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 $5$.

  • $\begingroup$ 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 $N$. $\endgroup$ – antkam Oct 20 '18 at 18:25
  • $\begingroup$ BTW typo: should be 22/64 & 42/64 instead of 21/64 & 43/64. :) $\endgroup$ – antkam Oct 20 '18 at 18:28
  • $\begingroup$ @antkam: you are right about the typo. Fixed. Thanks $\endgroup$ – 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.


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.