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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?

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Evidently this approach is correct, as I received full marks on the assignment. Just posting this answer to close off the question.

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  • $\begingroup$ Receiving full marks is not always indicative of correctness. It depends on the patience, focus, and skill of the person determining the marks. Your equation (1) looks correct to me. The final answer is so simple that it makes one wonder if there isn't another method to reach it. Indeed, $p_1$ is the chance that the first answer is type 1, and $\frac{p_2}{1-p_1}$ is the chance that the next answer-that-isn't-type-1 is type 2, so there you have it. I would guess the person giving out the marks simply checked the final answer, and not your logic or equations (1) or (2). $\endgroup$ – Matt Jun 18 '18 at 15:57

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