Thursday, June 04, 2009

John Shelby Goes 0- for- 10

Wednesday, June 03, 2009

Today is the 20th anniversary of a 22-inning game between the Los Angeles Dodgers and Houston Astros (box score). Houston ultimately won, 5-4. From a streakiness perspective, the thing I've always remembered about this game is L.A.'s John Shelby going 0-for-10.

Looking at Shelby's career statistics, he clearly had a bad year with the bat in 1989, hitting only .183. (63-for-345). A .183 batting average translates into a .817 (i.e., 1 - . 183) failure rate on each at-bat. Raising .817 to the 10th power, for the probability of 10 successive failures (assuming independence of events), yields .133 as the likelihood of Shelby's going 0-for-10. We have not, of course, taken into account the quality of the opposing pitchers or any other factors, so this will have to be a rough estimate.

A 13% chance of Shelby going 0-for-10 is not astronomically small by any means. It's still fairly rare, however. The following figure shows Shelby's probabilities of getting 0, 1, 2, 3, etc., hits out of 10 at-bats. Not surprisingly, the likeliest scenarios were for him to get 1 or 2 hits. For the probability of 1 hit (and 9 failures), for example, we would take .817 to the 9th power, then multiply the result by .183, thus yielding .0297. There are 10 different ways to get exactly 1 hit out of 10 (i.e., in the first at-bat, or in the second, ... , or in the 10th), so we multiply .0297 X 10, yielding .297 (which is shown in the figure).

Shelby also had a higher probability of getting 3 hits in the game than 0 hits, but the chances start tailing off once we get to 4 hits. The above probabilities were obtained from the Vassar Binomial Calculator.

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