Answer to Question #146248 in Statistics and Probability for Alvin

Question

A bag contains 5 white balls and 3 red balls. Two players, A and B take turns at drawing one ball from the bag at random, and the balls are not replaced. The player who first gets two red balls is the winner, and the drawing stops as soon as either two players has drawn two red balls. Player A draws first. Find the probability that A is the winner, given that the winning player wins on his second draw.

Answer 1

Lets denote R for drawing a red ball and W for drawing a white ball

Possible results given that the winning player wins on his second draw:

RRR or RWR or RRWR or WRWR or WRRR

where RRR or RWR are possible results when A wins.

So

P(RRR or RWR)=$\frac{3}{8}\cdot\frac{5}{7}\cdot\frac{2}{6}+\frac{3}{8}\cdot\frac{2}{7}\cdot\frac{1}{6}=\frac{3}{28}$

P(RRR or RWR or RRWR or WRWR or WRRR)=$\frac{3}{8}\cdot\frac{5}{7}\cdot\frac{2}{6}+\frac{3}{8}\cdot\frac{2}{7}\cdot\frac{1}{6}+\frac{3}{8}\cdot\frac{2}{7}\cdot\frac{5}{6}\cdot\frac{1}{5}+\frac{5}{8}\cdot\frac{3}{7}\cdot\frac{4}{6}\cdot\frac{2}{5}+\frac{5}{8}\cdot\frac{3}{7}\cdot\frac{2}{6}\cdot\frac{1}{5}=\frac{3}{14}$

Therefore

P(A is the winner, given that the winning player wins on his second draw)=$\frac{\frac{3}{28}}{\frac{3}{14}}=\frac{1}{2}$

Answer: $\frac{1}{2}$