# Conitional Probability

• Apr 6th 2010, 06:14 AM
jjbrian
Conitional Probability
A pair of dice is thrown repeatedly until the total obtained on the first throw is obtained again. What is the expected number of throws necessary?

The answer from book is 12.

Here is my reasoning

There are 11 different totals for a pair of dice (you can't roll a 1)
(2,3,4,5,6,7,8,9,10,11,12)

11 possibilities plus the 1 original throw = 12

Is there a formal way to do this?(Can this be solved by using a specific formula?)
• Apr 6th 2010, 12:01 PM
Failure
Quote:

Originally Posted by jjbrian
A pair of dice is thrown repeatedly until the total obtained on the first throw is obtained again. What is the expected number of throws necessary?

The answer from book is 12.

Here is my reasoning

There are 11 different totals for a pair of dice (you can't roll a 1)
(2,3,4,5,6,7,8,9,10,11,12)

11 possibilities plus the 1 original throw = 12

Is there a formal way to do this?(Can this be solved by using a specific formula?)

What about the following: let $N$ be the number of throws until the first throw appears again, then the expected value of N is
$\mathrm{E}(N)=\sum_{i=2}^{12}\mathrm{E}(N|\text{fi rst throw was i})\cdot \mathrm{P}(\text{first throw was i})=\ldots$
As you can see, it gets a little more complicated than you might have expected...