another riddle with a stopping rule

A puzzle on The Riddler last week that is rather similar to an earlier one. Given the probability (1/2,1/3,1/6) on {1,2,3}, what is the mean of the number N of draws to see all possible outcomes and what is the average number of 1’s in those draws? The second question is straightforward, as the proportions of 1’s, 2’s and 3’s in the sequence till all values are observed remain 3/6, 2/6 and 1/6. The first question follows from the representation of the average

as the probability to exceed n is the probability that at least one value is not observed by the n-th draw, namely

3+(1/2)ⁿ+(2/3)ⁿ+(5/6)ⁿ-(1/6)ⁿ-(1/3)ⁿ-(1/2)ⁿ

which leads to an easy summation for the expectation, namely

3+(2/3)³/(1/3)+(5/6)³/(1/6)-(1/3)³/(2/3)-(1/6)³/(5/6)=73/10

Checking the results hold is also straightforward:

averages <- function(n=1){
    x=matrix(sample(1:3,100,rep=TRUE,prob=1:3),100,3)
    x[,1]=as.integer(x[,2]<2) x[,3]=as.integer(x[,2]>2)
    x[,2]=1-x[,1]-x[,3]
    y=apply(apply(x,2,cumsum),1,prod)
    m=1+sum(y==0)
    return(apply(x[1:m,],2,sum))}

since this gives

mumbl=matrix(0,1e5,3)
for (t in 1:1e5) mumbl[t,]=averages()
> apply(mumbl,2,mean)
[1] 1.21766 2.43265 3.64759
> sum(apply(mumbl,2,mean))
[1] 7.2979
> apply(mumbl,2,mean)*c(6,3,2)
[1] 7.30596 7.29795 7.29518

(The puzzle this week is not intelligible for someone foreign to baseball rules so I will not mention it next week!)