Bernoulli Batters, Markov Coaches — Modeling Games by Plays

by Alex Nelson, 5 July 2015

Abstract. We review probability distributions related to discrete “success-or-fail” trials (“coin flips”), and show how it applies to an individual batter’s plate appearance. We briefly discuss how to improve approximating probabilities for a batter’s success. Finally we introduce Markov models, which lets us model games and make predictions.

Contents

Introduction

We will model baseball as a series of interactions between batters and pitchers. The batter will either hit the ball or strike out, just as a flipping a coin results in either heads or tails. A game boils down to a series of coin flips.

We will iteratively refine the model in question, starting with something comically simple. But we can get arbitrarily complex, taking into factors like the batter’s age and abilities. Eventually, we will produce a model simulating games—this is the so-called “Markov Chain” model.

Again, we assume familiarity with probability theory at the level of my notes. The code related to this post can be found on github (checkout version v0.2.0).

Mathematics First: Bernoulli and Geometric Distributions

Puzzle 1.

Given a coin which lands on heads with probability p, what is it expected to land on? What is its variance (standard deviation squared)?

Solution. Well, lets denote the outcome of 1 for landing on heads, and 0 for tails. Then the expected result of flipping a coin would be $p(1) + (1-p)0 = p$. Does this look odd? Well, if p were small, meaning the coin is loaded to always land tails most of the time, then we should expect the outcome to be tails (i.e., closer to zero). If p were large, then the expected outcome should be close to one (i.e., more likely heads). So the expected outcome should be directly proportional to p.

For the variance, there are two clever ways to compute it. One is to consider the expected value of the outcome-squared minus the square of the expected outcome: $v(X) = E(X^{2})-E(X)^{2}$. For the coin, $E(X^2)=E(X)$, so we get the variance be $p-p^{2}=p(1-p)$. This is one valid proof.

The other is to consider $E[(X-E(X))^2]$ where I use square brackets for syntactic clarity. We know the expected outcome of a coin flip $E(X)=p$, so we have $E[(X-E(X))^2]=E[(X-p)^2]$. The computation boils down to similar considerations as before. (End of Puzzle)

Definition. The probability described by flipping a biased coin is called a “Bernoulli Distribution”. (End of Definition)

Puzzle 2.

Consider a coin that will be heads with probability p, and tails with probability 1 - p. What’s the probability of getting exactly 3 heads in 5 flips? Equivalently, this is the probability of flipping exactly 2 tails in 5 flips.

Notation. We write a string to indicate the “history” of the coin flips, read from left to right. So “HT” is “First I got a heads, then I got a tails”. The history “THT” is “First a tails, then a heads, then a tails”. (End of Notation)

Solution. We can draw a probability tree, then find the paths describing the desired states. But that’s hard to do in HTML, so I will leave it for the reader!

We can also use a combinatorial identity, since each flip is independent of each other. We are considering permutations of the string “HHHTT”, since these are all the possible ways to get exactly 3 heads out of 5 flips. How many ways can we permute this? We can recall the number of ways to choose k items from n possibilities is the binomial coefficient. The permutations of “HHHTT” are precisely choosing 3 heads of 5 flips, or “5-choose-3”.

We then have to note each permutation has the same probability, since each flip is independent. The probability we flip HTHTH is the same as HHHTT: $p^{3}(1-p)^{2}$ (the cube of the probability of heads, times the square of the probability for tails). Combining it all together, the probability of 3 heads given 5 flips is precisely

This is the solution we sought. (End of puzzle)

Definition. The probability described by flipping a varying number of k heads out of a fixed number of n flips with the probability p of getting a heads is the “Binomial Distribution”. Observe when n = 1, we recover the Bernoulli distribution.

Puzzle 3.

If our loaded coin has probability p of landing on heads, how many times do we have to flip before we get a heads?

Solution. We have some self-similarity here, because each trial (coin flip) is independent of each other. So the first flip is either a heads (and we’re done), or a tails (and we need to keep going). Generating the list of outcomes, we have H, TH, TTH, TTTH, TTTTH, etc. The probability of heads on the k-th flip would be

The expected outcome would be

If we remember our calculus, this looks like the derivative of a geometric series. This tells us

The smaller the probability of getting a heads p, the longer it will take. This makes intuitive sense, and concludes the puzzle. (End of Puzzle)

Definition. The probability described by flipping a biased coin until we succeed is the “Geometric Distribution”.

