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 $E[XY]$?


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 $\frac{13}{4}$.


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.

  • $\begingroup$ sorry, but I still don't understand your hints. Can you give me a calculation formula or something like that for me to start? $\endgroup$
    – user59036
    Mar 24 '13 at 19:42
  • $\begingroup$ @user59036 : I have looked at the setup for a valid calculation of $E[XY]$ but it's quite involved and there are a lot of separate cases. One has to keep track of whether the ace of spades is in the hand or not, and also make sure the sum $X+Y$ doesn't conflict with the hand having only 13 cards. I see no simple way to do it. $\endgroup$
    – coffeemath
    Mar 24 '13 at 22:26

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.

  • 1
    $\begingroup$ This argument looks dubious to me. For instance, if I know you have three Aces, then that tells me that you can have at most eleven Spades. Or if I know you have twelve Spades, then that tells me that you must have at most two Aces. What am I missing here? $\endgroup$
    – TonyK
    Mar 24 '13 at 12:09
  • $\begingroup$ You are looking at this "after the fact". You are posing a different question: given I have 3 Aces, what is my probability of 11 spades. These are indeed correlated BUT it is not the same problem. $\endgroup$
    – Dale M
    Mar 24 '13 at 20:00
  • $\begingroup$ Put another way, the problem as stated requires the 13cards to be revealed all at once, not one by one. That is why draw and stud poker have different odds. $\endgroup$
    – Dale M
    Mar 24 '13 at 20:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.