Answer:
As someone pointed out in the answer to the five-game problem, this is great as long as you assume that we’re just doing a dressed up probability problem. The answers would be the same if you were asking about the probability of flipping four heads (and stopping when you do) out of seven flips with an unfair coin that came up heads 60% of the time.
Real baseball doesn’t work that way, not least because each game’s starting pitchers have a huge effect on that game. You might look at the teams overall and decide that the Sox have a .6 probability of winning a game in general, but if Walter Buehler spins a gem that night, you can toss the probabilities out the window. It is at best a rough approximation.
Explanation:
Sox win in 4
This can only happen one way: WWWW. Odds are .64≈13.0%
Sox win in 5
This can happen four ways, all of which are mathematically the same as LWWWW. Odds are .4×.64×4≈20.7%
.
Sox win in 6
This can happen ten ways, all of which are mathematically the same as LLWWWW. That’s .42×.64×10≈20.7%
. It’s an interesting coincidence that this is the same as the odds of winning in 5, but it’s only a coincidence.
Sox win in 7
This can happen in 20 ways, all of which are mathematically the same as LLLWWWW. That’s .43×.64×20≈16.6%
.
So the total chances of the Sox winning are just the sum of those, or 71.0%.