Answer:
79% - almost 4 times out of 5.
The key to this is realizing that the number of games will not always be 5.
If A wins in a sweep - you have 2/3*(2/3)*2/3 percent chance of that happening - 8/27, or 29.63%
If A wins in 4, now we have 2/3*(2/3)*2/3*(1/3)*3 - the 1/3 is the chance that B wins a game. Note - there are only 3 ways B can win a game, not 4. B cannot win Game 4 because Game 4 would not be played in case of a sweep. That is why you cannot use a straight Pascal’s triangle to get your coefficients - the 1–4–6–4–1 is not possible if B cannot win Game 4. Anyway, the math is the same as the above, a 29.63% chance of A winning in 4.
For a 5 game set, A could lose 2 games in 6 possible ways (lose 1&2, 1&3, 1&4, 2&3, 2&4 or 3&4). Again, A cannot lose Game 5 - it would not be played once A wins 3 games. So the odds become 2/3*(2/3)*2/3*(1/3)*(1/3)*6, or 19.75%.
Add them up and you get 79.01%
Step-by-step explanation:
Hope it helps<3
