If a trial consists of throwing an n-sided fair die having numbers a1,a2,a3…,an on its faces. What will be the expected number of trials required before we get atleast k1 times a1, k2 times a2,….kn times an.
I think it can be modelled as the expected value of negative multinomial distribution because each individual follows a multinomial distribution. In the simpler case where the trial is binomial, we can model "The expected number of trials required before we get k successes" as negative binomial.
An example for understanding…suppose that there is 3 sided dice with numbers 1,2 and 3 and I want to know the expected number of trials before I get to see say 4 1s, 5 2s and 6 3s.
PS: I cannot find any good free resource available on the net on Negative multinomial distributions
Best Answer
I'll just treat your example. I'm not sure how difficult it would be to express this in closed form; I haven't tried. We must distinguish between the cases where the last throw, the one that fulfills all the conditions, is a $1,2,$ or $3$. Suppose it is a $1$. Then we know that we threw $k\geq5$ $2$'s and $j\geq6$ $3$'s and that in the $k+j+3$ rolls prior to the last we rolled exactly $3$ $1$'s. We can make similar analyses when the last roll was a $2$ or a $3$. The expected number of rolls is $$\sum_{k=5}^\infty\sum_{j=6}^\infty(k+j+4){k+j+3\choose3,k,j}3^{-(k+j+4)}+\\ \sum_{i=4}^\infty\sum_{j=6}^\infty(i+j+5){i+j+4\choose4,i,j}3^{-(i+j+5)}+\\ \sum_{i=4}^\infty\sum_{k=5}^\infty(i+k+6){i+k+5\choose5,i,k}3^{-(i+k+6)} $$ where, of course, the first sum deals with the case where a $1$ is rolled last, the second where $2$ is last, and the third where $3$ is last.