Last week, I took a look at defensive shifts and how players could possibly fight back by understanding game theory. We used bunt probability research conducted by Tom Tango to give us a better idea of what types of outcomes could go into our game theory matrix. We then solved for equilibrium to give us a rough estimate of the ratio that batters should bunt or swing away, and the ratio that defenses should shift or not shift. In this post, I will demonstrate a different game theory structure that might be more suited for looking at bunting into the shift.
In the previous post, we used a matrix also known as “normal form” to play the game. This time we will use the extended form also called a “game tree.” This form of game is more conducive to visualizing a sequential game. In a sequential game the first player makes a move and the second player counters. In our example, player one is the defense and player two is the batter.
In some cases, player two does not know what move player one will make, but in our case the batter can easily see if the defense is in an over-shift; therefore we can say that all information is available to each player. This is important since an entirely different strategy would need to be applied if there was an information gap between the two players to begin the game. In many cases, player one would have the advantage by going first, but there is an intriguing fact about our situation between the defense and batter. Player two (the batter) contains all information needed for them to make the best decision possible, but player one is at a disadvantage because they do not simultaneously know the move that player two will make (will he swing away or bunt?). Players who often encounter the shift should take advantage of this. Especially if the defenses are shifting on them 100% of the time.
Let’s take another look at the 2X2 matrix below from the last post. We will use this matrix to transition into our game tree. Remember there are no pure strategies in this game meaning neither team can make a move that will best the other since all information is known. If the defense shifts the batter would lay down a bunt, but if the defense knows the batter will bunt then they would be more likely to not shift.
This matrix can be transposed into extended form (game tree) fairly easily. Take a look at the new structure which should seem logical.
Obviously, with how the payoffs are set up, it would seem that the defense would shift each time and the batter would bunt each time. In truth, the defenses are already shifting nearly every time on certain players. In a sequential game the batter should actually have the advantage, but they aren’t taking full advantage of their move. They are hitting into the shift each time instead of mixing it up, causing the defense to do the same
Now let’s go back to David Ortiz and his probabilities.
Take a look at the same game tree that includes David Ortiz’s probabilities of succeeding and failing for each possible move. For the swing away values, I chose to use the rate at which David Ortiz gets an extra base hit with the shift on and off. One could also use the BABIP of Ortiz with the same stipulations. (References to sites with the probabilities are listed at the end.)
This helps us understand that if the defense shift’s then it is in Ortiz’ best interest to bunt because it is his highest probability of succeeding. Ortiz is a power hitter though. The bigger payoff is located in the swing away branch located on the side where the defense does not shift. Keep this probability tree in the back of your mind as we will reference it later.
The question is how can Ortiz get a few more opportunities to swing away without a shift on? This question parlays into another question: What happens if Ortiz, or any shift prone player for that matter, starts to bunt more and becomes more efficient at doing so? How will defenses respond and how will that change Ortiz’ strategy?
Comparative statics can help us answer this question.
The definition of comparative statics is how a game’s equilibrium behaviors and outcomes change as a function of the game’s parameters. How do the outputs change as a function of the inputs? Let’s go back to our simple matrix, only this time, notice the top right cell contains the X variable. I have also included the probabilities which are the short equations next to each move. If you want to check the math see the breakdown below. (Note, this is assuming a simultaneous type of game. In truth the game should be more complex due to different types of shifts being used. For example if the defense sees the batter square around they might have the third baseman crash thus disrupting the structure of game theory to be applied making in more non-sequential then sequential.)
Solving for the defense:
EUSwingAway = σShift(-1) + (1-σShift)(2)
EUBunt = σShift(X) + (1-σShift)(-1)
σShift(-1) + (1-σShift)(2) = σShift(X) + (1-σShift)(-1)
σShift = 3/X+4
Solving for the batter:
EUShift = σSwingAway(1) + (1-σSwingAway)(-X)
EUNoShift = σSwingAway(-2) + (1-σSwingAway)(1)
σSwingAway(1) + (1-σSwingAway)(-X) = σSwingAway(-2) + (1-σSwingAway)(1)
σSwingAway = (1+X)/(X+4)
Without going into the gory details, comparative statics will show us how the batter should alter his strategy as he becomes more efficient at bunting with the defensive shift on. To calculate the comparative statics for this game, we must first identify the element (cell or box) we want to analyze. In this case we will be analyzing bunting with the shift on. The second step is to take the derivative of that element of interest, and we do this by using the probability which I calculated above. That probability is σBunt = 3/X+4. I’ll spare you the calculus and just cut right to the answer.
Probability of bunting in a shift = 3/(X+4)
Don’t get hung up on the goofy fractions. All we are really looking for is whether the derivative is positive or negative. The derivative we came up with is negative which means the function is always decreasing. A decreasing function for our bunting batter means that as he becomes more effective at bunting with the shift on, he should actually bunt less. This result seems a little strange and almost counterintuitive, but I can assure you it makes perfect sense. Recall that game theory is all about strategy. It is not just the batter’s strategy that matters, but also the defense’s strategy. If the defense knows a batter will always swing away, then they will simply counteract him by shifting. The batter should then compensate for this weakness by bunting more. However, as the batter improves his bunting skills causing his weakness to diminish, there would be no need to compensate for the weakness. This is why he will start bunting less frequently the stronger he becomes at it.
To prove this point let’s refer to our probability tree from earlier and transpose that into normal form (matrix). (Please note, to go back to matrix form and use a mixed strategy approach we are assuming a more simultaneous type of game. Sequential games are usually determined using backward induction.)
Remember the numbers in the cells represent the payoff for each player based on their moves. In this matrix we are using probabilities as the payoffs. Look at the top right cell. The numbers are telling us that David Ortiz has shown us he can get on base at a rate of 55% when he bunts with the shift on meaning the defense got Ortiz out the other 45% of the time. This is based on a small sample size as I explained in post one, but humor me for the sake of explanation. The comparative statics we calculated is telling us that if Ortiz, or any other player for that matter, can get on base by bunting when the shift is on at high rate, then he should actually be bunting less. So then the question becomes how often should Ortiz bunt when encountering the shift? Warning, more boring algebra ahead. Skip it if you want.
Solving for the batter (David Ortiz):
EUShift = σSwingAway(.913) + (1-σSwingAway)(.45)
EUNoShift = σSwingAway(.853) + (1-σSwingAway)(.9)
σSwingAway(.913) + (1-σSwingAway)(.45) = σSwingAway(.853) + (1-σSwingAway)(.9)
σSwingAway = 22/25 or 88%
Despite Ortiz hypothetically succeeding 55% of the time he bunts during a shift, our mixed strategy nash equilibrium calculation shows us that he should actually swing away around 88% of time he sees the shift and only bunt in 12% of those plate appearances. That 12% would become even smaller meaning Ortiz would bunt even less if he increased his rate of bunting success.
Alright, I am finally done rambling on about game theory. I set out to find a way for players to beat the shift and after boiling it all down I came up with a couple of conclusions. At best players could see some brief success by bunting when a shift is on. They might be able to grab a few easy bags, but eventually defenses will catch on and alter their strategy. In reality, both the batter and the defense will make their moves based on their probabilities of success and failure. With that being said, I would tell batters to keep an eye out for teams who do not respect the equilibrium theory suggested. Batters may have a chance to exploit these teams who do not alter their strategy based on previous or current information. .
Reference for the David Ortiz probability values:Webb-Albers Championship Update: EW Interview Commemorative Edition
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