Game theory and baseball, part 2: introduction to pitch selection

This article will be about pitch selection, which is probably the most important use of game theory in baseball. I will spend several articles discussing different ways of modeling pitch selection, but for today I will only consider pitch selection as a simultaneous move game between the batter and pitcher.

Obviously doing this analysis requires extreme oversimplification, but even this basic framework will suggest that pitchers and hitters are not maximizing their chance of winning. Using this framework may very well be able to provide teams with a solid opportunity to improve their odds of winning. See the previous article in this series for some definitions and concepts that I will use below.

If you ever hear a commentator talk about knowing what pitch should be thrown in any given situation, he is probably misinformed. If batters knew what pitches were coming in any given count, they would change their strategies. If it’s easy enough for the commentator to guess, then it’s easy enough for opponents to guess. This is especially true in two-strike counts where you often hear commentators proudly declare that the pitcher should have thrown the ball out of the strike zone to get the batter to chase. If pitchers never threw into the strike zone in two-strike counts, batters would react by simply taking the free ball and improving the count in their favor. Pitchers do sometimes throw strikes in these counts, and that is why hitters swing at balls in the dirt.

In reality, what we usually observe when watching games is an equilibrium outcome. Pitchers and hitters randomize their approaches to given situations to be less predictable, and we only see one such outcome during each pitch.

Dominant Strategies

Of course, sometimes there are dominant strategies. Take an example of a 3-0 count where the batter is a relatively average hitter. The pitcher has a pretty good chance of throwing a strike if he intends to, but it is not guaranteed. In 2012, batters making contact on 3-0 counts actually had wOBAs of just .390. When the count reached 3-1, batters collectively had a wOBA of .417 for the rest of the count. Of course, there are some selection biases present in these numbers, but it suggests that batters may be better off taking 3-0 pitches. This simplified example is an easy introduction to pitch selection.

We can rank the four outcomes from the hitter’s perspective, knowing that the pitcher’s perspective would obviously rank them in reverse order. The hitter’s preferences are:

(a) Ball/Take
(b) Strike/Take
(c) Strike/Swing
(d) Ball/Swing.

In this example, we can just call the payoffs to the hitter 4, 3, 2, and 1 for (a)-(d) above, and call the pitcher’s payoffs the negatives of these numbers.

Table 1A

Pitcher\Batter Swing Take
Strike -2,2 -3,3
Ball -1,1 -4,4

We can solve the equilibrium by finding the best response of each player, conditional on each of the other player’s strategies:

(a) When the pitcher chooses to throw a strike, the batter gets a payoff of 3 by taking, which is greater than his payoff of 2 if he chooses to swing. So, we bold and underline the “3” under Swing/Take.
(b) When the pitcher chooses to throw a ball, the hitter gets a payoff of 4 by taking, higher than his payoff of 1 if he swings. We bold and underline the “4” under Ball/Take.
(c) When the batter chooses to swing, the pitcher gets a payoff of -1 by throwing a ball, which is greater than -2 by throwing a strike. So, we bold and underline the “-1” under Ball/Swing.
(d) When the batter chooses to take, the pitcher gets a payoff of -3 by throwing a strike, which is greater than -4 by throwing a ball. So, we bold and underline the “-3” under Strike/Take.

Table 1B

Pitcher\Batter Swing Take
Strike -2,2 -3,3
Ball -1,1 -4,4

The equilibrium is obvious when we realize that the batter has a dominant strategy of taking, so knowing this, the pitcher’s preference is to throw a strike. Getting batters to actually guarantee taking 3-0 pitches in real life, challenging the pitcher to throw a strike, is obviously a tough sell. Effectively, the reason that this equilibrium exists is that the odds are high enough of a ball on a 3-1 or 3-2 count thereafter that it’s not worth letting the pitch move closer to a strikeout on a 3-0 pitch.

Mixed Strategies in Full Counts

There are not many situations above, where a player has a dominant strategy. In most cases, pitch selection and hitter responses will be solved by mixed strategies. As I demonstrated in the previous two articles, a mixed strategy entails a player selecting two (or more) strategies with probabilities between 0 and 1, and the Nash Equilibrium requires the player be indifferent between two (or more) strategies, conditional on his opponent’s (potentially mixed) strategy.

So let’s consider a new situation—a full count. The batter is better off not swinging if the pitcher throws a ball, since he will walk by taking, but will probably be out if he swings at a ball out of the strike zone. If the pitcher throws a strike, the batter is better off trying to make contact, rather than striking out, however. The pitcher naturally prefers the opposite of what the better prefers.

