I have a problem that is not stated so clearly, but the answer is given, I want to figure out how to solve it.
For sellers $A=\{1,2,\cdots,\alpha\}$ and user $i$, the corresponding probability that each seller encounter $i$ is $P_{i,1},~P_{i,2},\cdots,~P_{i,\alpha}$. Solve for the probability that the seller $n$ is the first seller who meets user $i$.
The answer is
$$P_{i,n}\times\sum_{S\in 2^{A\setminus \{n\}}}~~~\prod_{j\in S}\frac{P_{i,j}}{|S|+1}~~\prod _{\ell\in A\setminus(S\cup\{n\})}(1-P_{i,\ell})$$
The occurence of the first term $P_{i,n}$ is reasonable, since it means the first person is the seller $n$. However, I can't figure out what $|S|+1$ means, and why we should times $\prod _{\ell\in A\setminus(S\cup\{n\})}(1-P_{i,\ell})$.
I suspect there is a slight mistake in the formula and it should be
$$P_{i,n}\times\sum_{S\in2^{A\setminus\{n\}}}\frac1{|S|+1}\prod_{j\in S}P_{i,j}\prod_{\ell\in A\setminus(S\cup\{n\})}(1-P_{i,\ell}).$$
The idea here is that $S$ is the set of other sellers who also encounter $i$. For a particular $S$, we also need to multiply by the probability that those sellers in $S$, and no others, encounter $i$ - that is what the two products calculate. Then the total number of sellers who encounter $i$ is $|S|+1$, and presumably they are all equally likely to be the first one to encounter $i$, so $\frac1{|S|+1}$ is the probability that $n$ encounters $i$ before any of the other sellers in $S$.
Finally, we have to sum this over all possible $S$ to get the total probability (that for some $S$, $S\cup\{n\}$ are the sellers who encounter $i$, and $n$ is the first of these to do so).