Finding a joint probability mass function I have to find the joint probability mass function (pmf) of (X,Y) for the following problem: 
Roll a die repeatedly until a five or six appears, and let X be the number of rolls before a five or six appears. Then toss a coin X times, and let Y be the number of heads in these tosses.
So far, I have that X has a Geometric distribution with parameter 1/3. I think that I should then use the formula for a conditional pmf, which will allow me to get the joint pmf by rearrangement.
So I would then need to find an expression for the conditional pmf for (Y|X), which is where I am stuck. I have considered using that the conditional pmf (Y|X) has a Binomial distribution with parameters x and 1/2, but this lead to the following expression for the joint pmf which I am not sure is correct:
$$ p_{(X,Y)}(x,y) = \frac{1}{3^{x+1}}\binom{x}{y} $$
Is it correct to say that the conditional pmf has a Binomial distribution, or is it distributed otherwise?
 A: Verified.
Indeed $X$ is geometrically distributed and $Y$ is conditionally binomially distributed given $X$. $$X~\sim~\mathcal{Geom}_0(1/3) \\ Y\mid X ~\sim~ \mathcal{Bin}(X, 1/2)$$
Then you do have $p_X(x) = \dfrac {2^{x}}{3^{x+1}}$ and $p_{Y\mid X}(y\mid x) = \dbinom{x}{y}\dfrac 1{2^x}$, so indeed:
$$p_{X,Y}(x,y) = \dbinom{x}{y}\dfrac 1 {3^{x+1}} \quad\Big[x\in\{0,...\}, y\in \{0, .., x\}\Big]$$


From this, how would one proceed in finding the pmf for Y? I know that I need to sum this over all x, but I couldn't find a way to proceed, hence why I wasn't certain if what I found for the joint pmf was correct. 

Well, you could attempt to find a closed form for $p_Y(y) = \sum\limits_{x=y}^\infty \binom{x}{y}3^{-1-x}$
However this will give the same result as the following combinatorial argument:
A trial consists of a roll of the die and, if that doesn't end the run by showing 5 or 6, a coin toss.   So each trial has an equal chance of being an end, a head, or a tail event.   $Y$ is the count of heads before the (first) end.   However, if we simply ignore all the tails we will have an equal chance of heads or end.
$Y$ is the count of heads before the (first) end in all the not-tails trials.
$$Y~\sim~\mathcal{Geom}_0(1/2)$$
