Probability that someone will pick a red ball first? 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?
 A: 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.
A: Correct
An alternative approach is, out of all $\binom{5}{2}$ ways to place red (and green) balls in a line, count ways that place the second red ball when the first red ball is either the first or third ball in line.
$$\dfrac {\binom{4}{1}+\binom{2}{1}}{\binom{5}{2}}=\frac 3 5$$
A: Can this be interpreted as 
winning in first round OR not losing in second round AND winning in third round
which would be 
P(wr1) + P(nlr2) * P(wr3)
P(winning in round 1) = P(red ball)/P(total) = 2/5
P( not losing in round 2) = P(picking a green ball) * P(opponent picking a green ball from remaining) = 3/5 * 2/4
P( winning in round 3) = P(picking a red ball from remaining) = 2/3
Total probability = 2/5 + (3/5 * 2/4) * 2/3
Trying to create a recurrence would be:
Pw(2,3) + Pnl(2,3) * Pw(2,1)
which is Probability of choosing red with 2 red, 3 green + Probability of not loosing * Probability of winning with 2 red and 1 green.   
