**From time to time, I will be posing a question, mostly something you hear on TV all the time from the talking heads who corrupt our children’s baseball minds. Feel free to comment in the comments section. There are no guarantees that my math is correct, since much of it is done at 2 in morning!**

You often hear from a TV commentator something like this: “They can afford to be more aggressive since they have the lead.”

A typical situation is a 1 or 2 run lead with a stolen base threat on first or a base runner potentially going first to third on a single. The implication is that if you are down, or perhaps if the game is tied, that it is not wise to attempt a “risk,” whereas with a lead, it is acceptable to do so.

The *thinking *(or, often, *non-thinking*) that goes along with that, is, “The team already has the lead, so it is not big deal if they make an out on the bases – they’ll still have the lead. However, if behind, an out on the bases is costly.”

While that is generally true, the missing part of the equation is that if behind, a success can be quite advantageous, even though a failure can be devastating. And just because something “feels” right, doesn’t mean that it is – in fact, strategies which “feel” right are often incorrect and vice versa, because human beings tend to be risk averse. We typically would rather have a guaranteed 10 dollars in our pocket than risk a 50% chance of losing $10 and a 50% chance of winning $40 (which is a net gain of $15).

Anyway, rather than guessing whether it is in fact *more *beneficial to go for a risk when up by a run or two, as opposed to being down by a run or two, let’s see if we can figure it out. It should be easy.

Let’s say that it is the bottom of the 7^{th} and the MGL Braniacs (haha) have a 1 run lead with a runner on 1^{st}and 2 outs. What is the break-even point for a successful steal assuming that everything about this game is major league average? To answer that, all we need are some league average win expectancy charts, which Tango generously provides (http://www.tangotiger.net/welist.html). All you aspiring saberists should bookmark that web address. In this situation, our WE is 77.2% (we don’t need that number for this calculation). On the average, we will win the game 77.2% of the time, assuming league average hitting, pitching, defense, etc., for the remainder of the game. If we execute a successful steal, the WE is now 78.4% and if we are thrown out, the inning is over and our new WE is 75.3%.

First of all the gain in WE from a stolen base is only 1.2% and the loss from a caught stealing is 1.9%. In a second, we’ll see how those numbers compare when we are down by a run. So what must be our success rate for the average result for the average result to be a net gain of zero? That is our *break-even *(BE) point. We must solve this equation for X: 1.2 * X – 1.9 * (1-X) = 0. The answer is X = .613. We have to be successful more than 61.3% of the time for a stolen base attempt to have a positive expectation, i.e., increase our chances of winning. So, it is true that this is a good time to take a risk!

But, before we close the door, let’s see what the break-even point is when we are down by a run. The original “theory” says that this is *not* a good time to take a risk; therefore the BE point must be higher. With 2 outs and a runner on first with a 1 run *deficit*, our WE is 29.6%. After a stolen base, it is 32.2%, a gain of 2.6% (win a 1 run lead, the gain was only 1.2%!). If our runner is erased, the WE falls to 24.6%, a loss of 5%, quite a bit more than our 1.9% loss when we had the lead. So, it is a bigger risk, but there is also a bigger gain, so it is not surprising that managers would be averse to attempting a steal when down by a run, especially with 2 outs (if you fail, the inning is over – bang!). But, all we really care about is the BE point. Again, that is easy to figure: Solving for X in this equation, 2.6 * X – 5 * (1-X) = 0, we get .658. We have to be successful more than 65.8% of the time in order for the SB attempt to have a positive expectation. Before, it was 61.3%. So, it *is* correct that we can be slightly more aggressive in attempting a steal, at least with 2 outs, when we are up by a run rather than down by a run. It’s a small difference, but a difference nonetheless. I’m not sure the narrative (e.g., “They can afford to take a risk now that they are winning.”) is justified, but technically it is correct, at least as compared to being down in the game. In other words, they *should* say something like, “They can afford to take *more* of a risk, now that they are winning.”

What about in a tie game? Conventional TV commentator wisdom (and managers) would put that in the same category as being behind in the game I would think. Maybe not though. Let’s see where it actually stands in terms of the BE point.

The equation in a tie game is 2.5 * X – 4 * (1-X) = 0, which yields a BE point of 61.5%, slightly more “risky” than when up by a run.

Do we get the same pattern with 0 outs? Up by a run, the BE point is 66.2%, down by a run, it is 69.2%, and in a tie game, it is 65.5%. So, in this case, while being down by a run is still a more risky situation to attempt a steal, a tied game is actually *less* risky than when up by a run – by a hair. So, I think we can safely put the tie game situation into the “up by a run” bucket and call it a *low* risk situation for attempting a steal, at least as compared to being down by a run. Again, from an absolute perspective, who would think to call a requisite success rate of 65 or 66% *low risk* and one of 69% *high risk*?

What about with a 3 run lead? Most commentators, I think, would call this a, “What the heck, go for it situation – you have nothing to lose.” If you attempt a steal and get thrown out, you only lose .59%, so, sure, what the heck, who cares? You are going to win the game over 93% of the time either way. If you succeed, you gain only .31%, so, now it’s more like, ”Why try it in the first place? The runner might get hurt.” In any case the BE point is 65.6%, so it is really no more or less “risky” than anytime else unless you want to define risk by the downside. In other words, it would be silly to recklessly attempt a steal when the BE point is a thoroughly typical 66%. If you do attempt a steal with a poor base stealer, you are needlessly giving away win expectancy, albeit a very small amount (and risking an injury). So really, rather than, “What the heck, we have nothing to lose,” the mantra should be, “We have almost nothing to gain, so why risk an injury?”

Let’s try one more thing. What about going from 1^{st} to 3^{rd}, with one out (which is always the number of outs with the lowest BE point when trying for third base – of course)? Is it better to take a risk with a lead or a deficit. We found that with the stolen base (of second), it is slightly better to take the risk with a lead where “risk” is defined by the BE point – the lower the BE point, the lower the risk.

If we go from 1^{st} to 3^{rd} on a single with a 1 run lead and 1 out in the bottom of the 7^{th}, we increase our chances of winning by 5.66%. If we get thrown out at third, it goes down by 2.38%. If we don’t send him, we gain 3.75%. The BE point is therefore 63.8%. (If any third base coaches are reading this, keep that in mind – your runner needs at least a 64% chance of being safe at third for you to send him in this situation!)

Same thing, but down by a run? We pick up 14.86% with a successful stretch, and lose 6.14% when he gets thrown out, and gain 7.46% when we hold the runner on second. So the BE point is 68%.

So, similar to the stolen base, we need a higher success rate when down by a run as compared to leading by a run, but not by *that* much.

In a tie game, the BE point is 63%, the lowest of the 3 run differentials, by a hair, again, similar to the stolen base situation (a tie game is essentially the same as being up by a run).

So there you have it. The first Question of the Day, asked and answered!