- #1

AllRelative

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## Homework Statement

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.

## Homework Equations

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 E

_{1}: Number of rolls until the first type appears.

Let E

_{2}: Number of rolls until the second type appears.

Let E

_{3}: Number of rolls until the third type appears.

1)

Here p = 1. The first roll to eliminates a first type.

Mean(E

_{1}) = 1

2)

Since a first type is out, The new p for either of the other two types of appeating are 4/6= 2/3.

Therefore,

Mean(E

_{2}) = 1 / p = 1 / (2/3) = 1.5

3)

The new p is now 2/6 = 1/3.

So

Mean(E

_{3}) = 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.