Let $X$ be the number of aces and $Y$ be the number of spades. Show that $X$, $Y$ are uncorrelated. A deck of 52 cards is shuffled, and we deal a bridge of 13 cards. Let $X$ be the number of aces and $Y$ be the number of spades. Show that $X$, $Y$ are uncorrelated.
Here is what I did:
$Cov(X,Y) = E[XY]-E[X]E[Y]$ 
uncorrelated means $Cov(X,Y) = 0$, hence  $E[XY]=E[X]E[Y]$ 
$E[X] =  \sum_{k=0}^{k=4} k \frac{\dbinom{4}{k} \dbinom{48}{13-k}}{\dbinom{52}{13}} $ 
$E[Y] =  \sum_{k=0}^{k=13} k \frac{\dbinom{13}{k} \dbinom{39}{13-k}}{\dbinom{52}{13}} $ 
Are the summations above correct? and how do I calculate $E[XY]$?
 A: We show that although $X$ and $Y$ are not independent, the conditional expectation of $Y$, given $X=x$, is equal to the plain expectation of $Y$.  
Given that $x=0$ (no Aces), we are choosing $13$ cards from the $48$ non-Aces. The expected number of spades is then $13\cdot \frac{12}{48}=\frac{13}{4}$.
Given that $x=1$ (one Ace), there are two possibilities: (i) the Ace is a spade or (ii) it is not. 
(i) If the Ace is a spade (probability $\frac{1}{4}$), then we have $1$ assured spade. In addition, we are choosing $12$ cards from the $48$ non-spades, so the expected number of additional spades is $12\cdot\frac{12}{48}$. Thus (i) makes a contribution of $\frac{1}{4}\cdot\left(1+12\cdot\frac{12}{48}\right)$ to the conditional expectation.   
(ii) If the Ace is a non-spade (probability $\frac{3}{4}$), the expected number of spades is $12\cdot \frac{12}{48}$. Thus
$$E(Y|X=2)= \frac{1}{4}\cdot\left(1+12\cdot\frac{12}{48}\right)+\frac{3}{4}\left(12\cdot \frac{12}{48}  \right).$$
This simplifies to $\frac{13}{4}$.
A similar analysis works for $X=2$, $3$, and $4$. For instance, if $x=2$, then with probability $\frac{1}{2}$ the Ace of spades is included among the two Aces, and with probability $\frac{1}{2}$ it is not. The analogue of the calculation we made for $x=1$ yields
$$E(Y|X=2)= \frac{1}{2}\cdot\left(1+11\cdot\frac{12}{48}\right)+\frac{1}{2}\left(11\cdot \frac{12}{48}  \right),$$
which again simplifies to $\frac{13}{4}$.
A: The sums for $E[X]$ and $E[Y]$ look correct. To get $E[XY]$ we multiply $k\cdot t$ by the probability of $k$ aces and $t$ spades, and sum over the $5\cdot 14=70$ pairs $(k,t)$ with $0 \le k \le 4$ and $0 \le t \le 13.$ If that calculation turns out to be $kt$ multiplied by your density for $k$ times the density for $t$ obtained by replacing $k$ by $t$ in the right side of your $E[Y]$ formula (without the initial $k$ before the binomial fraction), then you are finished, since one may then factor the summand into two factors, one depending only on $k$ and the other only on $t$, making the double sum equal the product $E[X]E[Y]$.
I think the hard part will be actually showing the probability of $k$ aces and $t$ spades is indeed the above product expression. 
A: An argument from logic rather than mathematics.
If you tell me a card is a spade, I gain no improvement in my knowledge that it is an Ace and vice-versa, therefore they are urcorrelated.
Contrast this with telling me if a person is male or female on my knowledge as to if they are a mother or a father and vice versa.
hint for the maths
Treat the Ace of spades as a special case, there are therefore 3 other Aces and 12 other spades.
