Discrete Probability Question With No Named Distribution I've got a real world probability problem that I have been able to solve easily using simulation but that I am struggling to find (out of pure interest) an analytic solution to. For privacy reasons, I have changed both the context and the actual numbers.
A vendor is selling tickets. Customers either purchase 1 ticket with probability 60%, 2 tickets with probability 30%, or 3 tickets with probability 10%. If the vendor needs to sell 10 tickets, how many customers on average does he need to sell to (assuming customer purchases are iid)?
A simulation produces an average of 6.6 customers, but I am curious what an analytic solution to this might look like (if there exists a convenient one) as there is no named distribution I can think of that corresponds to this problem.
 A: One analytical approach is find the probability that you have not sold $10$ tickets after seeing $n$ customers. This is the sum of the coefficients of $x^k$ with $0\le k \le 9$ in the expansion of $$\left(0.6x+0.3x^2+0.1x^3\right)^n$$
For example with $n=2$ you get $0.36x^2+0.36x^3+0.21x^4+0.06x^5+0.01x^6$ and adding up the coefficients with $0\le k \le 9$ gives $1$ as you might have thought anyway; it becomes interesting when $4 \le n\le 9$ as then you may or may not have sold $10$ tickets. For various values of $n$ this gives
n customers      0 1 2 3  4      5       6       7          8           9           10+
P(tickets < 10)  1 1 1 1  0.9909 0.90126 0.64152 0.3055968  0.0839808   0.010077696  0

and if you add these probabilities up then you get the expected number of customers when you have sold at least $10$ tickets of $6.933335296$, as lulu found
A: To write out the recursive solution sketched in the comments:  Let $E_n$ denote the expected number of customers needed to sell $n$ tickets.  Then, of course, $E_0=0, E_1=1$  We compute $E_2=.6\times 2+.4\times 1 =1.6$.
Recursively, we see that, for $n≥3$,  $$E_n=1+.6E_{n-1}+.3E_{n-2}+.1E_{n-2}$$
Since the associated cubic does not have pleasant roots, the closed form for the solution is not pleasant.  It is, however, easy to compute $E_n$ for small $n$ and one gets $$\boxed {E_{10}=6.933335296}$$  matching the result obtained by @Henry.
