# Find the probability that A is the winner of the series given that A won the first game.

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 :)

• Your try is not quite correct, because there can be less than 5 games. Commented Sep 4, 2014 at 2:04
• An easy test of your reasoning is to check if $P(E_A)+P(E_B)=1$. Commented Sep 4, 2014 at 2:55

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}

Given that A wins the first game, the probability that A will win the series is when A wins two additional games straight, A wins two additional games in three more games(B has won 1 game), A wins two additional games in four more games ( B has won 2 games). This translates into the following expression.

$$E(A) = p^2 + {2\choose1} p^{2}(1-p) + {3\choose2} p^{2}(1-p)^{2}$$

Consider that while the series actually stops after 3 games are won, it could hypothetically continue to 5 games without further wins or losses affecting the outcome.

So for A to win, A must win at least 3 of a hypothetical 5 in any order.

For B to win the converse must happen: A only wins at most 2 of a hypothetical 5 in any order.

By similar argument we have the conditional. Given the first game, A must then win at least 2 of the next 4 games to be ultimately victorious.

\begin{align}\therefore P[E_A] & = {5\choose 3}p^3(1-p)^2 + {5\choose 4}p^4(1-p) + p^5 \\ & = 6 p^5-15 p^4+10 p^3 \\[1ex] P[E_A\mid E_1] & = {4\choose 2}p^2(1-p)^2+{4\choose 3}p^3(1-p)+p^4 \\ & = 3 p^4-8 p^3+6 p^2 \end{align}

Since that is all you want, there's no need to apply Baye's Theorem.