Set of n dice are thrown...

Question

I need help with this question:

A set of n dice is thrown. All those that land on six are put aside and the others are again thrown. This is repeated until all the dice have landed on six. Let N denote the number of throws needed. Let $m_{n} = E[N]$.

Let $X_{i}$ denote the number of dice rolled on the ith throw. Find $E[\sum_ {i=1}^{N} X_{i}]$.

Here is a solution I was provided with but do not understand the reasoning behind it:

$E[X_{i}] = E[E[X_{i}|X_{i-1}]] = E[X_{i-1} - (X_{i-1}*(1/6))] = (5/6)*E[X_{i-1}] = n(5/6)^{i-1}$

Then, $E[\sum_ {i=1}^{N} X_{i}] = \sum_ {k=1}^{n} ([1-(5/6)^{k}]/(1/6))(k)P(N=k)]$.

Where, $P(N=1)=(5/6)^{n}$ and $P(N=2) = \sum_ {X=0}^{n-1} \binom {n}{x} (1/6)^{n} (25/36)^{n-x} = (31/36)^{n} - (1/6)^{n}$.

Could someone please explain this too me? I appreciate any help. Thanks

Answer 1

This is a double counting question. Note that $$\sum_{i=1}^N X_i = \sum_{i=1}^n N_i,$$ where $N_i$ is the number of times the $i$th die is rolled.

Try to finish the solution from here.

The answer is $6n$.