A father and son take turns picking red and green balls from a bag. There are 2 red balls and 3 green balls. The first person to pick a red ball wins. There is no replacement. What is the probability that the father wins if he goes first?
I drew a binary tree to solve this. The father can only win the first round and the third round.
P(father wins first round) = $\frac25$
P(father wins third round) = $\frac35 * \frac24 * \frac23 = \frac15$
P(father wins first round) + P(father wins third round) = $\frac25+\frac15 =\frac35$
Is this correct?
Best Answer
Your proposed solution is exactly correct. Nice work.
To check your answer, you can use the same method to calculate the second player's probability of winning; it ought to be $1-\frac35 =\frac 25$. Let $P_i$ be the probability that the game ends in round $i$; you have calculated $P_1 + P_3 = \frac 35$. Then $$\begin{align} P_2 & = \frac35\cdot \frac 24 & = \frac 3{10}\\ P_4 & = \frac 35\cdot \frac24\cdot\frac 13\cdot \frac22 &= \frac1{10} \end{align} $$
So $P_2 + P_4 = \frac25$ as we expected.