Puzzle 4: Coupon Collector’s Problem.

Wheaties has a sport’s card in their cereal box. If there are n different sports cards, how long will it take to collect all of them?

Sketch of Solution. We need to rephrase the problem to take advantage of geometric distributions. The probability of getting a new coupon given j coupons is $p_{j+1} = (n-j)/n$, which is a geometric distribution. We can then take the sum of the expected number of trials to succeed in getting j=0, …, n coupons. (End of sketch)

Puzzle 5: Expected Number of Flips.

Suppose we flip a coin until we get 3 tails. How many heads should we expect to flip? Assume the coin is heads with probability p.

(This puzzle is important because, well, 3 tails and you’re out. We can model a half-inning using this puzzle!)

Solution. We see the probability of flipping k heads for this situation would be

$$\Pr(k) = {k+3\choose k}p^{k}(1-p)^{3}.$$

The expected number of heads would simply be

$$E[H] = \sum_{k=0}k\Pr(k) = \sum_{k=0}k{k+3\choose k}p^{k}(1-p)^{3}.$$

We can note

$$\sum_{k=0}k{k+3\choose k}p^{k}(1-p)^{3} = \frac{(1-p)^{3}p}{3!}\frac{\mathrm{d}^{4}}{\mathrm{d}p^{4}}\sum_{k=0}p^{k+3}$$

where the 3! in the denominator comes from the binomial coefficients. Using the geometric series, we find

$$E[H] = \frac{(1-p)^{3}p}{6}\frac{24}{(1-p)^{5}}=\frac{4p}{(1-p)^{2}}.$$

If p = 0.3, for example, we would expect there to be about 2.44898 heads. (End of Puzzle)

Definition. If we demand there are exactly k failures when considering a Bernoulli trial (“flipping a coin”), while allowing a variable number of successes, then the resulting probability distribution is called a “negative Bernoulli distribution”. (End of Definition)

Folklore (Everything is a Coin Flip).

In probability, we can consider various different distributions. But everything boils down to flipping coins at the end of the day. So, really learn to love flipping coins, because that’s all a probability distribution is in some appropriate sense. (End of Folklore)

First Model: Ignore the Pitcher

Suppose we wanted to model a half-inning, i.e., the period of an inning where one team is “at bat”. The rules state the team keeps sending batters to the plate until they receive 3 outs. What does this mean? We keep performing “trials” until we get “3 failures”.

Lesson: Simplify. One lesson we must emphasize for pure mathematicians is for the first model, always make it simple. Why not treat the batter hitting the ball as a coin flip? Intentionally ignore other factors, and just use some statistic which approximates “how often the batter hits the ball”. We can successively refine this model. (End of lesson)

Simplification: We suppose the probability for a “trial” to “succeed” (i.e., for the batter to hit the ball, and get on a base safely) is independent of the pitcher. We also assume it is independent of the batter. It depends only on the team. (End of Simplification)

This simplification is wrong, but helpful in simplifying a lot of calculations.

Now, we will suppose the batter succeeds with probability p. How can we approximate p? One crude method is

$$ p \approx \frac{(\mbox{Hits})+(\mbox{Walks})}{\mbox{Plate Appearances}}$$

For us, we will perform computations supposing $p=7/20$ (or 0.35).

Observation. From Puzzle 5, we should expect there to be roughly 3.31 successes (well, exactly 560/169, but who’s counting?). (End of Observation)

Scoring a Run

We make a further approximation, based on Pankin’s research from the early ’90s, we have the following aggressive behavior for runners:

We can get arbitrarily technical, but this is good enough for now.

Example (Dummy Data).

Lets consider the fictional probability for the following events:

Event Probability Pr(Event | Success)
Walked by Pitcher 7.8% 39/175 = 22.2857%
Hit a Single 21.2% 106/175 = 60.5714%
Hit a Double 3.3% 33/250 = 9.42857%
Hit a Triple 0.7% 7/250 = 2.8%
Homerun 2.0% 2/25 = 8%

Sanity Check. Observe when we add up the probabilities, we find the probabilities sum to 35% or 7/20, i.e., the probability of success. This is good, because it means the dummy data is consistent with our assumptions so far. Also observe, since one of these events describes a success, the conditional probability Pr(Event | Success) = Pr(Event)/Pr(Success). (End of Sanity Check)

The problem statement now: given 3 successes (that’s the expected number for our model), what’s the expected number of runs scored?

