The only "events" to you are innings in which a man is on 3rd with < 2outs. Can happen multiple times in an inning.
Other than just having the exact data, there is no way to do the math on this because so many situations that can arise.
This is from '08 and just single ABs.
[SIZE=-1]Runner on third, less than two outs -- MLB, 2008:[/SIZE]
[SIZE=-1]
[SIZE=-1]Opportunities[/SIZE] | [SIZE=-1]9,296[/SIZE] |
[SIZE=-1]Driven in by Batter[/SIZE] | [SIZE=-1]5,156[/SIZE] |
[SIZE=-1]Percentage[/SIZE] | [SIZE=-1]55.5%[/SIZE] |
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So if the guy is scoring 55% of the time with less than 2 outs per AB, we can probably assume he scored like 75-80% of the time before all 3 outs occur.
(.45)^2 (2 ABs) is 80% of the time they get the guy home. 3 ABs is 91.1% of the time but then with 2 outs I'm sure the odds go down for that AB significantly. You got double plays, forceouts that change the dynamics of the inning.
So even though I couldn't answer it, probably in the 75%-80%ish range. [/SIZE]
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