The following problem is a variation from one of Ross's Stochastic Methods textbook problems: team 1 and 2 are playing a match. The teams score points according to independent Poisson processes with rates 4 and 1, respectively. The match ends when one of the teams scores 5 more points than the other. Find the probability that team 1 wins.
I have approached the problem as follows:
$$P(N_1\ge N_2+5)=P(N_1-N_2\ge 5)=\sum_{i=0}^{\infty} P(N_1=i+5,N_2=i)=\sum_{i=0}^{\infty} P(N_1=i+5)P(N_2=i)=\sum_{i=0}^{\infty} e^{-4}\frac{4^{i+5}}{(i+5)!}e^{-1}\frac{1^{i}}{i!}=e^{-5}4^{5}\sum_{i=0}^{\infty}\frac{4^{i}}{i!(i+5)!}$$
Where $N_1,N_2$ are the random variables representing the scores of team 1 and team 2, respectively. Calculating this with Maplesoft gives me a probability of $0.1088$.
However, if I use the gambler's ruin problem, where a gambler starts with 5 points, wins if it gets to 10 points and losses if it goes to 0 points, with the probability of team one winning one point equal to $\frac{4}{5}$ and losing one point equal to $\frac{1}{5}$, the probability of team one winning is $$\frac{1-(1/4)^5}{1-(1/4)^{10}}\approx1$$
Any ideas of what I am doing wrong? Thank you!