Naive Simple Calculation of the Improbability of an Eighteen Inning Game
I am really calculating the odds of an eighteen inning game after a tie nine inning game has already gone into extra innings.
This is motivated by trying to figure out how improbable it is for both teams to go eight innings without either scoring even one run. First, we examine the probability of getting runs by singles or on-base at bats. Then we look at the effect of home runs.
First we will start with the simplest scoring method of getting a run by hitting three singles or on-bases in an inning. For simplicity, we will take all of the batters of having an on-base average of s = 0.4 for the Dodgers and Red Sox, although it varies from .3 to .5, generally.
On average, we say it takes three on-bases to get a run, before you commit three outs. How many sequences are there that does that, at the start, not considering multiple run innings? Then the probability of scoring a run will be s^3 = s x s x s, times the number of ways this can occur.
Here are the run sequences: s is a single or on-base, and o is getting an out:
S s s o o o
S s o s o o
S s o o s o
S o s s o o
S o s o s o
S o o s s o
O s s s o o
O s s o s o
O s o s s o
O o s s s o
So there are 10 successful sequences, and the probability of a run is
10 x s^3. (S cubed). That is the probability of a team scoring by three on-bases per inning is Ps = 10 x (0.4)^3 = 0.064 x 10 = 0.64. For 9 innings, this would give a game scoring average of 5.76.
Let’s check this against the Dodgers regular year. They played 163 games (92-71) and scored 804 runs. This is 4.93 runs per game. Considering the ignorance of a thousand other possibilities to score or innings played, this could be considered close. Also the 0.4 on base average is just a rough average of that which should not be averaged.
That would mean that the probability of one team not scoring in an inning by three on-bases is Po = 1 – Ps = 1 – 0.64 = 0.36.
The probability of a team not scoring in nine innings in a game is
Pns = (0.36)^9 = 10^(-4), or one in ten thousand. This is way too small. The number of Dodgers shutouts was 8 out of 163, or a probability of 0.049.
So calculating the probability of being shut out for eight extra innings cannot be calculated this way.
The Red Sox had a 108-54 record, playing 162 games. They had 876 runs for an average of 5.41 runs per game. That is closer to the 5.76 naive average calculated at the start. The Red Sox were shut out in 7 games, which is a probability of 0.043.
While the longest World Series game went 9 extra innings, we can’t rigorously use the actual nine inning shutout seasonal results, but we will calculate it as an approximation. So we multiply 0.049 x 0.043 = 0.0021, or one chance in 500.
While the Dodgers scored 235 home runs, their best regular season home run hitter was Max Muncy, with 35. Of the 163 games, his probability of hitting a home run in nine innings was 0.215 or one in 5. If you take the top 9 hitters as a team, they scored 205 out of the 235 home runs. These are pretty close. So the odds of a home run per game are 235/163 = 1.44 per game.
Out of the 804 runs scored, 235/804 = 0.292, or at least 30% of the runs are scored by home runs. Of course runs batted in can be scored by outs or with home runs as runs batted in.
The odds of a home run per inning is roughly 235/(163×9) = 0.160, or about one in six. So the odds of not having a home run in an inning is 1 – 0.16 = 0.84.
The odds of not having a home run in eight innings by both teams is then roughly 0.84^16 = 0.0614. So only one out of sixteen games should have such a streak. Multiply this by the 0.16 odds for a home run for a team in the last inning, by two teams, gives 0.0614 x 0.16 x 2 = 0.0196, or about 0.02, or 1 in 50, for the odds of how the nine-inning overtime of the third game went.
