We have a normal 6 sided dice marked from 1 to 6. There is an equal chance to get each number at every roll. Let's put 1&2 as A type, 3&4 as B type and 5&6 as C type.
We roll the dice over and over until we get a number of every type.
Let X be the number of rolls.
We are interested in the mean of X.
I see three geometric distributions in there.
Let Y be a Geometric Random Variable and p the probability of success:
P(Y=x) = p(1-p)^(x-1)
Mean(Y) = 1/p
The Attempt at a Solution
I am not sure if my reasoning is legit. I split an experiment into 3 smaller ones and added their mean.
I split the problem into three steps.
Let E1: Number of rolls untill the first type appears.
Let E2: Number of rolls untill the second type appears.
Let E3: Number of rolls untill the third type appears.
Here p = 1. The first roll to eliminates a first type.
Mean(E1) = 1
Since a first type is out, The new p for either of the other two types of appeating are 4/6= 2/3.
Mean(E2) = 1 / p = 1 / (2/3) = 1.5
The new p is now 2/6 = 1/3.
Mean(E3) = 1 / p = 1 / (1/3) = 3
We can say that the Mean(X) = 1 + 1.5 + 3 = 5.5
We wait on average 5.5 rolls before the three types apprears.