The Chicago Cubs are playing a best-of-five-game series (the first team to win 3 games win the series and no other games are played) against the St. Louis Cardinals. Let X denotes the total number of games played in the series. Assume that the Cubs win 59% of their games versus their arch rival Cardinals and that the probability of winning game is independent of other games.
(b) calculate the conditional mean and standard deviation for X given that the Cardinals win the first game.
I asked a similar question the other day and I learned about the negative binomial distribution here: Can you use combinatorics rather than a tree for a best of five match?
Now in this part of the question, the conditions are slightly tweaked in that the Cardinals won the first game so it has to be modified. I tried doing this using the previous post but couldn't get the same answer. This is my approach:
Let $X$= number of games for any $k$ in the interval $w\le X\le 2w-1$ where $w$ equals the number of wins to win the series (in this case $3$). Also, let $p$ equal the probability the cubs win which is $59\%$.
Cardinals win:
k=4
The key inference to make is the following: In order to have the series of length k, the winning team must have a win in its last game. This is because if there are $w$ wins in the $(k-1)$ interval, then the series would be over before $k$ games are played.
So for cardinals winning in $k=4$, we have: W _ _ W. So there are $2$ remaining spots left which is represented as $(k-2)$ and we have to place $1$ win. So I would write this as:
$\Pr(\text{Cardinals Win})=\binom{k-2}{w-2} \cdot (1-p)^w \cdot p^{k-w} \text{ for } k=3,4,5\tag{1}$
We could test this with k=5 and get the same derivation: W _ _ _ W. Here we have to place 1 win in 3 spots.
Cubs Win:
k=5
The cubs lost the first game, must win the last game, and place 2 wins in the remaining spots:
L _ _ _ W.
I wrote this as: $\Pr(\text{Cubs Win})=\binom{k-2}{w-1} \cdot (p)^w \cdot (1-p)^{k-w} \text{ for } k=4,5\tag{2}$
Note that for the case k=3, the Cubs lost the first game and they cannot win for that case.
Since the event that the teams win are disjoint, the probability for the number of games played is simply the sum of equations (1) and (2).
I then solved the equations for each value of $k$:
$Pr(\text{Cubs win in 4 games})=p^3 \cdot (1-p)= (.59)^3(.41)$
$Pr(\text{Cubs win in 5 games})=3 p^3 \cdot (1-p)^2=3(.59)^3(.41)^2$
$Pr(\text{Cardinals win in 5 games})=3 (1-p)^3 \cdot (p)^2=3(.59)^3(.41)^2$
This is not the same thing as the book. What's even more confusing is that the term numbers don't even add up to the value of $k$. If I drew a tree diagram, the probability of an event is the product of the nodes of the tree. So if I have $k=4$, I would expect to have a product of 4 values, but they have 1 less.
Can someone please tell me what I am doing wrong? Thank you again in advance.
Best Answer
Cardinals winning the first game complicates things, since the situation becomes less symmetrical. Let $p=0.59$.
If we can find the probability distribution function of $X$, the number of games (conditional on Cards winning the first game), then finding the mean and variance is staightforward.
Easiest is to find the (conditional) probability that $X=3$. the Cards have to win the next $2$ games. This has probability $(1-p)^2$.
Next we find the probability that $X=4$. This can happen in $2$ different ways: (i) Cards win in $4$ and (ii) Cubs win in $4$.
(i) Cards win in $4$ if they lose exactly one of Games 2 or 3, and win the rest. This has probability $(2)p(1-p)(1-p)$.
(ii) Cubs win in $4$ if they win Games 2, 3, 4. This has probability $p^3$.
For the probability that $X=4$, add the numbers obtained in (i) and (ii).
Next we find the probability that $X=5$. We could do a cases analysis. But we know the answer! It is $1-\Pr(X=3)-\Pr(X=4)$.
Remark: In the post, there is, among other things, the assertion that the probability Cubs win in $4$ games is $p^3(1-p)$. This is true for unconditional probabilities. But the Cards won the first game, that's a fact. So the probability the Cubs win in $4$ games, given that fact, is simply $p^3$.