# Probability of winning a 7-game series if you win game no. $j$

This question is inspired by the ongoing baseball playoffs, but pertains to any tournament where 2 teams play a 7-game series, where the first to win 4 games is the overall (series) winner.

In times like these, the news coverage is full of useless statistics like "in the past 25 years, the team that wins game 3 (i.e. the 3rd game) has gone on to win the series 68% of the time". However, this does get me thinking...

• In a symmetry sense, every game is equally important toward victory.

• However, a series can end early, s.t. games 5, 6, 7 might not even be played. Given this asymmetry, it is correct to say that the winner of game 7 (if it is played) is always the series winner, but the winner of game 6 (if it is played) is not always the series winner.

Let $$j \in \{1,2,...,7\}$$ and $$A_j$$ be the event that game number $$j$$ is played AND team $$A$$ wins that game. Let $$A_s$$ be the event that team $$A$$ wins the series. My Question: What is $$P(A_s | A_j)$$?

Further thoughts:

For this question assume the chance of team $$A$$ winning any single game is $$1/2$$ and each game is independent.

Obviously each $$P(A_s | A_j)$$ can be calculated with a little bit of effort e.g. by (exhaustive) combinatorial counting. For something small like 7 games, this can be done by hand or using a small program. Moreover, the following are obvious:

• $$P(A_s | A_7) = 1$$

• $$P(A_s | A_6) = 3/4$$, because there is $$1/2$$ chance team $$A$$ had a 3-2 lead, in which case it wins after winning game 6, and a $$1/2$$ chance team $$B$$ had a 3-2 lead, in which case team $$A$$ has a $$1/2$$ to win the series (after winning game 6).

• $$P(A_s | A_1) = P(A_s | A_2) = P(A_s | A_3)$$ by symmetry, since sudden ending cannot happen before game 4. [The symmetry exploited here is one can arbitrarily permute the results of these 3 games.]

However, is there a clever way to calculate $$P(A_s|A_j)$$ without resorting to (too much) explicit counting / chasing down the "tree-of-possibilities"?

Generalization to odd $$N$$ beyond $$N=7$$ would also be interesting.

The team that loses game $$1$$ now has to win at least $$4$$ out of $$6$$ games if we imagine all the games are played. We can read off from Pascal's triangle that happens $$\frac {1+6+15}{2^6}=\frac {22}{64}$$ of the time, so $$P(A_s|A_1)=\frac {42}{64}$$. This applies to game $$4$$ as well by your permutation argument for games $$1,2,3$$.

I don't see an easy way for game $$5$$.

• Good point that the symmetry extends to $P(A_s|A_4) = P(A_s|A_1)$, and indeed that value can be calculated by counting as you showed. For a 7-game series this leaves only $j=5$. But I'm still hoping someone comes up with a clever way to calculate these, esp. for arbitrary $N$. – antkam Oct 20 '18 at 18:25
• BTW typo: should be 22/64 & 42/64 instead of 21/64 & 43/64. :) – antkam Oct 20 '18 at 18:28
• @antkam: you are right about the typo. Fixed. Thanks – Ross Millikan Oct 20 '18 at 19:04

I don't see a clever way here that doesn't require enumeration and is faster to come up with than enumeration takes to solve.

Visualize the state space as a five by five square with one missing corner, with states (m,n) signifying the score at that time, e.g. (2,1) would mean team A has won two games so far and team B has won one. You can have any combination of the five numbers from 0 to 4, except (4,4), hence 25 - 1 = 24 states. (0,0) is the initial state, and any state (4,n) or (m,4) is a terminal state. Terminal states have no outgoing edges, all non-terminal states (m,n) have one edge to (m+1,n) and one to (m,n+1).

To solve the problem, we need to assign two attributes to each state: the probability to reach and the probability to win. Both can be determined inductively:

$$P_R(m,n)=\cases{1 & if m,n = 0,0 \cr P_R(3,n)/2 & if m=4 \cr P_R(m,3)/2 & if n=4\cr (P_R(m-1,n)+P_R(m,n-1))/2 & otherwise}$$ $$P_W(m,n)=\cases{1 & if m=4 \cr 0 & if n=4 \cr (P_W(m+1,n)+P_W(m,n+1))/2 & otherwise}$$

Now we need to locate the subsets on which the conditional proabilities to win shall be calculated. Within layer $$j$$, identified by $$m+n=j$$, these are only those states which have an incoming edge in the direction of team A winning, i.e. from a parent (m-1,n), not (m,n-1). We can't use their probability to reach directly, however, since the condition requires us to reach these states via a team-A-winning edge only. We therefore also need a slightly modified probability to reach with a team A win: $$P_R'(m,n)=\cases{P_R(m-1,n)/2 & if n<4 and m>0 \cr 0 & otherwise}$$ We now have all the tools required to calculate the demanded probabilities: $$P(A_s|A_j)=\frac{\sum_{m+n=j}P_R'(m,n)*P_W(m,n)}{\sum_{m+n=j}P_R'(m,n)}$$ This yields the following results:

1. P1 = 21/32 ~= 65.6%
2. P2 = 21/32 ~= 65.6%
3. P3 = 21/32 ~= 65.6%
4. P4 = 21/32 ~= 65.6%
5. P5 = 19/28 ~= 67.9%
6. P6 = 3/4 ~= 75%
7. P7 = 1/1 ~= 100%

This didn't feel very elegant, but at least it's a definitive answer. It also confirmed your intuitive observations.