# Getting Lucky in a Playoff Series

Sports have a constant uncertainty and randomness in every aspect of the game including determining champions. This is one area you wouldn’t expect to have a lot of variability, since you would want the team that has the best roster composition and played the hardest to win the championship. This concept is usually brought up in the arguments against the one-game Wild Card round that MLB introduced in 2012 saying there’s too much that can happen in one game to determine the fate of a season. [The counter-argument to this specifically is the division winners now have a reward for winning the division, besides having cool sweatshirts.]

The Sports Side

The basis for championship series in MLB, NBA, and NHL is the an odd-numbered of games series with the champion being the team that wins the majority of games in those series. Most sports use a 7-game series; so for example the Boston Red Sox had to win 4 games to win the World Series last year. Using an example of randomness I got from Leonard Mlodinow’s The Drunkard’s Walk: How Randomness Rules Our Lives, I can illustrate how a team that’s clearly an underdog can win a playoff series against a superior opponent. Mlodinow has a recorded lecture where he explains what he wrote in his book. [It’s a good book, you should read it.]

Let’s use two teams; one is the Favorite, and it is assumed they will beat the Underdog 55% of the time [given enough games]. This also means that the Underdog will win 45% of the time. This represents win probabilities more uneven that you are likely to find in a playoff game since teams are typically much more evenly matched [at least in baseball]. The last assumption of this example is that the teams win probabilities don’t change with a different starting pitcher or home field/court/ice advantage. These are terrible assumptions if you wanted to project real playoff series, but the underlying principle of random sequencing will still hold true.

In order to win the playoff series, a team has to win a certain number of games before the opponent wins that number. To model this distribution based on pure randomness, you can use the negative binomial distribution to determine the probability that the Favorite will win a 7-game playoff series in 4, 5, 6, or 7 games. If you wanted to design a playoff series to minimize the chances that the underdog will win, you’d want to choose a number of games which would have the smallest probability of the underdog winning the series.

This chart shows the probability for all 8 possible outcomes of a 7-game playoff series based of a 55/45 winning percentage split and pure randomness with no home-field advantage. As you can see there’s a substantial chance [39% probability] that the Underdog will win a 7-game series. 39% is rather large, and this is a 7-game series. Baseball also employs a 5-game series for their division series (LDS) and a one-game playoff for the the Wild Card round (WC). The chances of an upset becomes more likely as the number of games decreases. I’ve also added another set of teams (60/40 split — greater disparity) for comparison’s sake.

It should be obvious that the 1-game series has the greatest chance of an upset, hence the objections to its use in baseball. Though my contention would be that a 3-game series does not offer much more certainty that the best team will win.

The Math Side

I first calculated these probabilities by writing out all the possible combinations then adding up those probabilities. I have since realized there was a much easier way to determine these probabilities, and that is by using the negative binomial distribution (NBD). If you want to familiarize yourself with what the distribution represents please read the count data primer. In short, the NBD will determine the probability that a team will lose a certain number of games [0-3] before the other team wins 4 games. The NBD is defined by the following function:

$P(X=k) = {{r+k-1}\choose{k}} p^{r} (1-p)^{k}$

where X is the random variable whose probability we are calculating, k is number of Team A losses [this will vary], r is the number of Team A wins [for the 7-game series, it will be 4 games], and p is the probability of Team A winning. In the case of this example we will be determining the probability of Team A winning a 7-game series, when Team A has a 55%/45% advantage over Team B.

