Problem introduction
Let us consider the following problem.
We have 2 players Alice and Bob and they play a game which is about getting points. The player with most points wins. The points player get depend on their and opponent's strategies.
We know the Alice's and Bob's points that they get over a game come from Poisson distribution with parameters $w_a$, $w_b$.
Moreover we know that points what both players lose are also from a Poisson distribution with parameters $l_a$ and $l_b$ respectively.
We wish to know what are the probabilities that a player wins before the game starts.
Possible solution
Let $W_a$ and $W_b$ denote points that Alice and Bob get and $L_a$ and $L_b$ denote points that Alice and Bob lose.
P(Alice wins) = P($W_a > L_a$) = $\sum_{k=0}^{\infty}P(W_a = k) P(L_a < k)$
P(Alice wins) = P($W_a > W_b$) = $\sum_{k=0}^{\infty}P(W_a = k) P(W_b < k)$
P(Alice wins) = P($L_b > W_b$) = $\sum_{k=0}^{\infty}P(L_b = k) P(W_b < k)$
P(Alice wins) = P($L_b > L_a$) = $\sum_{k=0}^{\infty}P(L_b = k) P(L_a < k)$
For example if the means for $W_a$, $W_b$, $L_a$ and $L_b$ are $8$ $5$ $7$ and $2$, the above formulas give us following results: $0.55, 0.76, 0.09$ and $0.03$.
Question
Is there any way to combine all these approaches so that we take into account all $4$ variables and the best prediction?
PS: code to calculate numbers:
The event Alice wins occurs when $W_a-L_a>W_b-L_b$ which is the same as $W_a+L_b>W_b+L_a$. For two id Poisson RV $X_1,X_2$ with parameters $\lambda_1,\lambda_2$, we know that $X_1+X_2$ is Poisson RV with parameter $\lambda_1+\lambda_2$, so your question is to compute that probability that one Poisson RV (with parameter $w_a+l_b$) is bigger than the other (with parameter $w_b+l_a$).
As you can see here, Probability of one Poisson variable being greater than another , there isn't a nice answer.