Although it sounds simple, this is not quite so. Why? Order matters: two singles followed by a triple is not the same as a triple followed by two singles. The first case results in 2 runs, the second case produces a single run.

There are 53 = 125 different configurations, producing different outcomes. The code for producing the expected number of runs for a given sequence of plays is available online, but we’ll produce a summary table for the case of n* plays (we want _n = 3, of course):

Number of Successful Plays n Expected number of Runs
0 0
1 0.057142857
2 0.2670204
3 1.090936
4 2.090936

If this is the case, we would expect the number of runs at the end of the game to be 9×1.090936 = 9.818424. But 9 runs per game is, well, astoundingly large! What gives?

The answer is unsurprising: the data we were working with is too generous. A more accurate table would be (drawing upon the average values from 2014):

Event Probability Pr(Event | Success)
Walked by Pitcher 8.6% 86/315 = 27.30159%
Hit a Single 15.7% 157/315 = 49.84127%
Hit a Double 4.4% 44/315 = 13.968255%
Hit a Triple 0.5% 1/63 = 1.5873071%
Homerun 2.3% 23/315 = 7.3015876%

But if we naively plug this into the program, we would get 1.17 runs per half-inning…still too large! We need to also observe the probability of success is now 31.5%, which means we’d expect 2.685279 successes per half-inning (which amounts to 0.87352395 runs per half-inning, or 7.86 runs per team in one game—still too large, but at least we’re approaching the average number of runs per game).

Inadequacies of Ignorance

We made a number of simplifying assumptions for this first model. But this model demonstrates the basic nuts and bolts of what lays ahead. Improving prediction amounts to optimizing the computation for the probability of a play’s success, and moreover the probability for each outcome.

We never took into account anything unique about the batter. Put another way, we assumed each batter was identical on a given team. So what? The number of hits per game would be the same, in this instance. To get a more realistic model, we need to modify the probability of success to vary. Factors we could incorporate may include the home advantage, the pitcher’s abilities, the individual batter’s abilities, etc.

As I said, refining this model amounts to making more accurate estimates of p, the probability of success when the batter’s at plate. Just to stress again, the basic model where we consider different histories, the probability & runs scored for a given history, use the expected number of runs underpins the entire approach. People call this method the “Markov Chain” model.

Improving the Probability by Including the Pitcher

So, our basic model is correct (an inning at least 3 coin flips), but the probability of success crucially depends on the pitcher. We can use a variation of the Log5 model to determine the probability of success from the characteristics of the hitter and the pitcher. More precisely, we use the Odds Ratio method, which—to the best of my knowledge—Dan Levitt first applied to baseball batting. (Of course, Bill James invented the Log5 method, but James applied it to whether a team would win a game.)

Basic Example

I’ll borrow the basic examples from elsewhere.

Hitter Statistics. Suppose we have a hitter, Buck Bokai, with the following statistics:

  AB Hits 2B 3B HR BB K AVG
Buck Bokai 3720 1087 251 48 174 441 631 .292

We have the plate appearances (PA) for Bokai be PA = AB + BB = 4161. If we wanted to be more precise, we would have included sacrifice hits (SH + SB) and hit by pitch (HBP)…but these are normally small in comparison to at bats and balls. We can use these statistics to estimate the probabilities

Signal Formula Statistic
Probability of walk HBB = BB/PA 0.10598414
Probability of Single H1 = (Hits - (2B + 3B + HR))/PA 0.14756069
Probability of Double H2 = 2B/PA 0.06032204
Probability of Triple H3 = 3B/PA 0.011535688
Probability of Homerun H4 = HR/PA 0.04181687
Combined Probability Hcomb 0.36721942

The probability of making an out would simply be 1-0.36721942=0.63278055, ignoring rounding errors.

Pitcher Statistics. Now suppose we have a pitcher, Eddie Newsom, with the following statistics:

  IP Hits HR BB K
Eddie Nesom 200 210 15 50 80

So the pitcher threw 200 innings, which would be roughly 600 batters faced. We can be more precise, computing the batters faced as also including those on base as determined by the BB and “Hits” statistic. Then we may compute the probability for the following events

Signal Formula Statistic
Batters Faced BF = (IP×3) + Hits + BB 860
Probability of Homerun P4 = HR/BF 0.017441861
Probability of hit Ph = H/BF 0.23255815
Probability of Walk PBB = BB/BF 0.058139537

We modify the table, to only include a subset of all possible statistics.

