Toss a die until SIX appears. E (number of ONEs | at leat one TWO) =? A fair die is tossed repeatedly until a Six appears. Let $X$ denote the number of One's
that are thrown in this game/experiment. Let $A$ denote the event that at least one Two was thrown. Compute$ E(X | A)$.
This questions puzzles me because the number of tosses itself is a random variable. I was trying to use $E(X) = E(X|A)Pr(A) + E(X|A^C)Pr(A^C) $. This is how I compute $Pr(A^C)$:
$Pr(A^C) = \sum_{k=1}^\infty Pr(A^C|N=k) Pr(N=K) = \sum_{k=1}^{\infty} (1-\frac{2}{6})^{k-1}(\frac{5}{6})^{k-1}(\frac{1}{6}) = \frac{3}{8}$. However, my notes says $Pr(A^C) = \sum_{k=1}^\infty \frac{1}{6} (1 - \frac{2}{6})^{k-1}=\frac{1}{2} $. Where did I (or less likely, the TA) get it wrong? Thanks!
 A: Actually this answer is more than you ask for. Maybe you can use it to check your own answer.
Let $N$ denote the number of tosses before the first Six appears.
We will find $\mathbb{E}\left(X\mid A\right)$ on base of: 
$$\mathbb{E}X=\mathbb{E}\left(X\mid A\right)P\left(A\right)+\mathbb{E}\left(X\mid A^{c}\right)P\left(A^{c}\right)=\mathbb{E}\left(X\mid A\right)\left(1-P\left(A^{c}\right)\right)+\mathbb{E}\left(X\mid A^{c}\right)P\left(A^{c}\right)$$
If we manage to calculate $\mathbb{E}X$, $P\left(A^{c}\right)$ and
$\mathbb{E}\left(X\mid A^{c}\right)$ then we are actually ready. 
$N$ takes values in $\left\{ 0,1,2,\dots\right\} $ with $P\left[N=n\right]=\left(\frac{5}{6}\right)^{n}\frac{1}{6}$.
If $B$ denotes the event that the first toss gives a Six then
$\mathbb{E}N=\mathbb{E}\left(N\mid B\right)P(B)+\mathbb{E}\left(N\mid B^{c}\right)P(B^{c})=0\times\frac{1}{6}+\left(1+\mathbb{E}N\right)\times\frac{5}{6}$
leading to $\mathbb{E}N=5$. 
Looking at $X$ under condition $N=n$ we deal clearly with a binomial
distribution having parameters $n$ and $\frac{1}{5}$ so $\mathbb{E}X=\mathbb{E}\left(\mathbb{E}\left(X\mid N\right)\right)=\mathbb{E}\frac{1}{5}N=1$.
Looking at $\left(X\mid A^{c}\right)$ under condition $N=n$ again
we deal with a binomial distribution, this time with parameters $n$
and $\frac{1}{4}$ wich likewise leads to $\mathbb{E}\left(X\mid A\right)=\frac{5}{4}$. We can calculate $P\left(A^{c}\right)$ by:
$$P\left(A^{c}\right)=\sum_{n=0}^{\infty}P\left(A^{c}\mid N=n\right)P\left(N=n\right)=\sum_{n=0}^{\infty}\left(\frac{4}{5}\right)^{n}\times\left(\frac{5}{6}\right)^{n}\frac{1}{6}=\frac{1}{2}$$
However, there is a simpler route. Realize that $A^c$ stands for the event that  'the Six appears before the Two appears'. Off course it has the same probability as its complement $A$: 'the Two appears before the Six appears'.
So $P(A)=P(A^c)$ and $P(A)+P(A^c)=1$ leading to $P(A^c)=\frac{1}{2}$. Easy isn't it? :-)
Based on $1=\mathbb{E}\left(X\mid A\right)\times\frac{1}{2}+\frac{5}{4}\times\frac{1}{2}$
we find: 

$$\mathbb{E}\left(X\mid A\right)=\frac{3}{4}$$