$P(X=2) = {{4+2-1}\choose{2}} 0.55^{4} (1-0.55)^{2}$

$P(X=2) = {{4+2-1}\choose{2}} 0.55^{4} (1-0.55)^{2}$

$P(X=2) = 10 * 0.0915 * 0.2025 = 18.53\%$

This is the probability for just one possible outcome, Team A wins the series in 6 games. To determine the probability that Team A wins the series, you add the probabilities for each outcome Team A wins in 4, 5, 6, or 7 games. So this calculation then repeated for every loss possibility:

$P(WinningSeries) = P(X=0) + P(X=1) + P(X=2) + P(X=3)$

$P(WinningSeries) = 9.15\% + 16.47\% + 18.53\% + 16.68\% = 60.83\%$

From these calculations, there is a 60.83% chance that the Team A wins just by randomness. Conversely, there is a 39.17% [100% – 60.83%] chance that Team B, the inferior team, wins because of random sequencing.

Conclusion

The MLB Wild Card game rightfully gets criticized for being too susceptible to having a bad day or getting a bad bounce. I wanted to illustrate that any playoff series has a lot of randomness in it. Beyond the numbers, people remember the bad bounces way more than they remember the positive or neutral events that occur [negativity bias]. A bad bounce or a pitcher having a bad day could easily benefit the team you are rooting for. The only real way to root out the randomness you would need to play hundreds of games, and somehow I don’t think that is feasible.

# MLB — Run Distribution Per Game & Per Inning — Negative Binomial

This is an extension of an earlier post I wrote about the runs per inning distribution. In this post I use the negative binomial distribution to better model the how MLB teams score runs in an inning or in a game. I wrote a primer on the math of the different distributions mentioned in the post for reference.

The Baseball Side

A team in the American League will average .4830 runs per inning, but does this mean they will score a run every two innings? This seems intuitive if you apply math from Algebra I [1 run / 2 innings ~ .4830 runs/inning]. However, if you attend a baseball game, the vast majority of innings you’ll watch will be scoreless. This large number of scoreless innings can be described by discrete probability distributions that account for teams scoring none, one, or multiple runs in one inning.

Runs in baseball are considered rare events and count data, so they will follow a discrete probability distribution if they are random. The overall goal of this post is to describe the random process that arises with scoring runs in baseball. Previously, I’ve used the Poisson distribution (PD) to describe the probability of getting a certain number of runs within an inning. The Poisson distribution describes count data like car crashes or earthquakes over a given period of time and defined space. This worked reasonably well to get the general shape of the distribution, but it didn’t capture all the variance that the real data set contained. It predicted fewer scoreless innings and many more 1-run innings than what really occurred. The PD makes an assumption that the mean and variance are equal. In both runs per inning and runs per game, the variance is about twice as much as the mean, so the real data will ‘spread out’ more than a PD predicts.

The graph above shows an example of the application of count data distributions. The actual data is in gray and the Poisson distribution in yellow. It’s not a terrible way to approximate the data or to conceptually understand the randomness behind baseball scoring, but the negative binomial distribution (NBD) works much better. The NBD is also a discrete probability distribution, but it finds the probability of a certain number of failures occurring before a certain number of successes. It would answer the question, what’s the probability that I get 3 TAILS before I get 5 HEADS when I continue to flip a coin. This doesn’t at first intuitively seem like it relates to a baseball game or an inning, but that will be explained later.

From a conceptual stand point, the two distributions are closely related. So if you are trying to describe why 73% of all MLB innings are scoreless to a friend over a beer, either will work. I’ve ploted both distributions for comparison through out the post. The second section of the post will discuss the specific equations and their application to baseball.

Runs per Inning

Because of the difference in rules regarding the designated hitter between the two different leagues there will be a different expected value [average] and variance of runs/inning for each league. I separated the two leagues to get a better fit for the data. Using data from 2011-2013, the American League had an expected value of 0.4830 runs/inning with a 1.0136 variance, while the National League had 0.4468 runs/innings as the expected value with a .9037 variance. [So NL games are shorter and more boring to watch.] Using only the expected value and the variance, the negative binomial distribution [the red line in the graph] approximates the distribution of runs per inning more accurately than the Poisson distribution.

