# Help with Bayes Theorem (possibly?)

So let's suppose that there is a baseball player with a .300 batting average (i.e., the probability that he gets a hit is 30%) and he is facing a pitcher who has a .400 avg against (the probability that he gives up a hit is 40%). What would be the probability that this batter will get a hit off this pitcher? I tried to use Bayes theorem, but I got an answer of .222, which doesn't really make sense. Would the answer just be .350 (the average of the two), or is there some Bayesian way to get an answer?

As @jlimahaverford said, there is not enough information. You are trying to determine

$$P(H|B=b,P=p)$$

where $H$ is the event "a hit", $B=b$ indicates the batter is $b$, and $P=p$ indicates the pitcher is $p$. But the information you have is

$$P(H|B=b) \quad \mbox{and} \quad P(H|P=p)$$ and there is no obvious way to get from these two pieces of information to what you are trying to determine.

One approach you might want to look into is logistic regression where the response is the binary outcome of each at bat (hit or no hit) and the explanatory variables are the IDs of the batter and hitter in that at bat. Then you could learn the quality of each batter and pitcher. From this information, you could predict what would happen for a particular batter-pitcher combination.

• You would probably need a lot of data for the logistic regression? – user30490 Sep 30 '15 at 3:57
• This is baseball...there exists lots of data at least for estimating main effects. – jaradniemi Sep 30 '15 at 13:46

As you stated your question, it's a bit ambiguous. Based on my understanding of your question, it doesn't provide sufficient information to approach this problem in a Bayesian manner.

Suppose $A$ is the event that the batter succeeds, and $B$ is the event that the bowler fails. You are given $p(A) = 0.3$ and $p(B) = 0.4$ and you are asked what $p(A|B)$ is. Without any more information or constraints (i.e., whether $A$ depends on $B$ or not), the best you can say is $A$ is independent of $B$ and conclude that $p(A|B)=p(A)=0.3$.

• I don't quite think this is right. What are $A$ and $B$ here? I agree there's not enough information. But what two events are you assuming are independent. I think we have two conditional distributions for the same event (hit) conditioned on two different conditions. $P(A|B), P(A|C)$. I agree that Bayes theorem doesn't apply but you said "it's not Bayesian" which means something very different (though also true). – jlimahaverford Sep 30 '15 at 1:53
• Are they really independent of each other though? If a batter has hit above average against a variety of pitchers, wouldn't I expect him to hit even better against a pitcher who has allowed above average offense against him? – Ryan G Sep 30 '15 at 2:33
• I was probably a bit terse in my reply. Here, A is the event that the batter succeeds. And B is the event that the bowler fails. The first statement "is a baseball player with a .300 batting average" without any more information is just $p(A)$. The same goes for the bowler as well. Right? – Vimal Sep 30 '15 at 12:29
• @RyanG: Yes, it does sound intuitively correct, but I think statements in English are pretty easy to get confused about unless they are stated unambiguously. Here, I think we all agree the statements are a bit ambiguous, so any deduction that makes the fewest assumptions makes sense? – Vimal Sep 30 '15 at 12:31
• @RyanG: I see where you are coming from. In the extreme, if you have a batter with 100% success and a certain bowler with 0% failure, then will the batter hit or not? Is it even possible to have the above statistics? It's hard to say unless we are given how $p(A)$ is computed and what it is conditioned on. – Vimal Sep 30 '15 at 12:37