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.
 A: 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 $5$.
A: 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:


*

*P1 = 21/32 ~= 65.6%

*P2 = 21/32 ~= 65.6%

*P3 = 21/32 ~= 65.6%

*P4 = 21/32 ~= 65.6%

*P5 = 19/28 ~= 67.9%

*P6 = 3/4 ~= 75%

*P7 = 1/1 ~= 100%


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