# Moneyball question

1. Jun 29, 2008

### cgrady

Hi, I'm reading moneyball a book about stats and baseball. In the book it says when you consider two batters one with a .275 batting average and one with .300 batting average and you watch them over 15 games there is a 40% chance the .275 batter will get more hits. Could someone explain to me how they are calculating this. Thank you.

2. Jun 29, 2008

### matt grime

Either very inefficiently or by breaking it up into a sum over cases, of one hitter getting 0,1,2,... hits, and the other getting more.

3. Jun 29, 2008

### Crosson

The only thing you need to calculate this is the rule that the probability of two independent events A and B is the product of the individual probabilities:

Prob( A & B ) = Prob(A) * Prob(B)

What is the probability of flipping a coin twice and getting heads both times?

Prob( Heads & Heads ) = Prob(Heads) * Prob(Heads) = (0.5)*(0.5) = 0.25

What is the probability of a batter with a 0.300 batting average getting a hit at 10 out of his next 10 at bats?

(0.3)^10 = 0.3*0.3*0.3*0.3*0.3*0.3*0.3*0.3*0.3*0.3 = 0.000006

Which is the same as 0.0006% (very unlikely, but there is a chance).

Other than this concept of multiplying probabilities, they had to consider a lot of specific cases, as Matt Grime said above. The best way to do this is with a computer.

4. Jun 29, 2008

### cgrady

so basically it would be something like the probability 275 hitter gets 1 or more hits * the probability 300 hitter get less 1 hit. Plus probability 275 hitter gets 2 or more hits * probability 300 hitter gets less than 2 hits. Etc.

5. Jun 29, 2008

### matt grime

Those are not disjoint events, so you can't just add the probabilities. Just think about the 275 hitter having 5 hits, and the 300 hitter having 1 - you've accounted for that twice.

Try 'the 300 hitter has precisely 1 hit'.

6. Jun 29, 2008

### cgrady

Ah, thank you, I understand now. It would probability 275 hitter has 1 hit *probability 300 hitter has 0. Etc

7. Jun 29, 2008

### CRGreathouse

The batting averages tell you how likely a person is to hit the ball each time he's at-bat, but the number of times the player bats can vary too. If both players were at bat 55 times in those 15 games, the chance that the one with the lower batting average would have more hits is about 34.6%. But variability in the number of hits per game favors the weaker player: if he happens to be at bat more, he can get more hits even if his average in that period is lower.

8. Jun 29, 2008

### cgrady

.3461 is the number I got too, when I used 55 at bats and plugged the numbers into a summation, which makes me feel happythat i kinda know how to do it now. I also tried 45 at bats, because I think the average at bats per game is close to 3, and the probability was 35%, so I'm guessing the authors just choose to round up or used a lower number of at bats.

Anyway, thank everybody for their help, now I can get back to reading the book

9. Jun 30, 2008

### CRGreathouse

There's no fixed number of at bats that will give you as high as 40%, which is why I mentioned the possibility of the batters hitting a different number of times. That could get you up to 40%.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook