A coin is tossed as long as it shows heads. Let $X$ be the number of heads before the first tail. After that you throw a dice $X$ times. Let $Y$ be the number of sixes.
calculate
$E[Y]$
I've conditioned on X; in this manner $E[E[Y\mid X]]$ . where i found that $Y\mid X = x$ is binomial distributed - $bin(x,1/6)$ . And so $E[Y\mid X = x] = x \times 1/6$. Letting $x$ be random we get $E[X \times 1/6]$. finally i obtain $E[Y] = E[X]\times 1/6$. Where $X$ is a geometric random variable. , However this seem to be wrong according to my book where they have used that $X+1$ is geometric with parameter $0.5$. i don't really se where they get that extra $1$ from?
You did not show your full computation, since you omitted showing your calculation of $E(X)$. So I do not know whether the calculation was correct.
It turns out that $E(X)=1$. To do this, we use the standard fact that the number of trials until the first "success" (including the trial when we got the success) has mean $\frac{1}{p}$, where $p$ is the probability of success on any trial. So the number of tosses up to and including the first tail has mean $2$, and therefore the number of heads before the first tail has mean $2-1$.