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Suppose we have Series 1 and Series 2, and four teams $A,B,C,D$. $A$ and $B$ play in Series 1 and $C$ and $D$ play in Series 2. In each series, the first team to win 3 games wins the series. The home team has a $3/4$ chance of winning in each game. The winner of each series plays each other in a similar format. What is the probability of $A$ defeating $D$ in the final series $3-2$?

The home team is based on:

 Game #              Home Team
 Game 1                 A/C
 Game 2                 A/C
 Game 3                 B/D
 Game 4(if necessary)   B/D
 Game 5(if necessary)   A/C

Can the probability be decomposed to something like: $$P(\text{A defeats D 3-2}) = P(\text{A defeats D 3-2}|\text{A wins Series 1 and D wins Series 2})$$

Added In the $A/C$ notation, Game 1 of Series 1 is played in A's hometown and Game 1 of Series 2 is played in C's hometown, etc.

Games 1,2, and 5 will be played in the Series 1 winner's hometown, Games 3 and 4 will be played in Series 2 winner hometown.

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  • $\begingroup$ Your notation for the Home Team isn't clear. In, say, the $A/C$ series...is $A$ always the home team? Does it alternate? In the final series between the two winners...how does home team advantage sort out? $\endgroup$ – lulu Jan 16 at 20:18
  • $\begingroup$ To your closing question: No. What you want isn't that conditional probability since the answer you want depends (heavily) on the probability that $A$ and $D$ win their respective series. You have to multiply your conditional by the probability that both $A,D$ win. $\endgroup$ – lulu Jan 16 at 20:21
  • $\begingroup$ @lulu: Games 1,2, and 5 will be played in the Series 1 winner's hometown, Games 3 and 4 will be played in Series 2 winner hometown.In the $A/C$ notation, Game 1 of Series 1 is played in A's hometown and Game 1 of Series 2 is played in C's hometown, etc.. $\endgroup$ – Damien Jan 16 at 20:30
  • $\begingroup$ Regardless of the confusion: because of the home team advantage, it's probably easiest to simply write out all the cases. It's tedious, but not at all difficult. $\endgroup$ – lulu Jan 16 at 20:33
  • $\begingroup$ @lulu: See my addition. $\endgroup$ – Damien Jan 16 at 20:34

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