Repeat Problem 21 if the players toss a pair of dice trying to get a double (that is, both dice showing the same number).

A pair of dice are rolled and the person who gets same numbers first is the winner.

The events are said to be independent when the occurrence or non-occurrence of any event does not have any effect on the occurrence or non-occurrence of the other event.

There are many cases where the first player can win. He can win in the first try, third try and so on. This implies that on the second, fourth and even numbered tries, second player loses.

Find the probability that the first player wins.

$$\begin{gathered} \mathit{P}\mathbf{(}\mathbf{first}\mathbf{}\mathbf{player}\mathbf{}\mathbf{wins}\mathbf{)}=\frac{1}{6}+\frac{1}{6}\times \frac{5}{6}\times \frac{1}{6}+\frac{1}{6}\times \frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}+\otimes \\ =\frac{{\displaystyle \frac{1}{6}}}{1-{\displaystyle \frac{25}{36}}} \\ =\frac{6}{11} \end{gathered}$$

Thus the probability that the first player wins is$\frac{6}{11}$.

Find the probability that the second player wins by subtracting the obtained probability from 1.

$$\begin{gathered} P\left(secondPlayerWins\right)=1-\frac{6}{11} \\ =\frac{5}{11} \end{gathered}$$

Thus the probability that the second player wins is $\frac{5}{11}$.