $P\left(Awins\right)=\frac{P\left(S\right)}{1-P\left(\overline{S}\right)P\left(\overline{S}\right)}=\frac{1}{9}\times \frac{81}{81-64}=\frac{9}{17}$