Components of Poisson Distribution mean

Question

Suppose that we have a game between Team A and Team B. Team A has a 75% probability of winning and Team B has a 25% probability of winning. Draw cannot occur. I also know, that there is calculated average of points scored in this game and it is 3.5. This average consists of average of points scored by Team A and average of points scored by Team B. How can I calculate those averages per team?

Answer 1

Let $X$ be the score of A and $Y$ the score of B. Define $E[X] = a, E[Y] = b$. We know that $$ P(X > Y) = P(X - Y > 0) = 0.75 $$ and that $E[X + Y] = a + b = 3.5$ so, $b = a - 3.5$. Define $Z = X - Y$, then $$ P(X > Y) = P(Z > 0) = 1 - P(Z \le -1) $$ If $X$ and $Y$ were poission the probability of the difference being zero would be $p_0 = e^{-(a + b)} I_{0}(2\sqrt{ab})$ where $I$ is the https://en.wikipedia.org/wiki/Bessel_function#Modified_Bessel_functions:_I%CE%B1,_K%CE%B1.

Since draws are not allowed, we have to truncate the distribution so: $P(Z \le -1) = \frac{F(-1)}{1-p_0}$ where $F$ is the cumulative distribution function for the non-truncated distribution given by: $F(-1; a, b) = \sum_{k = -\infty}^{-1} e^{-(a + b)} \big(\frac{a}{b}\big)^{k/2}I_{k}(2\sqrt{ab})$.

In your case, $$0.75 = 1 - \frac{F(-1; a, 3.5 - a)}{1-p_0}.$$ Solving this gives $a = 2.2577$ and $b = 1.2423$ which agrees with your suggested answer.