Conditional Expectation

[marcmtlca]

Suppose we have a di and we let X be the number of rolls before you get a 5. and Y be the number of rolls before you get a 6. I am supposed to find E[X].

Now, the problem is that X and Y are not independent, so I can't just use the geometric distribution properties and say it is 6. What I tried to do was use E[X]=E[E[X|Y]]. I am pretty sure there is an easier way, and also I think I may have done it wrong since my answer is too complicated. my writeup is here:

http://marcandre.ifastnet.com/stochas1.pdf

If anyone wants to give it a try, or take a look at what I did, it would be appreciated.

THanks.

 

[pka]

Suppose that N is the number of tosses before a 5 appears.
Then \(\displaystyle P(N) = \left( {\frac{5}{6}} \right)^{N - 1} \left( {\frac{1}{6}} \right) = \frac{{5^{N - 1} }}{{6^N }}.\)
To find E(N) we note: \(\displaystyle E(N) = \sum\limits_{k = 1}^\infty {kP(k)}.\)

From calculus:
\(\displaystyle \begin{array}{l}
\sum\limits_{k = 1}^\infty {x^k } = \frac{x}{{1 - x}}\quad ,\quad \left| x \right| < 1 \\
\sum\limits_{k = 1}^\infty {kx^{k - 1} } = \frac{1}{{\left( {1 - x} \right)^2 }} \\
\sum\limits_{k = 1}^\infty {kx^k } = \frac{x}{{\left( {1 - x} \right)^2 }} \\
\end{array}\)

Now let complete the problem.

 

[marcmtlca]

yeah I am probably overthinking this problem.