<s>[INST] Roll a fair die and denote the outcome by $Y$ . Then flip $Y$ fair coins and let $X$ denote the number of tails observed. 
Roll a fair die and denote the outcome by $Y$. Then flip $Y$ fair coins and let $X$ denote the number of tails observed.

*

*Find the probability mass function of $X$

*Compute the expectation of $X$

My Attempt
The number of possible combinations of $y$ coins is $2^y$  and the number of combinations with $x$ tails is $\binom{y}{x}$. Thus, the probabilities are $p_{X,Y}(x,y)=\frac{\binom{y}{x}}{2^y}$. Then the joint p.m.f. is
\begin{array}{c|c|ccc}
& & & & Y & & & \\ \hline
& & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline
& 1 & 1/2 &\ 1/2 & 3/8 & 1/4 & 5/32 & 3/32 \\ 
& 2 & 0 & 1/4 & 3/8 & 3/8 & 5/16 & 15/64 \\ 
X& 3 & 0 & 0 & 1/8 & 1/4 & 5/16 & 5/16 \\ 
& 4 & 0 & 0 & 0 & 1/16 & 5/32 & 15/64\\ 
& 5 & 0 & 0 & 0 & 0 & 1/32 & 3/32\\ 
& 6 & 0 & 0 & 0 & 0 & 0 & 1/64\\ 
\end{array}
This is my thinking but I know that it is wrong because the sum of all the probabilities exceeds $1$. Can I get some pointers on where I went wrong?
After I find the joint p.m.f., I can sum the values in the rows to find the p.m.f. for $X$.
 [/INST] If Y represents the outcome of a fair die and X represents the number of tails observed then,
$ \ 0 \leq x \leq y$ and $1 \leq y \leq 6 \ $
$ \displaystyle P(X = x|Y=y) = {y \choose x} \cdot \frac{1}{2^y}$
So the table that you have put together is correct but you missed $x = 0$, probability of which is same as for $x = y$.
Edit:
To find pmf of X,
$ \displaystyle P(X = x) = \frac{1}{6} \sum \limits_{y=1}^6 P(X = x|Y=y) \ \ (0 \leq x \leq 6)$
To find expectation of X,
$\mathbb{E}(X) = \sum \limits_{x=0}^6 x \cdot P(X = x)$
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