0
$\begingroup$

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 :)

$\endgroup$
  • $\begingroup$ Your try is not quite correct, because there can be less than 5 games. $\endgroup$ – Mateus Sampaio Sep 4 '14 at 2:04
  • $\begingroup$ An easy test of your reasoning is to check if $P(E_A)+P(E_B)=1$. $\endgroup$ – Graham Kemp Sep 4 '14 at 2:55
2
$\begingroup$

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:

$$\begin{align} AA&: q=p^2\\ ABA&: q=p^2(1-p)\\ ABBA&: q=p^2(1-p)^2\\ BAA&: q=p^2(1-p)\\ BABA&: q=p^2(1-p)^2\\ BBAA&: q=p^2(1-p)^2\\ \text{Total}&: q=p^2(1+2(1-p)+3(1-p)^2) \end {align}$$

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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.

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

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.