It’s clear that there are a lot of scoreless innings, and very few innings having multiple runs scored. This distribution allows someone to calculate the probability of the likelihood of an MLB team scoring more than 7 runs in an inning or the probability that the home team forces extra innings down by a run in the bottom of the 9th. Using a pitcher’s expected runs/inning, the NBD could be used to approximate the pitcher’s chances of throwing a no-hitter assuming he will pitch for all 9 innings.

Runs Per Game

The NBD and PD can be used to describe the runs scored in a game by a team as well. Once again, I separated the AL and NL, because the AL had an expected run value of 4.4995 runs/game and a 9.9989 variance, and the NL had 4.2577 runs/game expected value and 9.1394 variance. This data is taken from 2008-2013. I used a larger span of years to increase the total number of games.

Even though MLB teams average more than 4 runs in a game, the single most likely run total for one team in a game is actually 3 runs. The negative binomial distribution once again modeled the distribution well, but the Poisson distribution had a terrible fit when compared to the previous graph. Both models, however, underestimate the shut-out rate. A remedy for this is to adjust for zero-inflation. This would increase the likelihood of getting a shut out in the model and adjust the rest of the probabilities accordingly. An inference of needing zero-inflation is that baseball scoring isn’t completely random. A manager is more likely to use his best pitchers to continue a shut out rather than randomly assign pitchers from the bullpen.

Hits Per Inning

It turns out the NBD/PD are useful in many other baseball statistics like hits per inning.

The distribution for hits per inning are slightly similar to runs per inning, except the expected value is higher and the variance is lower. [AL: .9769 hits/inning, 1.2847 variance | NL: .9677 hits/inning, 1.2579 variance (2011-2013)] Since the variance is much closer to the expected value, the hits per inning has more values in the middle and fewer at the extremes than the runs per inning distribution.

I could spend all day finding more applications of the NBD and PD, because there are really a lot of examples within baseball. Understanding how these discrete distributions will help you understand how the game works, and they could be used to model outcomes within baseball.

The Math Side

Hopefully, you skipped down to this section right away if you are curious about the math behind this. I’ve compiled the numbers used in the graphs for the American League above for those curious enough to look at examples of the actual values.

The Poisson distribution is given by the equation:

$P(X = x) = \frac{e^{-\lambda}\lambda^x}{x!}$

There are two parameters for this equation: expected value [$\lambda$] and the number of runs you are looking to calculate the probability for [$x$]. To determine the probability of a team scoring exactly three runs in a game, you would set $x = 3$ and using the AL expected runs per game you’d calculate:

$P(X = x) = \frac{e^{-4.4995}4.4995^3}{3!} = 16.87\%$

This is repeated for the entire set of $x$ = {0, 1, 2, 3, 4, 5, 6, … } to get the Poisson distribution used through out the post.

One of the assumption the PD makes is that mean and the variance are equal. For these examples, this assumption doesn’t hold true, so the empirical data from actual baseball results doesn’t quite fit the PD and is overdispersed. The NBD accounts for the variance by including it in the parameters.

The negative binomial distribution is usually symbolized by the following equation:

$P(X=k) = {{r+k-1}\choose{k}} p^{r} (1-p)^{k}$

where $r$ is the number of successes, $k$ is the number of failures, and $p$ is the probability of success. A key restriction is that a success has to be the last event in the series of successes and failures.

Unfortunately, we don’t have a clear value for $p$ or a clear concept on what will be measured, because the NBD measures the probability of binary, Bernoulli trials. It’s help to view this problem from the vantage point of the fielding team or pitcher, because a SUCCESS will be defined as getting out of the inning or game, and a FAILURE will be allowing 1 run to score. This will conform to the restriction by having a success [getting out of the inning/game] being the ultimate event of the series.

