How to find the probability of one score over other if they follow a normal distribution?

Question

I am predicting scores over a golf tournament, what I want is to find the probability of one player bets another player on the same tournament. For example, the predicted score of player A is 63, while the predicted score of player B is 65, how can I convert this into probabilities that let me decide how likely is that player A win over player B? I know from the distribution of predicted and actual scores that the scores follow either a normal or Poisson distribution.

Answer 1

If $X_1$ and $X_2$ are independent normal RVs with means $\mu_1$ and $\mu_2$ and variances $\sigma_1^2$ and $\sigma_2^2$, then $X_1-X_2$ is normal with mean $\mu_1 - \mu_2$ and variance $\sigma_1^2 + \sigma_2^2$, so $$P(X_1 < X_2) = P(X_1-X_2 < 0) = P\left(\frac{X_1-X_2 - (\mu_1 - \mu_2)}{\sqrt{\sigma_1^2 + \sigma_2^2}} < - \frac{\mu_1-\mu_2}{\sqrt{\sigma_1^2+\sigma_2^2}} \right) = \Phi\left(- \frac{\mu_1-\mu_2}{\sqrt{\sigma_1^2+\sigma_2^2}} \right).$$

The missing pieces of information you need are the variances of the two players' predicted scores, and the assertion that their scores are independent. If they are not independent but you know the correlation $\rho$ of their scores, then the variance of $X_1-X_2$ will $\sigma_1^2 + \sigma_2^2 - 2 \rho \sigma_1 \sigma_2$ instead.