Suppose that voters arrive to a voting booth according to a Poisson process with rate $\lambda = 100$ voters per hour. The voters will vote for two candidates, candidate $1$ and candidate $2$ with equal probability $\left(\frac{1}{2}\right)$. I am assuming that we start the voting at time $t=0$ and it continues indefinitely. I am trying to find probability of candidate $1$ receiving $k$ votes in the first $4$ hours of voting, CONDITIONAL on $1000$ voters arriving during the first $10$ hours. My calculation keeps coming up as $\frac{1}{2}$ due to the symmetry involved.
My idea is to define $A_t =$ number of votes for candidate $1$ at time $t$, then find $P(A_4 = k | N(10) = 1000)$, where $N(t)$ is the number of voters who have arrived at time $t$. Could someone tell me if this the right approach? Thanks.
You have :
$$\mathbb{P}(A_4=k|N_{10}= 1000)=\frac{\mathbb{P}(A_4=k,N_{10}=1000)}{\mathbb{P}(N_{10}=1000)}$$
$$\mathbb{P}(N_{10}=1000)=\frac{(10\times 100)^{1000}}{1000!}e^{-10\times 100}$$
You have : $$A_t = \sum_{i=1}^{N_t} \xi_i$$ where $\xi_i$ are independent and identically distributed $$\mathbb{P}(\xi_i=1)=\mathbb{P}(\xi_i=0)=\frac{1}{2}$$
$$\mathbb{P}(A_4=k,N_{10}=1000)=\sum_{k\leq n\leq 1000}\mathbb{P}(\sum_{i=1}^n\xi_i=k,N_{10}-N_4=1000-n)$$ using independence of increments of $N$ you get: $$\mathbb{P}(A_4=k,N_{10}=1000)=\sum_{k\leq n\leq 1000}\mathbb{P}(\sum_{i=1}^n\xi_i=k)\mathbb{P}(N_{10}-N_4=1000-n)$$ using that increments are stationnary $N_{t+h}-N_t\stackrel{\mathcal{L}}{=}N_h$ and that $\sum_{i=1}^n\xi_i$ is a Binomial $\mathcal{B}(n,\frac{1}{2})$ you get:
$$\mathbb{P}(A_4=k,N_{10}=1000)=\sum_{k\leq n\leq 1000}\binom{n}{k}\frac{1}{2^n}\frac{(6\times 100)^{1000-n}}{(1000-n)!}e^{-6\times 100}$$
and endly you get :
$$\mathbb{P}(A_4=k|N_{10}= 1000)=\frac{\sum_{k\leq n\leq 1000}\binom{n}{k}\frac{1}{2^n}\frac{(6\times 100)^{1000-n}}{(1000-n)!}e^{-6\times 100}}{\frac{(10\times 100)^{1000}}{1000!}e^{-10\times 100}}$$