Problem with expectation value and dice

[MrAnonymous]

Problem:
Let X be the number of D6 throws needed to get 13 times the number 6 in a row.
Find E(X).

My work so far:
X must be larger than or equal to 13, as this is the minimum of throws needed to satisfy the condition.
The first three possible values of X are 13, 14, 15, with their probabilities listed below:
P(X=13)=(1/6)^13
P(X=14)=5/6 x (1/6)^13
P(X=15)=6/6 x 5/6 x (1/6)^13 = 5/6 x (1/6)^13
All values for X up to and including X=26 have this last probability.

Then comes X=27.
Here we need to make sure that the first 13 dice rolls are not all sixes. To do this, simply add one factor of 5/6:
P(X=27)=(5/6)^2 x (1/6)^13
We then continue:
P(X=28)=(5/6)^2 x (1/6)^13
...
P(X=39)=(5/6)^2 x (1/6)^13

For the next one, we need to include another factor of 5/6, as there are now more than 3 sequences of 13 sixes possible:
P(X=40)=(5/6)^3 x (1/6)^13

Note: I'm not entirely sure if X=40 is the next one to include another factor of 5/6 - it might be X=40. This means, from here on out, I'm not sure at all if what I'm doing is correct.

The sequence like this.

We can conclude a general formula:
P(X=13n+k)=(5/6)^n x (1/6)^13, where n is a positive integer and 1 <= k <= 13.

And this is where I'm stuck. How do I get to E(X) from here?
Please help.

 

[Steven G]

Problem:
Let X be the number of D6 throws needed to get 13 times the number 6 in a row.
Find E(X).

My work so far:
X must be larger than or equal to 13, as this is the minimum of throws needed to satisfy the condition.
The first three possible values of X are 13, 14, 15, with their probabilities listed below:
P(X=13)=(1/6)^13
P(X=14)=5/6 x (1/6)^13
P(X=15)=6/6 x 5/6 x (1/6)^13 = 5/6 x (1/6)^13
All values for X up to and including X=26 have this last probability.

Then comes X=27.
Here we need to make sure that the first 13 dice rolls are not all sixes. To do this, simply add one factor of 5/6:
P(X=27)=(5/6)^2 x (1/6)^13
We then continue:
P(X=28)=(5/6)^2 x (1/6)^13
...
P(X=39)=(5/6)^2 x (1/6)^13

For the next one, we need to include another factor of 5/6, as there are now more than 3 sequences of 13 sixes possible:
P(X=40)=(5/6)^3 x (1/6)^13

Note: I'm not entirely sure if X=40 is the next one to include another factor of 5/6 - it might be X=40. This means, from here on out, I'm not sure at all if what I'm doing is correct.

The sequence like this.

We can conclude a general formula:
P(X=13n+k)=(5/6)^n x (1/6)^13, where n is a positive integer and 1 <= k <= 13.

And this is where I'm stuck. How do I get to E(X) from here?
Please help.

You must get at the end, not 6, followed by 13 6's or just 13 6's in 13 throws. 13 6's in 13 throws can be done in (1/6)¹³ ways. Not 6 followed by 13 6's can be done in (5/6)(1/6)¹³ ways..

So what happens if you need more than 14 throws?. Let K>13, then P(K throws) = (6/6)^K-14(5/6)(1/6)¹³ =(5/6)(1/6)¹³