# X = # of times u need to roll to get the first 5. E(X ∣ X > 2) = (1/(1/6)) + 2 Why?

• Aug 22nd 2010, 05:46 PM
CaptainLorax
X = # of times u need to roll to get the first 5. E(X ∣ X > 2) = (1/(1/6)) + 2 Why?
Hello, this is my first time posting here. I am studying for the P1 actuarial exam
Thank you for any help :)

X = # of times you need to roll to get the first 5.
E(X ∣ X > 2) = (1/(1/6)) + 2

How do I get this ^ solution ^
• Aug 22nd 2010, 07:31 PM
CaptainBlack
Quote:

Originally Posted by CaptainLorax
Hello, this is my first time posting here. I am studying for the P1 actuarial exam
Thank you for any help :)

X = # of times you need to roll to get the first 5.
E(X ∣ X > 2) = (1/(1/6)) + 2

How do I get this ^ solution ^

Lets assume you are talking rolling a fair die. The prior history of rolls does not effect the future outcomes, so:

$E(X|X>2)=E(X)+2$

and:

$\displaystyle E(X)=\sum_{k=1}^{\infty} k \left(\frac{5}{6}\right)^{k-1}\frac{1}{6}=\frac{1}{6} \sum_{k=1}^{\infty} k \left(\frac{5}{6}\right)^{k-1}$

CB
• Aug 22nd 2010, 08:36 PM
CaptainLorax
thank you very much
• Aug 22nd 2010, 10:10 PM
CaptainLorax
actually I still don't understand how the end of your equation = 1/(1/6)
• Aug 24th 2010, 02:19 AM
CaptainBlack
Quote:

Originally Posted by CaptainLorax
actually I still don't understand how the end of your equation = 1/(1/6)

To get the first 5 on the n-th roll requires that you roll n-1 non-fives with probability 5/6 and the last roll is a 5 with probability 1/6

CB
• Aug 24th 2010, 11:16 AM
CaptainLorax
now I just have to figure out how to send a virtual thanks your way
edit-found it