I have been having some trouble with the following problem, and I have finally reached a solution that I think is correct but I would like to double check. The problem is the following:
Consider a sequence of multinomial trials with three outcomes: type 1, type 2, and type 3, each occurring with probability $p_1,p_2,p_3$, respectively (so $p_1+p_2+p_3=1$). Let $N_i$ be the first trial which had outcome $i,~i=1,2,3$. Compute $P(N_1<N_2<N_3)$.
There are other parts to the problem, and one of them is to describe the distribution of each $N_i$, which I said is geometric with success probability $p_i$. In order for $N_1<N_2<N_3$ we must have $N_1=1$, because there must be an outcome for the first trial. In another part of the problem I showed that, for $1<j<k$, we have $$(1)\qquad P(N_1=1,N_2=j,N_3=k)=p_1^{j-1}p_2(1-p_3)^{k-j-1}p_3$$ since the first outcome can occur $j-1$ times before outcome two, then any outcome other than three occurs $k-j-1$ times before outcome three. So I would think that the desired probability is just the sum over $j\geq 2$ and $k\geq j+1$ since $P(N_1<N_2<N_3)=P(N_1=1,~1<N_2<N_3)$. So using the geometric series formula, we have: $$(2)\qquad \begin{align}P(N_1<N_2<N_3)&=\sum_{j=2}^\infty\sum_{k=j+1}^\infty p_1^{j-1}p_2(1-p_3)^{k-j-1}p_3\\&=\frac{p_2p_3}{p_1(1-p_3)}\sum_{j=2}^\infty\left(\frac{p_1}{1-p_3}\right)^j\sum_{k=j+1}^\infty (1-p_3)^k\\&=\frac{p_2p_3}{p_1(1-p_3)}\sum_{j=2}^\infty\left(\frac{p_1}{1-p_3}\right)^j\left(\frac{(1-p_3)^{j+1}}{p_3}\right)\\&=\frac{p_2}{p_1}\sum_{j=2}^\infty p_1^j\\&=\frac{p_1p_2}{1-p_1}\end{align}$$
My questions: Is my logic correct, and mainly: are equations $(1),(2)$ correct?
