End of the line?
According to numbers cruncher Clay Davenport of Baseball Prospectus, there is a 1 in 194 chance that Chase Utley will catch Joe DiMaggio's hitting streak and a 1 in 244 chance of Utley passing Joltin' Joe.
At the same time, there is a better than 20 percent chance that Utley's streak ends tonight. Or, for the more postive folks out there, a 20 percent chance of the streak reaching 40 games.
How did Davenport arrive at these numbers? Well, let's allow him to explain:
Hitting streaks are notoriously difficult beasts for a statistician to deal with, even though they seem to be a simple application of probability. You would think that it would work like this: take some representation of his ability to get a hit, say his current average of .328, and then invert it to .672 (1 minus his average), his chance of not getting a hit in any AB. If he gets four ABs a game, his chance of not getting a hit in any game is just .672 raised to the fourth power, or .204. You have to reverse it again to get .796 (his chance of getting a hit in a game), and raise that to the number of games needed--currently 23--so you have .796 raised to the 23rd power, or 0.05%, about 1 chance in 190.
Gotta love math. There are other variables to this, but all those numbers make me start to feel dizzy. It's better if the author explains it.
Here's the story from Davenport on Baseball Prospectus. You need a subscription to read it.
At the same time, there is a better than 20 percent chance that Utley's streak ends tonight. Or, for the more postive folks out there, a 20 percent chance of the streak reaching 40 games.
How did Davenport arrive at these numbers? Well, let's allow him to explain:
Hitting streaks are notoriously difficult beasts for a statistician to deal with, even though they seem to be a simple application of probability. You would think that it would work like this: take some representation of his ability to get a hit, say his current average of .328, and then invert it to .672 (1 minus his average), his chance of not getting a hit in any AB. If he gets four ABs a game, his chance of not getting a hit in any game is just .672 raised to the fourth power, or .204. You have to reverse it again to get .796 (his chance of getting a hit in a game), and raise that to the number of games needed--currently 23--so you have .796 raised to the 23rd power, or 0.05%, about 1 chance in 190.
Gotta love math. There are other variables to this, but all those numbers make me start to feel dizzy. It's better if the author explains it.
Here's the story from Davenport on Baseball Prospectus. You need a subscription to read it.
posted by Motorboat Jones at 7:36 PM
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