A cardinal rule of base-running is hilariously incorrect

Sometimes I watch Major League Baseball games. I’ve noticed that the following is never questioned:

Unwritten Rule: With fewer than 2 outs, a base-runner on first base shall be certain that a ball hit in the air will fall in for a hit before straying too far away from first base.

This is a natural consequence of a written rule allowing for double plays when the ball is caught and the base-runner doesn’t  “tag up” before advancing.

So…the purpose of this post is to identify one situation where the unwritten rule is wrong—so wrong that it costs the game.

One runner, on first base with 1 out

Imagine there is a runner on first base with 1 out. It’s a tie game in the bottom of the ninth inning and the outfielders are playing back to prevent a double. The ball is hit in the air, but off the end of the bat—a blooper into right field. The right fielder runs; no one can tell if he’ll get there in time to catch it, but if he does the base-runner better get back to first base to prevent a double play. If the ball falls in for a hit there will only be 1 out but the runner only makes it to second base, since he was standing and waiting to see if the outfielder would catch it.

Should the base-runner stand and wait to see if the ball is caught?

The answer is “no”, and it’s not even close. This is the risk/reward analysis that is apparently not understood. But it’s a really simple argument. When the ball is hit in the air, let’s say the runner makes a snap judgment that it’s “better than even” that the blooper will fall in for a hit. In other words, let p be the probability that the blooper will be a hit and the runner quickly decides p>0.5.  Our analysis involves two possible base-runners, A and B. Player A (A for “aggressive”) breaks protocol and just runs without waiting to see if the ball is caught. His aggressiveness either gets him to third base or else gets him thrown out. Player B (for “baseball professional”) follows protocol, and waits.

Who wins more often, A or B?

Remember it’s tied in the bottom of the ninth, so one run wins the game. If the probability of scoring from third base with 1 out is T (T for “third”) then the product pT is the probability that player A scores. This product captures the situation that the blooper falls in for a hit, and then somehow the base-runner eventually scores. (We don’t care how this happens, we just assume it happens with probability T). He won’t score if the blooper is caught since he’ll be doubled up and the inning will be over.

For player B there are two cases. The first case is that the blooper is a hit, putting him at second base. He could eventually score from second—again we don’t care how he might score from second, but we assume he does so with probability S. The second case is that B could score if the blooper is caught (although it would be much harder since there would be two outs and he would be at first base). His probability of scoring is therefore pS+(1-p)F where F is the probability of eventually scoring from first base with TWO outs.  So what’s bigger pT or pS+(1-p)F?

The punchline

The quantities T,S and F above can all be calculated! (We know how likely it is that a runner will score from third base with 1 out, etc). The short answer is that

pT>pS+(1-p)F whenever p>0.38.

So if you think there is a 38% chance or better that the blooper will be a hit, RUN!

Many more details are available from our preprint:


Rally killers and Mariano Rivera

Here is the general justification for running like base-runner B (from  http://baseball.isport.com/baseball-guides/cardinal-rules-of-base-running):

“There’s no worse rally killer than a double play. Moreover, there are few plays more devastating for an offense than a line drive with runners on that turns into a double play. To avoid such a huge momentum shift, you need to make it a priority not to get doubled off on the bases.”

Well I don’t think it’s much of a rally if there are 2 outs and only one runner, on first base. In other words the rally is on life-support as soon as the blooper is caught. Furthermore, if it’s a tie game in the bottom of the ninth, you just need one run, not a rally.  In http://arxiv.org/pdf/1505.00456.pdf we consider “high-leverage innings” defined as the 8th or 9th inning where the difference in score is at most 1. Between the years of 2003 and 2008, Mariano Rivera pitched in 58 different high-leverage innings where there were 2 outs and one runner, on first. None of those runners scored. Unlike most pitchers (where 38% certainty is enough) one only needed to be 14% sure that the blooper would be a hit before running aggressively when Rivera pitched. This is not just because Rivera’s pitching record is better than other pitchers (although it certainly is!) but more importantly it is because he was only slightly better than average at preventing a runner scoring from third with 1 out. This is the point—you don’t need a hit to score from third with one out. In high-leverage situations, one is often facing the pitchers against whom hits are most difficult to obtain, and these are the situations that determine the outcomes of games.

Why don’t they do it this way?

The simple answer is that there is no direct data to show that player A wins more games, because no one ever does it that way. If a team tried it this way, then data could be collected to bear out the conclusions of our indirect arguments. We continue to wait for someone brave…

UPDATE: I’ve written a new post about this topic, in part because we’ve improved it enough that it ended up published in the April 2016 edition of the journal Math Horizons. The new blog post, and even better, the published paper, do a better job of focussing on the main point. Enjoy!


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One Response to A cardinal rule of base-running is hilariously incorrect

  1. Actually, I think you might have undervalued aggression slightly. Let E be the probability of the home team winning in extra innings. Then in scenario A the probability of a win is pT + (1-p)E, while in scenario B it is pS + (1-p)F + (1-p)(1-F)E. Since (1-p)E > (1-p)(1-F)E, the break-even value of p is actually smaller than what you found.

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