Question

The probability of winning on a single toss of the dice is p. A starts, and if he fails, he passes the dice to B, who then attempts to win on her toss. They continue tossing the dice back and forth until one of them wins. What are their respective probabilities of winning? expectation

Answer 1

Answer:

The probability of A winning is $\frac{1}{2-p}$ and the probability of B winning is $\frac{1-p}{2-p}$.

Step-by-step explanation:

The probability of winning is p.

The game stops when either A or B wins.

The sample space of A winning is as follows:

{S, FFS, FFFFS, FFFFFFS, ...}

The sample space of B winning is as follows:

{FS, FFFS, FFFFFS, FFFFFFFS, ...}

Compute the probability of A winning as follows:

P (A winning) = P (S) + P (FFS) + P (FFFFS) + ...

                      $=p+(1-p)(1-p)p+(1-p)(1-p)(1-p)(1-p)p+...\\=p\sum\limits^{\infty}_{i=0} {(1-p)^{2i}}\\=p\times \frac{1}{1-(1-p)^{2}}\\=\frac{p}{1-(1-p)^{2}}\\=\frac{p}{1-1-p^{2}+2p}\\=\frac{1}{2-p}$

Compute the probability of B winning as follows:

P (B winning) = P (FS) + P (FFFS) + P (FFFFFS) + ...

    $=(1-p)p+(1-p)(1-p)(1-p)p+(1-p)(1-p)(1-p)(1-p)(1-p)p+...\\=(1-p)p\sum\limits^{\infty}_{i=0} {(1-p)^{2i}}\\=(1-p)p\times \frac{1}{1-(1-p)^{2}}\\=\frac{(1-p)p}{1-(1-p)^{2}}\\=\frac{(1-p)p}{1-1-p^{2}+2p}\\=\frac{1-p}{2-p}$

Thus, the probability of A winning is $\frac{1}{2-p}$ and the probability of B winning is $\frac{1-p}{2-p}$.