Category Archives: probability

Statistics of fielding errors in baseball: BAL2019

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“An error is a statistic charged against a fielder whose action has
assisted the team on offense, as set forth in this Rule 9.12.”
– MLB Rule book, 2023

These questions seem natural:
1) Are the number of fielding errors correlated with a team’s winning percentage? If so, how were are they correlated?
2) Are errors uniformly distributed? That is, given the frequency of the 24 game states (as described in the game states post), do errors occur in the same rough frequency?

In the case of the 2nd question, the states with bases empty (0, 1, or 2 outs) are the most commonly occurring states, followed by the states with a runner on 1st (only). For a randomly selected game from the 81 home games of a given MLB team in a given season, are the proportions

“plays with bases empty (batter up, pitcher about to throw first pitch)”: “plays with runner on 1st (only)”

and

“plays with bases empty that have an error”: “plays with runner on 1st (only) that have an error”

roughly equal? (Short answer: I doubt it.)

To these and related questions, we use Retrosheet to compile data. The plays below use the Retrosheet event file notation, as explained in the Retrosheet documentation.

For example, what we examine these questions in the case where the team is the Baltimore Orioles and the (regular) season is 2019? The O’s had a low winning percentage that year: 33.3% in 2019 (the year Brandon Hyde was hired as the Orioles’ manager).
The Orioles finished the 2019 season with 54 wins and 108 losses. The team was also below the league average in offensive categories runs scored, batting average, on-base percentage, and slugging percentage. In the 81 home games, 111 errors were committed, 59 by a visiting team and 52 by the Orioles. The Os played 12 games with at least 3 errors (by both sides), and 6 games with 2 errors. A significant proportion of the errors were in those 12 games: about 15% of the games had over 1/3rd the total number of errors.

Here are three particularly interesting plays from that season of Orioles home games. Back-to-back errors that occur for consecutive batters, such as Alberto–Mancini below, are relatively uncommon. But so are put outs with an error modifier that didn’t get negated by that error, as in the 3rd play below.

  1. During the game TBA (Tampa Bay Rays) vs BAL (Baltimore Orioles) on 2019-08-22, the state before was (0,0,0,0,1,0,0,1). The batter came up to the play and a simple type of error occurred: play,1,1,albeh001,01,CX,E6/TH/G. The state after was (0,1,0,0,1,0,0,1).
    Summary: In the bottom of the 1st inning with bases empty, BAL batter Hanser Alberto hit a ground ball to short. A throwing error by the shortstop allowed the batter to reach 1st.
  2. Trey Mancini is next in the batting line-up. The state before is, as we know, (0, 1, 0, 0, 1, 0, 0, 1), where the runner on 1st is Alberto. Mancini’s play is: play,1,1,manct001,00,X,D7/G+.1-H(UR);B-H(TH)(E6/TH)(NR). The state after is (0, 0, 0, 0, 1, 0, 2, 1).
    Summary: In the bottom of the 1st inning with a runner on 1st, BAL batter Trey Mancini hit a hard ground ball double to left field. The runner on 1st base scored (1 RBI). A throwing error by the shortstop allowed the batter to stretch the double into a home run (no RBI).
    [Video showed more details of this unusual play: the left fielder attempted to throw out Alberto at home, but the throw was to the shortstop, not the 3rd baseman who could have held Mancini at 2nd. This is not an error. However, the shortstop’s errant throw home, hoping to stop Alberto form scoring, went into the camera well, so by MLB Rule 5.06(b)(4)(A), the Umpire awarded Mancini an extra base.]
  3. During the game SDN (San Diego Padres) vs BAL on 2019-06-25, the state before was (2, 1, 0, 0, 3, 5, 1, 1). The batter came up to the play and a simple type of error occurred: play,3,1,smitd007,00,X,D9/G+.1-H;BX3(95)(E9). The state after was (0, 0, 0, 0, 4, 5, 2, 0).
    Summary: In the bottom of the 3rd inning, BAL batter Dwight Smith hit a ground-rule double to right field. The runner on 1st base scored (1 RBI). The batter stretched the hit into a triple, due to an error by the right fielder. Here, the error (E9) does not negate the tag (BX3) in this case but the explanation is complicated: The runner reached 3rd but then (mistakenly?) stepped off and was tagged out. It was ruled that the error by the center fielder allowing the batter to reach 3rd was unrelated to the put out at 3rd. The batter was tagged out at 3rd base by the third baseman after safely reaching the bag. This out was the 3rd out, ending the half-inning.

