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Old 04-16-2014, 06:23 PM   #1
Mike Lowe
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Baseball Stat Nerds: Question

I'm a baseball nerd myself...relax!

So the common knowledge is that with the bases loaded and nobody out, you try and turn the double play versus getting the out at home.

Now, removing the variables like "what if the guy beats the throw?" or "misses the tag" etc., isn't this flawed logic according to simple mathematics?

While 2 outs are obviously better than one, you wouldn't take the 2 outs in a tie game in the 9th or perhaps even sooner, so how does probability change? Of course, it doesn't.

A successful hitter even with RISP position is still well below .400 (guestimate), and isn't that still better odds than giving up the run for 2 outs?

I guess I could lay this out a bit better, but I'm not trying to prove a point...just curious as to whether the "common knowledge" is flawed.

Thanks, dudes!
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Old 04-16-2014, 06:30 PM   #2
Sun Tzu
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This is less statistical, more situational strategy. The decision to throw home rather than step on third/throw to second and then turn two is entirely dependent on the score of the game, and, sometimes, the inning.

That's it, really. If it's a 1 run game, you go home, no questions asked. It's better to get 1 out and allow no runs (and maintain the lead) than it is to get two outs and lose the lead.
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Last edited by Sun Tzu : 04-16-2014 at 06:32 PM.
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Old 04-16-2014, 06:33 PM   #3
Mike Lowe
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Yeah, I get the "reasoning" behind it all--having coached for a number of years myself.

It just seems in the similar discussion as one of those silly old notions of small ball, just on a less noticeable scale.

I guess I'm just curious if there's anything out there in regards to the actual facts/numbers behind it. It was tough to Google search!

Thanks ST!
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Old 04-16-2014, 06:44 PM   #4
oykib
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There's also the fact that you generally have to bring the infield in to play to prevent the run. That drastically raises the hitter's chance of getting a hit. Shifting your normal infield to double play depth only nets a mild change in the hitter's chances.
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Old 04-16-2014, 06:45 PM   #5
Shkspr
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Join Date: Nov 2000
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Using data from the 2013 season:

The expected number of runs that would be scored with the bases loaded and no outs is 2.225. By taking the force at home, you change the expected number of runs that would be scored down to 1.526. If, however, you can turn a double play and move from having the bases loaded to a runner on third with two outs, you end up giving a run up but also lowering the expectations beyond that run to .353 additional runs.

The value lies in limiting the damage to a single run instead of the very real possibility of a hit plating more than a single run. The calculations change when, late in the game, you want to minimize not the number of runs that score, but the chances of even a single run scoring, which explains the difference in strategy.
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Old 04-16-2014, 07:08 PM   #6
Sun Tzu
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Ah, got it! No worries, man.
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Old 04-16-2014, 08:47 PM   #7
TroyF
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Someone mentioned above all of the variables. This is where having smart defensive players comes in. A ball is hit on a rope to the third baseman? Throw it home and prevent the run. The 8th place hitter hits a routine grounder to SS, a slow catcher is on third and the pitcher due up next? throw it home and get the out.

Situation is everything on that play. 90% of the time, the best play is to get the double play and concede the run.
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