# Probability one hitter is better

1. Nov 7, 2015

### KenNKC

Let's say baseball player A gets 90 hits in 300 at bats, and player B gets 25 hits in 100 at bats. The true ability of both players is unknown. What is the probability that player A has a higher true ability than player B? Would the Student's t-test be used to calculate this? Thanks.

2. Nov 7, 2015

### Hornbein

Student's t test is used on sample sizes of less than thirty.

Other than that, hits of baseball players follow which distribution? Once you know that, you can look up the test in a statistics text.

3. Nov 7, 2015

4. Nov 8, 2015

### Ronie Bayron

Don't you think that's obvious enough Player A hits (3/10) player B (2.5/10), so player A wins B. (0.5/10) difference B~A.
Even so that probability of hitting a ball per pitch is 50% and every event is independent with one over another. There's no way B would surpass A.

5. Nov 8, 2015

### Staff: Mentor

If player B is batting .250 in the major leagues and player A is batting .300 in their high school league, then player B is still probably better. The true ability required to hit depends also on the true ability of the pitcher.

Even within the same league, the difficulty of the pitchers must be accounted for. I use a similar approach for my pinewood derby.

6. Nov 8, 2015

### Ronie Bayron

Say, if we have to defeat the variability of the true ability of the pitcher for instance, and use one pitching machine for the two (A&B) instead, would that counts? I think B is defeated most probably.

7. Nov 8, 2015

### Staff: Mentor

Sure. If you are performing a controlled experiment then you could control for that. The OP wasn't clear what conditions the data were acquired under. But they sound like regular game play rather than experimentation. In regular play the skill of the competition matters greatly.

8. Nov 8, 2015

### KenNKC

Thanks for the replies. To clarify, I'm assuming both players faced equal quality pitching. Assuming a normal distribution, the true talent of player A with a 95% certainty is between .248 and .352. For player B the range is from .165 to .335. So player A's true talent could be .270 for example, and he just had a lucky 300 at bats. And player B's true talent could be .320 and he had an unlucky 100 at bats. So although Player A is more likely to be better than B based on the sample data, Player B could also be the better one. The part I am struggling with is coming up with a formula that gives this probability.

9. Nov 8, 2015

### Staff: Mentor

So the normal approximation to the binomial distribution is $\mu= np$ and $\sigma=np(1-p)$. You can take the PDF for two such normal variables and simply multiply them together to get a joint PDF.

Once you have the joint PDF, then you just integrate in the region $X>Y$ to get the probability.

I think.

10. Nov 8, 2015

### WWGD

Isn't this a difference of means test?

11. Nov 8, 2015

### Staff: Mentor

You could do it that way, but it would probably be better to do a one sided test for equal proportions.

Either way would be much simpler than my last suggestion.

12. Nov 8, 2015

### insightful

Also, remember that saying, "With 95% certainty we cannot reject the hypothesis that A is better than B," is not the same as saying, "There is a 95% probability that A is better than B."

13. Nov 8, 2015

### KenNKC

The test for equal proportions did the trick. I have a book with results of problems of the same type as my example, but the method to derive them isn't included. The equal proportions test got the right answers. For my example, the probability is about 83%. Thanks for everyone's input.

14. Nov 8, 2015

### Staff: Mentor

15. Nov 8, 2015

### Staff: Mentor

I think you are alluding to Bayesian vs frequentist methods? A frequentist approach would give the probability of the data given the hypothesis. The Bayesian approach would give the probability of the hypothesis given the data.