Two people are taking turns tossing a pair of coins; the first person to toss two alike wins. What are the probabilities of winning for the first player and for the second player? Hint: Although there are an infinite number of possibilities here (win on first turn, second turn, third turn, etc.), the sum of the probabilities is a geometric serieswhich can be summed; see Chapter 1 if necessary.

A pair of coins are toss and the person who gets same toss alike 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.

$$\begin{gathered} \mathit{P}\left(firstplayerwins\right)=\frac{1}{2}+\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}+\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}+\otimes \\ =\frac{{\displaystyle \frac{1}{2}}}{1-{\displaystyle \frac{1}{4}}} \\ =\frac{2}{3} \end{gathered}$$

Thus the probability that the first player wins is $\frac{2}{3}$.

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

$$\begin{gathered} \mathit{P}\left(secondPlayerWins\right)=11-\frac{2}{3} \\ =\frac{1}{3} \end{gathered}$$

Thus the probability that the second player wins is$\frac{1}{3}$.