League Averages. We also need to compute the league averages for all the various pitcher statistics. Using the 2014 team statistics, we find

League Average Formula Statistic
Batters Faced LABF = IPOuts + H + BB 186456
Walk LABB = BB/LABF 0.075192004
Hit Lh = H/LABF 0.22308213
Homerun LHR = HR/LABF 0.022450337

Remark (Doubles and Triples). We should, ideally, expand this table to include the number of doubles and triples. We can compute this using the number of doubles and triples from the batting table. With this data, we can approximate the number of doubles and triples our pitcher (in our case, Eddie Newsom) allowed by multiplying Ph by the ratio of L2B/LABF (the total number of doubles allowed in the league, by the total number of batters faced in the leage) or L3B/LABF, respectively. (End of Remark)

Computing the Probability. Now that we have all these statistics assembled, we can compute the probability the hitter will hit the ball. We multiply the action we’re interested in, say the hitter hitting the ball Hcomb, by the ratio of the pitcher’s corresponding statistic to the league’s average: Ph/Lh. This gives us some pseudo-Bayesian prior on the hitter hitting the ball. We have

Or Pr(H)=0.3950259, slightly higher than Hcomb. (“Does this make sense?” Well, the pitcher is slightly below average, so it stands to reason the batter would perform slightly better.) Observe the basic structure of the probability is

This is a common and recurring form that we use to compute probabilities.

Markov Model

We can use this to figure out the expected number of runs. As this is more a proof-of-concept than a rigorous implementation, we will simplify various aspects. The basic scheme would be to write out a probability tree of all the various “histories”, then add up the product of the probability of a given history times its score (if we think of the score as a 2-dimensional vector space, i.e., we can add scores “component-wise”).

The only problem is this approach suffers from combinatorial explosion, and eventually we’ll have histories which contribute negligibly to the expected runs. We’ll need a “pruning” method to make sure we keep only the “significant” histories.

It turns out the batting order is significant only when predicting the runs per season, and it’s on the order of magnitude of 5.5 runs per season (roughly 1% relative error).

As per the possible combinatorial explosion from the number of histories, we can avoid this. We represent a half-inning by a collection of 21 row vectors, each representing the possible number of runs (from 0 to 20 inclusive). There are 25 possible states, represented by the runners on base (8 possibilities: 0, 1, 2, 3, 12, 13, 23, 123), and the number of outs (0, 1, 2, 3). However, we ignore the runners when there are 3 outs, which gives us 3×8+1=25 possible configurations.

When a player hits the ball, he or she “evolves” the state of the game. We can represent this “evolution” by a transition matrix whose entries $(T)_{ij} = p_{i,j}=\Pr(j|i)$ are the probabilities of evolving from state i to state j.

Computing the geometric series $I+T+T^{2}+T^{3}+\dots$ for the transition matrix $T$ (and identity matrix $I$) amounts to computing $(I-T)^{-1}$. Hence we avoided combinatorial explosion, computations taking longer than the lifetime of the universe, with linear algebra.

There are several not-so-obvious issues with this model: how do we compute the expected number of runs? How do we populate the components of the transition matrix $T$? How do we handle “each player has their own transition matrix”?

Basic Structure of Transition Matrix

So, we will consider the basic structure of the transition matrix. The exact details are rather mundane, and can be found in the gambletron.batting code.

Whenever discussing the transition matrix, we should be clear about the states. We specifically have 25 states, specifically 1 “absorption state” (∗, 3 outs) which ends the half-inning for the team, and the remaining 24 are described by the ordered pair (runners on base, number of outs).

The transition matrix can be written in block form as

where the $A$, $B$, $C$ matrices are 8×8 block matrices, the $D$, $E$, $F$ are 8×1 column matrices.

The $A$ blocks represent events which do not increase the number of outs, the $B$ blocks are events increasing the number of outs by one, the $C$ block describe events which increases the outs by 2, and the remaining blocks describe events which take us to the “absorption state” where the team has 3 outs.

The first 8 columns indicate the transition to states with 0 outs, the next 8 columns indicate the transition to states with 1 out, the next 8 columns are for transitions to the states with 2 outs, and the last column is for transitions to the state with 3 outs. So the $D_{0}$ vector is the probability to go from zero outs to 3 outs.

