# Quantifying what a factor is worth

Looking for a little help here with a statistical question.

Let say I am trying to estimate how many hits a baseball player will hit in a game. Obviously its impossible to get it completely right, but we can find statistical factors (such as their usual production) to help us determine some sort of probability

I know there is an increased chance of getting a hit in the game each time a player has an AT Bat. The more chances you get to hit the ball, the more likely you are to hit the ball.

So if I have two players:

Player A will get 4 At Bats in the game Player B will get 3 At Bats in the game

However, lets say player B has a batting average of .300, while player A has a batting average of .280. So despite Player A being the worse hitter.. they have more opportunities to hit the ball.

From a statistical standpoint how do I determine what this extra at bat is worth? I am trying to do this with many different factors, but thought this example would be good to start.

One of them is a better hitter. Don't go off how much they hit the ball, go off how much them hitting the ball is worth. Runs are always be better than successful hits, as the score is what wins you the game. Go off how much they're worth, not how often they hit.

I'll leave the baseball strategy to others and try to answer just the probability question you asked. As you say, that seems a good place to start.

Binomial models. Player A will bat 4 times with probability 0.28 of getting a hit so his number is his will be distributed $$X \sim \mathsf{Binom}(n=4, p=.28).$$ Thus $$E(X) =1.12,\,$$ $$V(X) = 0.8064.$$

Player B will bat 3 times with probability 0.3 of getting a hit so his number is his will be distributed $$Y \sim \mathsf{Binom}(n=3, p=.3).$$ Thus $$E(X)= 0.9,\,$$ $$V(X) = 0.63.$$

Possible normal approximation. Assuming independent performances all around, you might use these means and variances to get a normal approximation to the distribution of $$X-Y$$ and use that to approximate various probabilities.

Exact approach: The marginal distributions of $$X$$ and $$Y$$ can be found in R as follows: [Ignore row numbes in brackets.]

x = 0:4;  PDF.x = dbinom(x, 4, .28);  cbind(x, PDF.x)
x      PDF.x
[1,] 0 0.26873856
[2,] 1 0.41803776
[3,] 2 0.24385536
[4,] 3 0.06322176
[5,] 4 0.00614656
y = 0:3;  PDF.y = dbinom(y, 3, .3);  cbind(y, PDF.y)
y PDF.y
[1,] 0 0.343
[2,] 1 0.441
[3,] 2 0.189
[4,] 3 0.027


Multiply as appropriate to make the $$4 \times 3$$ joint distibution table. And from that find probabilities $$P(X < Y), P(X = Y),$$ and $$P(X > Y).$$ Their approximate values are obtained in the simulation below.

Simulation. Using R to simulate a million such contests between the two, I got $$P(X < Y) \approx 0.274\;,$$ $$P(X=Y) \approx 0.324,\,$$ and $$P(X > y) \approx 0.402.$$ With this number of iterations, one can expect two place accuracy, probably three.

set.seed(714);  m = 10^6
x = rbinom(m, 4, .28)
y = rbinom(m, 3, .3)
mean(x < y)
[1] 0.274165
mean(x == y)
[1] 0.32378
mean(x > y)
[1] 0.402055


So it seems that A has a slight edge over B because of the additional chance to get a hit.