# Independent Poisson processes: Race to 5

I have this problem to solve:

Hockey teams 1 and 2 score goals at times of Poisson process with rates 1 and 2. Suppose that $N_1(0)=3$ and $N_2(0)=1$. What is the probability that $N_1(t)$ will reach 5 before $N_2(t)$ does?

I've re-worded this to: What is the prob that in the next 5 goals at least 2 of them are scored by team 1?

The only problem I have is finding out the probability of team one scoring a goal. Can we use the rates to work this out?

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You must assume that the two Poisson processes are independent. Goal-scoring (without regard to which team scores) is then a Poisson process $N(t) = N_1(t) + N_2(t)$ with rate $1+2=3$, and each individual goal that is scored has probability $1/3$ of coming from team 1.

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To find the probability of team 1 scoring a goal before team 2, you can consider the corresponding waiting times, which are exponentially distributed with means $1$ and $1/2$, respectively (and recall some previous exercise). Or, more simply, consider $\frac{{\lambda _1 }}{{\lambda _1 + \lambda _2 }}$ (and $\frac{{\lambda _2 }}{{\lambda _1 + \lambda _2 }}$). It is also worth considering here Thinning and Superposition of Poisson processes.

To find the probability of team 1 scoring a goal before team 2, you can consider the corresponding waiting times, which are exponentially distributed with means $1$ and $1/2$, respectively (and recall some previous exercise).

EDIT: Some details.

Let $X_i$, $i=1,2$, be independent exponential random variables with probability density functions $\lambda_i e^{-\lambda_i x}$, $x > 0$. It is an easy exercise to show that ${\rm P}(X_1 < X_2) = \lambda_1/(\lambda_1+\lambda_2)$. Since for $\lambda_1=1$ and $\lambda_2=2$ this probability is equal to $1/3$, it should be clear that the probability you were looking for is $1/3$ (consider inter-arrival times between goals). If, on the other hand, you consider the process $N=N_1 + N_2$, then the probability $\lambda_1/(\lambda_1+\lambda_2)$ can be accounted for as follows. First note that $${\rm P}(N_1 (t + \Delta t) - N_1 (t) \ge 1|N(t + \Delta t) - N(t) \ge 1) = \frac{{{\rm P}(N_1 (t + \Delta t) - N_1 (t) \ge 1)}}{{{\rm P}(N(t + \Delta t) - N(t) \ge 1)}}.$$ For small $\Delta t$ this gives $${\rm P}(N_1 (t + \Delta t) - N_1 (t) \ge 1|N(t + \Delta t) - N(t) \ge 1) \approx \frac{{{\rm P}(N_1 (t + \Delta t) - N_1 (t) = 1)}}{{{\rm P}(N(t + \Delta t) - N(t) = 1)}} ,$$ and hence $${\rm P}(N_1 (t + \Delta t) - N_1 (t) \ge 1|N(t + \Delta t) - N(t) \ge 1) \approx \frac{{e^{ - \lambda _1 \Delta t} \lambda _1 \Delta t}}{{e^{ - (\lambda _1 + \lambda _2 )\Delta t} (\lambda _1 + \lambda _2 )\Delta t}}.$$ Finally, letting $\Delta t \to 0$ accounts for the probability $\lambda_1/(\lambda_1+\lambda_2)$ you were looking for.

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