Simplification. We assume that the batter’s behaviour does not depend on the state of the game except when making sacrifice bunts/hits. Consequently, $A_{0}=A_{1}=A_{2}$, and the $B$ matrices are similar except $B_{1}$ takes into consideration the sacrifice bunts/hits. (End of Simplification)

We populate the $A$ matrix with the various probabilities:

Where Pr(1B)=(Number of Singles)/PA, etc. The only caveat here is, if we do use exotic figures like grounded into double play (GIDP) and sacrifice hits or bunts, then we need to modify PA to include them. For computational purposes, we just populate the matrix with the number of singles, or doubles, or GIDP, or whatever…then normalize the row (in the sense of making sure each row of the matrix sums to 1). This is important, we want the transition matrix to have probabilities add up to 1.

The $B$ matrix is similarly defined as $B_{0}=\Pr(Out)I_{8}$, simply the diagonal matrix whose entries are the probability of an out. The $B_{1}$ matrix has a perturbation due to the bunts and sacrifice hits, which may be considered negligible if desired.

Remark. Observe nothing we did, so far, cannot be computed for a single batter. That’s the plan: each batter gets their own transition matrix. We can even populate the matrix with relevant probabilities. If we knew the player was on the home-team, we could restrict our probabilities to be conditional $\Pr(-\mid\mbox{Home})$. Similarly, if we knew the weather, we could further refine our probabilities.

We now have to wonder about the ordering of transition matrices, i.e., the batting order. We also have yet to discuss how to get runs from the transition matrix, too. (End of Remark)

Expected Number of Runs

There appears to be two different schemes for computing the expected number of runs. One appears to be a simplified version of the other, so we will divide our attention to examine the simple model, then the complicated one.

Simplified Version

Following Tesar, we consider a half-inning. The first question we ask is “How many batters will come up during the half-inning?” Once we have this, we can compute the expected number of runs “in the obvious way”.

We can use the idea of Absorbing Chains to estimate the number of batters that will come to the plate for a half-inning. We take the 24-by-24 submatrix of the transition matrix (eliminating the 25th column and 25th row):

The “fundamental matrix” is $E=(I-Q)^{-1}$. If we know the state of the game corresponds to row $i$, then the expected number of batters to come to plate before the half-inning is over is $\sum_{j}E_{i,j}$.

So for one half-inning, from start to finish, the expected number of batters to come to plate is given by taking the first row of $E$, then summing up its components. Once we know the expected number of batters to come to plate, we know the expected number of plays. We can compute the expected number of runs given the expected number of plays.

Tesar’s approach is simpler. Construct a vector $\vec{r}$ whose components are the probability in state i a single run will be scored. Given this, Tesar takes the fundamental matrix and transforms this vector to get the expected number of runs given the state of the game. Intuitively this makes sense: $E$ represents how many batters are likely to come up before the half-inning is up, $\vec{r}$ represents the probability a run will be scored for any given batter. Their product $E\vec{r}$ would give us the expected number of runs for the half-inning, where each component describes a different state. The first entry $(E\vec{r})_{0}$ would describe the expected number of runs in a half-inning.

There is a serious error here. With Markov chains, as we are using them (i.e., following Bukiet, Harold, and Palacios—as Tesar follows them, too), the transition matrix acts on row vectors on the right. Tesar should have written either $\vec{r}^{T}E$ or $E^{T}\vec{r}$. The fact Tesar was able to obtain a reasonable result boils down to numerological accident, than careful analysis. (End of discussion on Error)

But, to quote Thucydides, “We bless your simplicity, but do not envy your folly.” This model is too simple. While it works great for estimating the average number of runs per game, it doesn’t seem to adequately capture the situation we’re interested in: who will win a given game, and what will the score be?

Complicated Version

For simplicity, we will suppose our team has 9 batters with transition matrices $T_{1}$, …, $T_{9}$. These are the full 25×25 transition matrices. We will sketch the algorithm presented in Bukiet, Harold, and Palacios.

For a Single Half-Inning.

For a single half-inning, we start with the state vector $u_{(0)}$ which has its only nonzero component be its first entry, which is set to unity. That is, $u_{(0)}=(1,0,\dots,0)$. Observe it is a row vector, and the subscript indicates how many batters have come up so far. We have $u_{(n+1)}=u_{(n)}T_{n}$ where the subscript on the transition matrix is taken modulo the number of players (i.e., it cycles through all the players).

The state vector $u_{n}$ gives us the probability distribution of states after n batters have come to the plate. Consequently, if we add up its components $\|u_{n}\|_{L^{1}}=1$ we get the probabilities must add up to unity.

