Three by five homerun derby
On July 25, 2015, Alex Rodriguez hit three homeruns in a single game, marking the fifth time in his career that he's hit three in one game. Over his entire career, he has hit 677 homeruns in 11,725 plate appearances. By random chance, how many three homerun games should we expect Alex Rodriguez to have?
Assumptions
- He has played 2,658 games in his career.
- For simplicity, assume that he gets four plate appearances in every game that he plays.
The answer is 1.92847.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
For any given at-bat, the probability that A-Rod hits a home run is p = 1 1 7 2 5 6 7 7 . So using the binomial theorem, for any given game with 4 at-bats, the probability of A-Rod hitting exactly 3 home runs is
( 3 4 ) × p 3 × ( 1 − p ) = 4 × ( 1 1 7 2 5 6 7 7 ) 3 ( 1 − 1 1 7 2 5 6 7 7 ) .
Multiplying this by 2 6 5 8 games, we find that the expected number of 3-homerun games is 1 . 9 2 8 .
The fact that he's actually had 5 such games seems anomalous, but in sports, once an exceptional player, (with or without steroids), gets in "the zone", hitting homeruns, scoring goals, etc., becomes almost automatic. However, 4-homerun games are extremely rare, and have only occurred 16 times in the 120-year history of MLB, with no single player having done it twice in their career. No one has ever hit more than 4 homeruns in a single game, (not even with extra innings).