Conditional Meeting Probability I have a probability problem. Basically, I am interested in how to compute a conditional probability for the following story. 
There are $N$ stores.
There is a buyer, who could be of different level of patient $i$ - He may visit $i$ stores before making purchasing decision. (In words, the more he visits, more patient he is.)
Denote $q(i)$ be the unconditional probability that a buyer visit $i$ stores where $i = 1,2,3,...n$. (And assume $n\ll N$)
My question is that from the perspective of a store, given it is visited by the buyer, what's the conditional probability that the buyer is of patient level $j$? (Visits $j$ stores in total)
Intuitively, given being visited, it is more likely that the visiting buyer is relatively patient. But how to compute the probability rigorously?
Thanks a lot!
 A: Suppose there are $N$ stores, and suppose further that a given customer visits $j$ stores with probability $p_j$, subject to
$$
\sum_{i=1}^N p_i = 1
$$
If we assume that all stores are equally likely to be visited by a given customer, then the probability that a store is visited by a customer, given that the customer visits $j$ stores, is
$$
P(\text{store visited} \mid \text{customer visits $j$ stores}) = \frac{j}{N}
$$
and the unconditional probability that the customer visits a given store is
$$
P(\text{store visited}) = \sum_{i=1}^N p_i \frac{i}{N}
$$
Using Bayes's law, we find that
\begin{align}
P(\text{customer visits $j$ stores} & \mid \text{store visited}) \\
    & = P(\text{store visited} \mid \text{customer visits $j$ stores})
        \frac{P(\text{customer visits $j$ stores})}
             {P(\text{store visited})} \\
    & = \frac{j}{N} \frac{p_j}{\sum_{i=1}^N p_i\frac{i}{N}} \\
    & = \frac{jp_j}{\sum_{i=1}^N ip_i}
\end{align}
For instance, suppose that $N \to \infty$, and $p_j = (1-\rho)\rho^{j-1}, j \geq 1$.  Then
$$
P(\text{customer visits $j$ stores} \mid \text{store visited})
    = \frac{(1-\rho)j\rho^{j-1}}{\sum_{i=1}^\infty (1-\rho)i\rho^{i-1}}
    = (1-\rho)^2j\rho^{j-1}
$$
