# 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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Yes, each scored goal has probability 1/3 of being scored by team 1 and probability 2/3 of being scored by team 2, and by the magics of Poisson proceses, each goal is colored independently from the others.

Hence you look for the probability of a random walk starting from (3,1) and with step (+1,0) with probability 1/3 and (0,+1) with probability 2/3 to reach the line (5,something) before reaching the line (something,5).

Writing u(x,y) for this probability starting from (x,y), you look for u(3,1) and you know that u(5,something)=1, u(something,0)=0 and 3u(x,y)=u(x+1,y)+2u(x,y+1). This shows that u(3,1) is part of the unique solution of a linear system of equations whose unknowns are the u(x,y) for (x,y) any of the points (3,1), (4,1), (3,2), (4,2), (3,3), (4,3), (3,4), (4,4).

Starting backwards, one gets u(4,4)=1/3, u(3,4)=u(4,4)/3=1/9, u(4,3)=1/3+2u(4,4)/3=5/9, u(3,3)=u(4,3)/3+2u(3,4)/3=7/27, u(4,2)=1/3+2u(4,3)/3=19/27, u(3,2)=u(4,2)/3+2u(3,3)/3=11/27, u(4,1)=1/3+2u(4,2)/3=65/81, and finally u(3,1)=u(4,1)/3+2u(3,2)/3=131/243.

Edit An alternative solution is, as suggested by the OP, to consider the five first goals and to note that team 1 loses the game if and only if team 1 scores one or zero of these five goals. The number of goals scored by team 1 amongst the five first goals is binomial (5,1/3), hence the probability that team 1 wins is 1-P(Bin(5,1/3)=0)-P(Bin(5,1/3)=1). This is 1-(2/3)^5-5(2/3)^4(1/3)=1-32/243-80/243=131/243.

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