# Picking off the runner

You gotta keep him close to first. Otherwise he might as well walk to second. Don’t give him anything for free.

But what’s the best way to do that? How often should one throw over to the first baseman to make sure that he doesn’t steal?

Let’s model this through the payoff to the pitcher and the runner. The runner, obviously, wants to steal second, or advance an extra base on a ball put in play. The pitcher wants the exact opposite: he wants to prevent that from happening. To do so, he will attempt to pick off the runner at first if he gets too greedy.

So let’s now assign values to this, as a function of how big a lead that the runner takes (x), and how often the pitcher throws the ball over to first (t). We say that the runner gets payoff  from gaining an extra base, and does so with probability $F(x)$, with $\frac{\partial F(x)}{\partial x}>0$. Meanwhile, the pitcher gets payoff v from picking off the runner, which occurs with probability $P(x,t)$, where $\frac{\partial P(x,t)}{\partial x} > 0$ and $\frac{\partial P(x,t)}{\partial t} > 0$ for $x>0$, while $P(0,t)=0$. In other words, the probability you get picked off is greater the bigger a lead you take, and the more often the pitcher throws over to first, assuming you’re not actually standing on first. Finally, since you win in baseball if and only if the other team loses, we model this as a zero-sum game.

If this were our entire expression, it would be obvious that the pitcher should throw to first base as much as possible. In other words, he should try to pick off the runner at first at literally every single moment, and ignore the batter at home. So why don’t we see this?

The reason is because the pitcher is only human. Every time he throws the ball to first, there’s a chance that it will get away from him and sail over the first baseman’s head. If so, the runner will get his extra base anyway as the team on the field scrambles to recover the ball. This means that by throwing more frequently, there are more opportunities to screw up. Assuming the probability of throwing the ball away is the same each time, this gives the runner an additional payoff of  given how often the ball is thrown. Thus, our expressions for the payoffs of the runner and the pitcher are:

$\pi_{r}=uF(x)+ut-vP(x,t)$

$\pi+{p}=vP(x,t)-uF(x)-ut$

A Nash equilibrium will occur whenever the pitcher cannot make it better for his team to throw more frequently, and the runner cannot improve his chances anymore of taking an extra base without risking himself too much. Thus, in equilibrium:

$u\frac{\partial F(x)}{\partial x}-\frac{\partial P(x,t)}{\partial x}=0$

$u-v\frac{\partial P(x,t)}{\partial t}=0$

Of course, it’s possible that the pitcher is so good at picking people off that the runner doesn’t dare step off the bag, in which case we have an equilibrium which doesn’t satisfy the above two equations. Alternatively, the runner could be so good that no matter how often the pitcher tries to pick him off, he gets the extra base anyway, in which case the equations don’t hold either. But these cases are not normally found in the big leagues, so we’ll leave it as is.