In order to make this work the NBD needs to be parameterized differently, for mean, variance, and number of runs allowed [failures]. The following equations are derived from the mean and variance equations of a negative binomial. $\alpha$ represents the ‘odds in favor‘ of getting out of the inning. And $r$ is the expected value multiplied by the ‘odds in favor’ which will yield a real, non-integer for the number of successes. The NBD can then be written as

$P(X=k) = \frac{\Gamma(k+r)}{\Gamma(k+1)\Gamma(r)} (\frac{\alpha}{1+\alpha})^{r} (\frac{1}{1+\alpha})^{k}$

where

$r = Expected Value * \alpha; \alpha = \frac{Expected Value}{Variance -Expected Value}$

So using the same example as the PD distribution, this would yield:

$r = 4.4995 * 0.8182 = 3.6815 ; \alpha = \frac{4.4995}{9.9989 - 4.4995} = 0.8182$

$P(X=3) = \frac{\Gamma(3+3.6815)}{\Gamma(3+1)\Gamma(3.6815)} (\frac{0.8182}{1+.0.8182})^{3.6815} (\frac{1}{1+0.8182})^{3}$

$= 14.18\%$

The above equations are adapted from this blog about negative binomials and this one about applying the distribution to baseball. The $\Gamma$ function is used in the equation instead of a combination operator because the combination operator, specifically the factorial, can’t handle the non-whole numbers we are using to describe the number of successes, and the gamma function is a continuous function from 0 to infinity.

Conclusion

The negative binomial distribution is really useful in modeling the distribution of discrete count data from baseball for a given inning or game. The most interesting aspect of the NBD is that a success is considered getting out of the inning/game, while a failure would be letting a run score. This is a little counterintuitive if you approach modeling the distribution from the perspective of the batting team. While the NBD has a better fit, the PD has a simpler concept to explain: the count of discrete event over a given period of time, which might make it better to discuss over beers with your friends.

The fit of the NBD suggests that run scoring is a negative binomial process, but inconsistencies especially with shut outs indicate elements of the game aren’t completely random. I’m explaining the underestimation number of shut outs as the increase use of the best relievers in shut out games over other games increasing the total number of shut outs and subsequently decreasing the frequency of other run-total games.

All MLB data is from retrosheet.org. It’s available free of charge from there. So please check it out, because it’s a great data set. If there are any errors or if you have questions, comments, or want to grab a beer to talk about the Poisson distribution please feel free to tweet me @seandolinar.

# Count Data Distribution Primer — Binomial / Negative Binomial / Poisson

Count data is exclusively whole number data where each increment represents one of something. It could be a car accident, a run in baseball, or an insurance claim. The critical thing here is that these are discrete, distinct items. Count data behaves differently than continuous data, and the distribution [frequency of of different values] is different between the two. Random continuous data typically follows the normal distribution, which is the bell curve everyone remembers from high school grade systems. [Which is a really bad way to grade, but I digress.] Count data generally follows the Binomial/Negative Binomial/Poisson distribution depending what context you are viewing the data; all three distributions are mathematically related.

Binomial Distribution:

The binomial distribution (BD) is the collection of probabilities of getting a certain number of successes in a given number of trials specifically measuring Bernoulli trials [a yes/no event similar to a coin flip, but it’s not necessarily 50/50]. My favorite example to understand the binomial distribution is using it to determine the probability that you’d get exactly 5 HEADS if you flipped a coin 10 times [it’s NOT 50%!].

It’s actually 24.61%. The probability of getting heads in any given coin flip is 50%, but over 10 flips, you’ll only get exactly 5 HEADS and 5 TAILS about 25% of the time. The equation below gives the two popular notations for the binomial probability mass function. $n$ is total number of trials. [the graph above used n=10]. $r$ is the number of successes you want to know the probability for. You calculate this function for each number of HEADS [0-10] for $r$ to get the distribution above. $p$ is the simple probability for each event. [$p$ = .5 for the coin flip.]

