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Wednesday, May 19, 2010
Chance of winning back-to-back games
H-H
H-A
A-H
A-A
“H” is home (or heads if you like). “A” is away (or ... well, a crude tail). If H=A=.50, then H-H plus A-A would be .50. If H=.6, and A=.40, then H-H plus A-A equals .52.
In baseball, if you win the first game, you probably had the better starting pitcher. That means in the second game, you probably have a worse starting pitcher. But you still have the home advantage. So, what might it look like?
If H1 wins, then chances are their presumed odds was say .560. If A1 wins, then chances are, their presumed odds was say .480. That is, given that you know that H1 wins, then there’s a decent chance they had an above average pitcher (and they might have been the better hitting team to begin with). On top of which, we also know that pitcher will not be available the next day.
So, H1-H2 = .56 x .53
A1-A2 = .48 x .45
Add the two together, and you get .513.
What were the empirical results?
The true answer: 51.3%, a little better than a coin toss.
I swear that I got lucky. I started with .540 for the home team, and .460 for the visiting team, and added +.005 for likely having the better hitting team and +.015 likely the better pitching team. That’s for game 1. For game 2, I used +.005 and -.015.
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