Question

Players A and B throw alternately with a pair of ordinary dice. Player A wins if he throws 6 before B throws 7, and player B wins if he throws 7 before A throws 6. If A begins, what is the chance that player A wins?

Answer 1

We know that ,

Probability of happening of an event = Favourable outcome/total outcomes

Total outcomes of rolling two dice = 36

Favourable outcomes for 6: (1,5), (2,4), (3,3), (4,2), (5,1) = 5

Favourable outcomes for 7: (1,6), (2,5), (3,4), (4,3), (5,2) ,(6,1)= 6

Probability of occurring 6 is
$p(A) = \frac{5}{36} \\ \\ p(\bar A) = \frac{31}{36}$
Probability of getting 7 is
$p(B) = \frac{6}{36} \\ \\ p(\bar B) = \frac{30}{36}$
As player A begins the game,so A wins in the following cases

$= p(A) + p(\bar A)p(\bar B)p(A) + p(\bar A)p(\bar B)p(\bar A)p(\bar B)p(A) + ...∞ \\ \\ = \frac{5}{36} + \frac{31}{36} \times \frac{30}{36} \times \frac{5}{36} + \frac{31}{36} \times \frac{30}{36} \times \frac{31}{36} \times \frac{30}{36} \times \frac{5}{36} + ...$

It forms an infinite GP

first term $a=\frac{5}{36}$

common ratio r
$r= \frac{30 \times 31}{36 \times 36} \\ \\ = \frac{155}{216}$

So r <1

Sum of infinite GP is

$S_{∞} = \frac{a}{1 - r} \: \: \: r < 1 \\ \\ S_{∞} = \frac{ \frac{5}{36} }{1 - \frac{155}{216} } \\ \\S_{∞}= \frac{ \frac{5}{36} }{ \frac{216 - 155}{216} } \\ \\ S_{∞}= \frac{5 \times 216}{36 \times 61} \\ \\S_{∞}= \frac{30}{61}$

Probability that A wins is

$\frac{30}{61}$

Hope it helps you