There are 108 others described in the full report “Game states via Retrosheet: Errors in BAL 2019 home games”. It’s available as a (approx 40-50 page) pdf for anyone who wants it. Just email me at wdjoyner@gmail.com.


Expected maximums and fun with Faulhaber’s formula

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A recent Futility Closet post inspired this one. There, Greg Ross mentioned a 2020 paper by P Sullivan titled “Is the Last Banana Game Fair?” in Mathematics Teacher. (BTW, it’s behind a paywall and I haven’t seen that paper).

Suppose Alice and Bob don’t want to share a banana. They each have a fair 6-sided die to throw. To decide who gets the banana, each of them rolls their die. If the largest number rolled is a 1, 2, 3, or 4, then Alice wins the banana. If the largest number rolled is a 5 or 6, then Bob wins. This is the last banana game. In this post, I’m not going to discuss the last banana game specifically, but instead look at a related question.

Let’s define things more generally. Let I_n=\{1,2,...,n\}, let X,Y be two independent, uniform random variables taken from I_n, and let Z=max(X,Y). The last banana game concerns the case n=6. Here, I’m interested in investigating the question: What is E(Z)?

Computing this isn’t hard. By definition of independent and max, we have
P(Z\leq z)=P(X\leq z)P(Y\leq z).
Since P(X\leq z)=P(Y\leq z)={\frac{z}{n}}, we have
P(Z\leq z)={\frac{z^2}{n^2}}.
The expected value of Z is defined as \sum kP(Z=k), but there’s a handy-dandy formula we can use instead:
E(Z)=\sum_{k=0}^{n-1} P(Z>k)=\sum_{k=0}^{n-1}[1-P(Z\leq k)].
Now we use the previous computation to get
E(Z)=n-{\frac{1}{n^2}}\sum_{k=1}^{n-1}k^2=n-{\frac{1}{n^2}}{\frac{(n-1)n}{6}}={\frac{2}{3}}n+{\frac{1}{2}}-{\frac{1}{6n}}.
This solves the problem as stated. But this method generalizes in a straightforward way to selecting m independent r.v.s in I_n, so let’s keep going.

First, let’s pause for some background and history. Notice how, in the last step above, we needed to know the formula for the sum of the squares of the first n consecutive positive integers? When we generalize this to selecting m integers, we need to know the formula for the sum of the m-th powers of the first n consecutive positive integers. This leads to the following topic.

Faulhaber polynomials are, for this post (apparently the terminology is not standardized) the sequence of polynomials F_m(n) of degree m+1 in the variable n that gives the value of the sum of the m-th powers of the first n consecutive positive integers:

\sum_{k=1}^{n} k^m=F_m(n).

(It is not immediately obvious that they exist for all integers m\geq 1 but they do and Faulhaber’s results establish this existence.) These polynomials were discovered by (German) mathematician Johann Faulhaber in the early 1600s, over 400 years ago. He computed them for “small” values of m and also discovered a sort of recursive formula relating F_{2\ell +1}(n) to F_{2\ell}(n). It was about 100 years later, in the early 1700s, that (Swiss) mathematician Jacob Bernoulli, who referenced Faulhaber, gave an explicit formula for these polynomials in terms of the now-famous Bernoulli numbers. Incidentally, Bernoulli numbers were discovered independently around the same time by (Japanese) mathematician Seki Takakazu. Concerning the Faulhaber polys, we have
F_1(n)={\frac{n(n+1)}{2}},
F_2(n)={\frac{n(n+1)(2n+1)}{6}},
and in general,
F_m(n)={\frac{n^{m+1}}{m+1}}+{\frac{n^m}{2}}+ lower order terms.

With this background aside, we return to the main topic of this post. Let I_n=\{1,2,...,n\}, let X_1,X_2,...,x_m be m independent, uniform random variables taken from I_n, and let Z=max(X_1,X_2,...,X_m). Again we ask: What is E(Z)? The above computation in the m=2 case generalizes to:

E(Z)=n-{\frac{1}{n^m}}\sum_{k=1}^{n-1}k^m=n-{\frac{1}{n^m}}F_m(n-1).

For m fixed and n “sufficiently large”, we have

E(Z)={\frac{m}{m+1}}n+O(1).

I find this to be an intuitively satisfying result. The max of a bunch of independently chosen integers taken from I_n should get closer and closer to n as (the fixed) m gets larger and larger.