Question

A and B throw a pair of dice alternately, till one of them gets a total of 10 and wins the game. Find their respective probabilities of winning, if A starts first.

Answer 1

Here, two persons are A & B and A is the first person to throw the pair of dice. Probability of getting a total of 10 in a throw of a pair of dice is success.

Now, probability of getting a total of 10 is:
P(total of 10) $=\frac{(4,6),(6,4),(5,5)}{36}=\frac{3}{36}=\frac{1}{12}$

P (not getting a total of 10)$=1-\frac{1}{12}=\frac{11}{12}$

P(A wins) $=\frac{1}{12}+{\left(\frac{11}{12}\right)}^2\left(\frac{1}{12}\right)+{\left(\frac{11}{12}\right)}^4\left(\frac{1}{12}\right)...\text{to}\infty$

$=\frac{1}{12}[1+{\left(\frac{11}{12}\right)}^2+{\left(\frac{11}{12}\right)}^4+...\text{to}\infty ]$

$=\frac{1}{12}\left[\frac{1}{1-{\left(\frac{11}{12}\right)}^2}\right]\left[\begin{array}{c}\text{for an infinite G.P.}\\ \text{sum}=\frac{\text{first term}}{1-r}\end{array}\right]$

$=\frac{1}{12}\left[\frac{1}{\frac{23}{144}}\right]=\frac{1}{12}\times \frac{144}{23}=\frac{12}{23}$

P(B wins)= 1-P(A wins)

$=1-\frac{12}{23}$

$=\frac{11}{23}$