# Probability a team wins a series, given they won the first game

Let teams A,B compete in a series and win if they win 3 match first.

Team A wins a match with a probability $p$. The match results are independent from each other.

Given that team A won the first game, what's the probability they won the series?

I know the answer can be computed without building a definite probability space, but I'm asked to define a random variable to compute this probability.

I tried defining 3 probability spaces - one for each of the possible outcomes; for a series of 3 games, of 4 games or of 5 games. Then computed the probability those occur, multiplied by the conditional probability that given that team A won the first match, the won the 3/4/5 game series, then added. My random variables then were $X$ for counting the number of wins by A, and $Y$ to indicate whether they won the first game (which is $p$?)

Either I got lost in the algebra or my answer is off. Do you see a simpler way to compute this with the use of an explicit random variable, and preferably with a probability space?

• Not sure I follow. There are at most four games left to play. The winning sequences for $A$ are $AA,ABA,ABBA,BAA,BABA,BBAA$. What else do you need to write out? – lulu Dec 11 '16 at 0:33

Two provide a little bit of insight, assume for simplicity that the team who wins two (not three) games wins the series.

# Simplification

Since you have three games with two possible outcomes each, it is enough to consider a probability with $2^3=8$ elements. Namely $$\Omega = \{\omega_{AAA}, \ \omega_{AAB}, \ \omega_{ABA}, \ \dots, \ \omega_{BBB}\}$$ For example, $\omega_{ABB}$ denotes the event that $A$ wins the first game and $B$ wins the second and third. Now define a random variable $X$ like this:

$$X=\begin{cases}1, & \text{A wins the series}, \\0, & \text{B wins the series} \end{cases}$$ Using the notation from above, we see that

$$X(\omega) =\begin{cases} 1,& \omega \in \{\omega_{AAB}, \ \omega_{ABA},\ \omega_{BAA}, \ \omega_{AAA} \}=:\Omega_A \\ 0, & \omega \in \{\omega_{BBA},\ \omega_{BAB},\ \omega_{ABB}, \ \omega_{BBB} \}:=\Omega_B\end{cases}$$

Finally we can calculate the desired probability, therefore we define the set of all the $\omega$, where $A$ wins the first game: $$C:=\{\omega_{AAA}, \ \omega_{AAB},\ \omega_{ABA}, \ \omega_{ABB} \}.$$ And now: $$\mathbb{P}(\text{A wins series} \:| \text{A wins first game})= \mathbb{P}(X=1\: |\: C) = \mathbb{P}(\Omega_A\: |\: C) =\frac{\mathbb{P}(C\cap \Omega_A)}{\mathbb{P}(C)} = \\ \frac{\mathbb{P}(\{\omega_{AAB}, \ \omega_{ABA},\ \omega_{AAA}\ \})}{\mathbb{P}(C)},$$ which can be calculated easily using the additivity of $\mathbb{P}$. For example $$\mathbb{P}(C) =\mathbb{P}(\omega_{AAA})+\mathbb{P}(\omega_{AAB})+\mathbb{P}(\omega_{ABA})+\mathbb{P}(\omega_{ABB})=\\ p^3+p^2(1-p)+p^2(1-p)+p(1-p)^2.$$

# General Case

For the general case you need to consider a probability space with $2^5 =32$ elements. Then $\Omega$ looks like this: $$\Omega= \{\omega_{AAAAA}, \ \omega_{AAAAB}, \dots, \ \omega_{BBBBB}\}.$$

The interpretation of the $\omega$s is the same as before: e.g., $\omega_{AABBA}$ describes the event that $A$ wins the first two games and the last one, and $B$ wins games 3 and 4. The random variable $X$ and the set $C$ can also be defined as before: the definition of $X$ yields a partition of $\Omega$ into two set $\Omega_A$ and $\Omega_B$, where $\Omega_A$ contains all events where $A$ wins more often than $B$ and $\Omega_B$ contains all events where $B$ wins more often than $A$.

$C$ is again the set of all events where $A$ wins the first game. $$C=\{\omega_{AAAAA}, \ \omega_{AAAAB}, \dots, \ \omega_{ABBBB}\}$$ And therefore $$\mathbb{P}(\text{A wins series} \:| \text{A wins first game})= \mathbb{P}(X=1\: |\: C) = \mathbb{P}(\Omega_A\: |\: C) =\frac{\mathbb{P}(C\cap \Omega_A)}{\mathbb{P}(C)}$$

# Final Coment

Finally, I want to say that for exercises like this, it is way faster to compute the desired probabilty directly, without going into to much detail about the probability space in the shadows. But it is a nice exercise after all.

• oh sorry, I thought the team who has the most wins in 3 games wins, but the methodology is the same, I will edit the response – Cettt Dec 11 '16 at 1:05