Let’s simplify the payoffs such that the payoff is equally good for the hitter when the outcome is strike/swing and when the outcome is ball/take. Similarly, the hitter is equally worse off when the outcome is ball/swing and when the outcome is strike/take.

Here is the normal form of this game:

Table 2

Pitcher\Batter Swing Take
Strike -1,1 1,-1
Ball 1,-1 -1,1

As we discussed in the last article, it is not hard to notice that no one has a dominant strategy here.

To solve the game, we consider the batter’s value from swinging and not swinging, conditional on a pitcher’s given strategy “p,” the probability of throwing a strike. The hitter’s value from swinging will be his payoff from Strike/Swing (1) times the probability of this outcome occurring if he swings (p), added to the product of his payoff from Ball/Swing (-1) times the probability of this outcome occurring if he swings (1-p). Therefore, we define the hitter’s value from swinging as:

V(s) = (1)*(p) + (-1)*(1-p) = 2p – 1

The batter’s value from taking is:

V(t) = (-1)*(p) + (1)*(1-p) = 1 – 2p

Conditional on the batter’s strategy “q,” his probability of swinging, the pitcher’s value from throwing a strike (we’ll call this V(g) to differentiate it from the hitter’s value V(s) above) is:

V(g) = (-1)*(q) + (1)*(1-q ) = 1 – 2q

The pitcher’s value of throwing a ball when the batter swings with probability “q” is:

V(b) = (1)*(q) + (-1)*(1-q) = 2q – 1

The batter will always prefer to swing if 2p – 1 > 1 – 2p (i.e. when p > 0.5), so in this case, his strategy is to always swing (i.e. set q = 1). Similarly, the batter will always take when p < 0.5, so his strategy would be to always take (i.e. set q= 0). However, the pitcher will always prefer to throw a strike when 1 – 2q > 2q – 1 (i.e. when q < 0.5), so in this case, his strategy is to always throw a strike (i.e. set p = 1). Similarly, the pitcher will always throw a ball when q > 0.5, so his strategy would be to throw a ball (i.e. set p = 0).

Taking these together:

(a) When p<0.5, the hitter will set q=0; but when q=0, the batter would set p=1. Since 1 is not less than 0.5, we know that there are no equilibria where p<0.5.
(b) Equivalently, if we set p>0.5, the hitter will set q=1; but when q=1, the pitcher would set p=0. Since 0 is not greater than 0.5, we always know there are no equilibria where p>0.5.
(c) However, when p=0.5, the pitcher is equally happy with any “q.”

So all equilibria will require p=0.5.

It’s not hard to see where this is going.
(a) When q<0.5, the batter would set p=0; but when p=0, the pitcher would set q=1, so there are no strategies where q<0.5.
(b) When q>0.5, the batter would set p=1, but when p=1, the pitcher would set q=0, so, there are no strategies where q>0.5.
(c) Therefore, the only possibility is for q=0.5, in which the hitter is equally happy with all “p”.

Therefore, the only Nash Equilibrium that exists is when p=0.5 and q=0.5. In words, the only possible outcome is for the pitcher to throw strikes in 50% of these full count situations, and the batter to swing in 50% of these situations. The expected values for the hitters and pitchers are both equal to 0 when this happens. No one can pick a strategy that increases their expected value above 0.

Mixed Strategies in 2-2 Counts

We do not know the exact payoffs for any hitter or pitcher above, so we cannot say that the exact probabilities that they should choose are 50% each. However, we can compare situations to figure out if the theory matches real baseball strategies, so let’s consider a 2-2 count. We will figure out what the optimal strategies would be in this situation and compare them to the 3-2 count. Are the probabilities of throwing strikes higher or lower, and are the probabilities of swinging higher or lower? Do these compare well with what we observe in real baseball?

We can pretty easily construct payoffs that are consistent with the ones above. We know that the payoffs of swinging at a strike on a 2-2 count should be the same as swinging at a strike on a 3-2 count. Either you strike out or you make contact, and whether there were two or three balls at the time won’t matter.

We also know that the payoffs of swinging at a ball in a 2-2 and 3-2 count should be the same. Taking strike three with a 2-2 count and a 3-2 count obviously are equivalent two, since both are strikeouts. The only difference is taking a ball, since the payoff in that case will be a 3-2 count—which means the expected value should be the 0 expected value of the equilibrium above.