Are we done yet? No, we need to keep track of how many runs have occurred. What to do? Well, we could just keep track of it mentally. This would be tedious and impossible to automate.

We could use a matrix $U_{(n)}$ to keep track of the number of runs. The rows of the matrix indicate the number of runs that have occurred. For simplicity, the original authors limited focus to a maximum of 20 runs per inning (since the probability more than 20 runs occurring in a single inning is incredibly rare, this is “good enough”), or 21 rows. When a run occurs, we simply move the probability from that component (of the given row) down appropriately one row. When 2 runs occur, we move the component down 2 rows.

To be explicit, we write $T=P_{0}+P_{1}+P_{2}+P_{3}+P_{4}$ where $P_{j}$ is the components describing j runs have occurred. So $P_{4}$ has nonzero components when the bases are loaded, and the batter hits a homerun. (Or, more rarely, if 4 errors have occurred.) For determining if j runs have occurred, we refer the reader to Appendix A of Tesar.

Then we have

If $j-k<0$, we simply use the zero row vector.

Iterating this “many times” (until the probability of 3 outs is, e.g., at least 99%), we get the final matrix $U_{(\infty)}$ whose 25th column gives the probability distribution for the runs that half-inning. We can compute the expected number of runs easily, once given the probability distribution.

For an Entire Game.

Once we have the half-inning done, the rest of the game should be a straightforward generalization: just use 9 times as many rows in these $U_{(n)}$ matrices, right? Each inning consists of a 21-row submatrix, and once a 3-out state has been reached, “move” the contents to the start of the next inning (21 rows down, because we use the rows to keep track of the runs we must move the contents down proportionally). Keep iterating until the probability there are 27 outs is at least 99.9%, and you’ve got the probability distribution for the runs per inning in the 25th column of the resulting matrix.

The expected number of runs can be found by examining the last inning (i.e., the last 21 rows of $U_{(\infty)}$). We obtain the probability distribution for the runs at the end of the game. Again, we suppose there are at most 20 runs per side, which appears valid. The last time a team had more than 20 runs was when the Phillies defeated the Cubs 23-22 in ten innings on May 17, 1979.

Probability of Winning the Game

So, for a given team, the 25th column in the last 21 rows forms a vector denoted $S(\mathrm{Team})$. The probability a team will win is given by the probabilities it scores more runs than the other team:

The probability the game goes to overtime is given by the probability the two teams score the same number of runs

What to do about overtime?

One solution is to ignore it as negligible. (Probably a good idea for a “first pass”.) But one possible solution is to modify “Tennis odds” to extrapolate the runs scored in overtime, which then allows us to modify our equations to include an extra term for the case when the game goes to overtime:

It turns out that it’s just as quick to compute the extra inning’s distribution, which we treat as just one inning from scratch. This is far from accurate, but it works.

Predictions

The part everyone is waiting for, the predictions.

Anaheim Angels vs Texas Rangers

Prediction: .@Angels 3 vs .@Rangers 4 for today's game.

— Alex Nelson (@anelson_unfold) July 5, 2015

The final score was Angels 12 - Rangers 6.

Oakland A’s vs Seattle Mariners

Prediction: .@Athletics 4 vs .@Mariners 2 for today's game.

— Alex Nelson (@anelson_unfold) July 5, 2015

The final score was Oakland 1 - Mariners 2.

LA Dodgers vs NY Mets

.@Dodgers 3 vs .@Mets 3 for today's game, projected to have an extra inning which Dodgers will win due to home advantage.

— Alex Nelson (@anelson_unfold) July 5, 2015

The final score New York Mets 8 - LA Dodgers 0.

Conclusion

We discussed the basic idea behind the Bernoulli trial, applied the concept to individual plays, and found the Markov chain was a natural extension of this idea. We discussed how to improve the probabilities by taking the pitcher into account, as compared to the league average.

We implemented the Markov model, which can be found on github (checkout version v0.2.0) but used naive probabilities for simplicity. We could improve our predictions using various techniques discussed earlier (e.g., take into account the pitcher, or organized the batters into an optimal batting order) or if we had more precise statistics (batting stats for away games, for home games, etc.). This requires another data source, however.

The low hanging fruit for us: picking an optimal batting order. Now that we have a rudimentary Markov chain model, we can easily determine the optimal batting order. But that’s the topic for another time.

References

Batting Order Optimization

Probability Batter will Hit the Ball

Markov Chain Models