$P(X=r) = {{n}\choose{r}} p^{r} (1-p)^{n-r} = \frac{n!}{r!(n-r)!} p^{r} (1-p)^{n-r}$

The equation has three parts. The first part is the combination ${{n}\choose{r}}$, which is the number of combinations when you have $n$ total items taken $r$ at a time. Combination disregard order, so the set {1, 4, 9} is the same as {4, 9, 1}. This part of the equation tells you how many possible ways there are to get to a certain outcome since there are many way to get 5 HEADS in 10 tosses. Since ${{10}\choose{5}}$ is larger than any other combination, 5 HEADS will have the largest probability.

There are two more terms in the equation. $p^r$ is joint probability of getting r successes in a particular order, and $(1-p)^{n-r}$ is the corresponding probably of also getting the failures also in a particular order. I find it helpful to conceptualize the equation as having three parts accounting for different things: total combinations of successes and failures, the probabilities of successes, and the probability of failures.

Negative Binomial Distribution:

While there is a good reason for it, the name of the negative binomial distribution (NBD) is confusing. Nothing I will present will involve making anything negative so, let’s just get that out of the way and ignore it. The binomial distribution uses the probability of successes in the total number of ATTEMPTS. To contrast this, the negative binomial distribution uses the probability that a certain number of FAILURES occur before the $r$th SUCCESS. This has many applications specifically when a sequence terminates after the $r$th success such as modeling the probability that you will sell out of the 25 cups of lemonade you have stocked for a given number of cars that pass by. The idea is that you would pack up your lemonade stand after you sell out, so cars that would pass by after the final success won’t matter. Another good example is modeling the win probability of a 7-game sports playoff series. The team that wins the series must win 4 games and specifically the last game played in the series, since the playoff series terminates after one team reaches 4 wins.

One of the more important restrictions on the NBD is that the last event must be a success. Going back to the sports playoff series example, the team that wins the series will NEVER lose the last game. With the 10 coin-flip example, the BD was looking for the probability of getting a certain number of HEADS within a set number of coin flips. Using the NBD, we will look for the probability of 5 HEADS before getting a certain number of TAILS. The total number of flips will not ALWAYS equal 10 and actually exceeds 10 as seen below.

The probability mass function that describes the NBD graph above is given below:

$P(X=k) = {{r+k-1}\choose{k}} p^{r} (1-p)^{k}$

The equation for the NBD has the same parts as the BD: the combinations, the success, and the failures. In the NBD case the combinations are less than the BD [for the same total number of coin flips]. This is because the last outcome is held fix at a success. The probability of success and failure parts of the equation are conceptually the same as the BD. The failure portion is written differently because the number of failures is a parameter $k$ instead of a derived quantity like [$n-r$].

Poisson Distribution:

The Poisson Distribution (PD) is directly related to both the BD and the NBD, because it is the limiting case of both of them. As the number of trials goes to infinity, then the Poisson distribution emerges. The graph for the PD will look similar to the NBD or the BD, and there is no example comparing the coin flip since there has to be some non-discrete process like traffic flow or earthquakes. The major difference is not what is represented, but how it is viewed and calculated. The Poisson distribution is described by the equation:

$P(X = x) = \frac{e^{-\lambda}\lambda^x}{x!}$

$\lambda$ is the expected value [or the mean] for an event and $x$ is the count value. If you knew an average of 0.2 car crashes happen at an intersection at a given day then you could solve the equation for $x$ = {0, 1, 2, 3, 4, 5, … } and get the PD for the problem.

One of the restrictions and major issues with the use of the PD is that the model assumes the mean and the variance are equal. In most real data instances the variance is greater than the mean, so the PD tends to favor more values around the expected value than real data reflects.

If you are interested in the derivations and math behind these I recommend this site: http://statisticalmodeling.wordpress.com/. I feel like they explain the derivation of the negative binomial better than most places I’ve found. It addresses why it’s called the NEGATIVE binomial distribution as well. The site also contains derivations of the PD being the limiting case of the BD and NBD.