| | I've never seen or heard this discussed before, but when it hit me, it seemed so obvious. Imagine that you're the 3rd base coach in the following two situations, both early in a game:
1. Player A is on 3rd, with 1 out. Player B hits a fly ball to left field which is about to be caught. You believe there is probability p that if player A tries to score, he'll make it safely.
2. Player A is on 2nd, with 2 outs. Player B lines a single into left field. You believe that there is probability p that if player A tries to score, he'll make it safely.
The question is: in order for you to send the runner home, does p have to be equally high in both situations? That is, should you be equally aggressive in sending the runner in the two situations? The answer is NO! You should be less aggressive in situation 2. Why? It's actually pretty simple, once you frame it the right way.
If you send the runner, there are two possible outcomes. Either he scores and the inning continues, or he's out and the inning ends. If you do not send the runner, he remains on 3rd and the inning continues. In scenario 2, you get an additional benefit from the inning continuing: you have another runner on base! This increases the average number of runs that you'd score in the rest of the inning. The safe option of holding the runner is now more desirable than it was in scenario 1. If you're not satisfied with the basic intuitive explanation, here's the math behind it.
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If you send the runner in scenario 1, you get (on average) 1+X runs for the rest of the inning if you're successful, and 0 if you're unsuccessful. X is the average number of runs you'd score starting with 2 outs and nobody on. If you do not send the runner, you get (on average) Y runs, where Y is the average number of runs you'd score starting with 2 outs and a runner on 3rd.
If you send the runner in scenario 2, you get (on average) 1+X' runs for the rest of the inning if you're successful, and 0 if you're unsuccessful. X' is the average number of runs you'd score starting with 2 outs and a runner on 1st. If you do not send the runner, you get (on average) Y' runs, where Y' is the average number of runs you'd score starting with 2 outs and runners on 1st and 3rd.
So, in scenario 1, you have the following expected #s of runs (p is the probability of success if you send the runner):
Send runner: p(1+X) + (1-p)0 = p(1+X)
Don't send: Y
In scenario 2:
Send runner: p(1+X') + (1-p)0 = p(1+X')
Don't send: Y'
So, the benefit of sending the runner in scenario 1 is p(1+X)-Y, and the benefit in scenario 2 is p(1+X')-Y'. What we want to know is whether or not the two benefits are equal. If we set p(1+X)-Y = p(1+X')-Y' and do some algebra, it works out to:
(Y'-Y) = p(X'-X).
If the left side is greater, that would mean that the benefit of sending the runner is higher in situation 1. But look back at what X, X', Y, and Y' are. There's no reason that Y'-Y and X'-X should be significantly different. Both of them roughly represent the probability of a runner scoring from 1st with 2 outs. However, when p is close to 1, the decision is very easy, you send the runner. It's only a tough decision for middle values of p, where the values of sending & not sending are close. But for middle values of p, it's very obvious that the above requirement is not satisfied. The left side is significantly greater, and thus the benefit of sending the runner home when there's nobody else on base is much, much higher.
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The lesson here? If there are two outs and you have a trailing runner on base, or especially two trailing runners, you should be MUCH LESS WILLING to send runners home from third. I have never seen this addressed anywhere, and I strongly suspect that major league 3rd base coaches are largely unaware of it. I have (many times) seen a 2-out single with runners on 1st and 2nd, only to witness the 3rd base coach foolishly send the lead runner home with a low chance of success, rather than let the next batter hit with the bases loaded.
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| | Posted 4/8/2009 4:33 PM - 7 Views - 0 eProps - 0 comments
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