Here is the normal form of the 2-2 count game then:

Table 3

Pitcher\Batter Swing Take
Strike -1,1 1,-1
Ball 1,-1 0,0

We can see that there are no pure strategy equilibria in this game as well. We can compute the value functions in each of these cases. The batter’s value of swinging is:

V(s) = (1)*(p) + (-1)*(1-p) = 2p – 1

The batter’s value of taking is:

V(t) = (-1)*(p) + (0)*(1-p) = -p

The pitcher’s value of throwing a strike is:

V(g) = (-1)*(q) + (1)*(1-q) = 1 – 2q

The pitcher’s value of throwing a ball is:

V(b) = (1)*(q) + (0)*(1-q) = q

The pitcher will prefer to throw a strike when 1 – 2q > q, which is when q < (1/3). The pitcher will prefer to throw a ball when 1 – 2q < q, which is when q > (1/3). The pitcher is indifferent when q = (1/3).

The batter will prefer to swing when 2p – 1 > p, which is when p > (1/3). The batter will prefer to take when 2p – 1 < p, which is when p < (1/3). The batter is indifferent when p = (1/3). The only equilibrium is going to occur when p = (1/3) and q = (1/3). In words, that means that the batter is swinging at a third of 2-2 pitches (less than he swings in 3-2 counts), and the pitcher will throw a strike a third of the time on 2-2 pitches (less than he throws over the plate in 3-2 counts). The pitcher’s decision to throw fewer strikes on 2-2 counts does seem to follow what we observe in real life, but I was surprised that the hitter should swing less often in 2-2 counts. It seems to me that batters smell a free base and take more pitches in 3-2 counts than 2-2. This framework strongly suggests that batters are not behaving optimally. They should be expanding the zone more on 3-2 pitches due to the pitcher’s increased willingness to throw a pitch that finds the corner, while they should be skeptical of pitchers with a ball to spare on 2-2 counts. In the next article, I will begin to explore what happens when we adjust for quality of pitchers, and consider whether pitchers with extraordinary skills should behave differently. The results may surprise you.


Matt writes for FanGraphs and The Hardball Times, and models arbitration salaries for MLB Trade Rumors. Follow him on Twitter @Matt_Swa.
3 Comments
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philosofool
11 years ago

It looks like you’re using ordinal scales utilities to generate mixed strategies, but how does that makes sense? Ordinal scales will reveal dominant strategies and saddle points, but they won’t help generate mixed strategies.

Matt Swartz
11 years ago

This is correct and it is why I said:
“We do not know the exact payoffs for any hitter or pitcher above, so we cannot say that the exact probabilities that they should choose are 50% each. However, we can compare situations to figure out if the theory matches real baseball strategies, so let’s consider a 2-2 count.”

In other words, we can compare 2-2 to 3-2 directionally but not compute the exact probabilities. When the ball/take box is less costly to the pitcher (less beneficial to the hitter), then the same point holds. The pitcher will have less incentive to throw a strike, so to find a mixed strategy that keeps the pitcher indifferent between throwing a strike and throwing a ball, the batter should swing less to lower the cost of throwing strikes in 2-2 counts.

Keep this in mind for the next few articles too: these are all ordinal, not cardinal, utilities, which means that we can only establish directional differences. We’d need actual run values for each outcome combined with many other data before we could give an actual suggestion for optimal probabilities. All we can do in these articles is figure out if batters and pitchers are making decisions that are consistent or inconsistent with good strategy. It appears to me they aren’t.

James macKay
8 years ago

in ” Beyond Numeracy” pg 92, J. A. Paulos discusses a game theory solution to a 3 x 3 matrix where the pitcher can throw three different pitchers and the batter can prepare to hit the same three. The batters averages for all 9 situations are known. The best solution is for the pitcher to never throw fastballs and the batter to prepare to hit mostly fastballs. Have you seen this solution. I’m trying with no avail to come up with the same numbers. I can get the .240 and the 60/40 for the pitcher using solver but I see indifference between fastballs and screwballs for the batter where he can select any percent from 0 to 100 for either and still hit .240. Data as follows:
no saddle point
A1 A2 A3
batter prepares for a:
curveball fastball screwball
curveball 0.400 0.300 0.000
fastball 0.200 0.400 0.300
screwball 0.000 0.200 0.400
solution: pitcher throws 60% screwballs and 40% fastballs. batter prepares for 80% fastballs and 20% screwballs. results .240 batting average

Jim