MrAnonymous
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- Dec 1, 2018
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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.
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.