Rolling dice until 6 appears, X is a number of trials, while Y is the number of odds in the resulting sequence.

Question

The main question, which I do not understand is that how we can get conditional expectation of $E(Y \mid X)$ and what is the probability $P(Y=y \mid X)$. Idea to use binomial for the probability $Y \sim \text{Binom}(X,3/6)$

Answer 1

Given that $X$ was realised as a natural number $x$, you rolled the die $x-1$ times and failed to get a six those $n-1$ times, i.e. all those rolls were from one to five. In that range are three odd and two even numbers, so $Y$ follows the binomial distribution with $x-1$ trials and success probability $\frac35$: $$P(Y=y\mid X=x)=\binom{x-1}y(3/5)^y(2/5)^{x-y-1}$$ The conditional expectation is $E(Y\mid X=x)=\frac35(x-1)$.