A and B play a series of games. Each game is independently won by A with probability $p$ and by B with probability $1-p$. They stop when the total number of wins of one of the players is $3$. Find the probability that A is the winner of the series given that A won the first game.
My attempt:
Let $E_A$ be the event that A wins the series and $E_1$ be the event that A wins the first game. Then I need to compute $$P[E_A|E_1]={P[E_1|E_A]P[E_A]\over P[E_1]}$$
$P[E_1]=p$ and $P[E_A]={\binom {5}{3}}p^3(1-p)^2$ this is because A and B can play a total of $5$ games: (for example A loses the first $2$ games but wins the last $3$ games) but we can do this in ${\binom {5}{3}}$ ways (becuase A needs to win 3 games out of 5)
But I don´t know how to compute $P[E_1|E_A]$
I would really appreciate if you can help me 🙂
Best Answer
You are over complicating this - there is no need for conditional probability here.
When A wins the first game you now have a different scenario - what is the probability of A winning 2 games before B wins 3 - this is small enough that you can enumerate it:
$$\begin{align} AA&: q=p^2\\ ABA&: q=p^2(1-p)\\ ABBA&: q=p^2(1-p)^2\\ BAA&: q=p^2(1-p)\\ BABA&: q=p^2(1-p)^2\\ BBAA&: q=p^2(1-p)^2\\ \text{Total}&: q=p^2(1+2(1-p)+3(1-p)^2) \end {align}$$