2 Cards are picked from a deck without replacement. Let X= number of aces, and Y= number of kings. Find the joint probability function (in a 3x3 table)

X and Y are both discrete random variables that can take on 0,1 and 2.

There are 5 choose 2 (1326) ways to pick 2 cards. The probability of getting an ace are $\frac{4}{52}$ (also the probability of getting a king).

How can I use this to find the joint probability function in a table? What kind of probability distribution can I use?

Edit: would the technique used here be similar to the question answered here: 5 cards / Joint Probability Function

I've also filled out three blocks in the table, since you can't have 2 aces and 2 kings, or 2 kings 1 aces, or 2 aces 1 king.

  • 2
    $\begingroup$ $\forall{x,y\in\mathbb{Z}}:x+y\in[0,2] \implies P([X=x]\wedge[Y=y]) = \frac{\binom4x\cdot\binom4y\cdot\binom{52-4-4}{2-x-y}}{\binom{52}{2}}$ $\endgroup$ May 25, 2016 at 15:09

1 Answer 1


Split up like $52=4+4+44$, i.e. $4$ kings $4$ aces and the other $44$ cards.

In total $N:=\binom{52}2$ different pair-picks are possible and equiprobable.

  • $\binom40\binom40\binom{44}2$ of them give $(K,A)=(0,0)$
  • $\binom40\binom41\binom{44}1$ of them give $(K,A)=(0,1)$
  • $\binom41\binom40\binom{44}1$ of them give $(K,A)=(1,0)$
  • $\binom40\binom42\binom{44}0$ of them give $(K,A)=(0,2)$
  • $\binom41\binom41\binom{44}0$ of them give $(K,A)=(1,1)$
  • $\binom42\binom40\binom{44}0$ of them give $(K,A)=(2,0)$

You can find the probabilities by dividing the outcomes by $N$.

edit (alternative):

Pick two cards one by one.

For $i=1,2$ let $A_{i},K_{i},E_{i}$ denote the events that the $i$-pick gives ace, king and other card respectively.

$P\left(A_{1}\cap A_{2}\right)=P\left(A_{1}\right)P\left(A_{2}\mid A_{1}\right)=\frac{4}{52}\frac{3}{51}$

$P\left(A_{1}\cap K_{2}\right)=P\left(A_{1}\right)P\left(K_{2}\mid A_{1}\right)=\frac{4}{52}\frac{4}{51}$

$P\left(A_{1}\cap E_{2}\right)=P\left(A_{1}\right)P\left(E_{2}\mid A_{1}\right)=\frac{4}{52}\frac{44}{51}$

$P\left(K_{1}\cap A_{2}\right)=P\left(K_{1}\right)P\left(A_{2}\mid K_{1}\right)=\frac{4}{52}\frac{4}{51}$

$P\left(K_{1}\cap K_{2}\right)=P\left(K_{1}\right)P\left(K_{2}\mid K_{1}\right)=\frac{4}{52}\frac{3}{51}$

$P\left(K_{1}\cap E_{2}\right)=P\left(K_{1}\right)P\left(E_{2}\mid K_{1}\right)=\frac{4}{52}\frac{44}{51}$

$P\left(E_{1}\cap A_{2}\right)=P\left(E_{1}\right)P\left(A_{2}\mid E_{1}\right)=\frac{44}{52}\frac{4}{51}$

$P\left(E_{1}\cap K_{2}\right)=P\left(E_{1}\right)P\left(K_{2}\mid E_{1}\right)=\frac{44}{52}\frac{4}{51}$

$P\left(E_{1}\cap E_{2}\right)=P\left(E_{1}\right)P\left(E_{2}\mid E_{1}\right)=\frac{44}{52}\frac{43}{51}$

As an example note that e.g. $$P(A=1,K=1)=P(A_1\cap K_2)+P(K_1\cap A_2)=\frac{4}{52}\frac{4}{51}+\frac{4}{52}\frac{4}{51}$$

  • $\begingroup$ Where does the 48 come from? Shouldn't it be 52-8 = 44 $\endgroup$
    – user300011
    May 25, 2016 at 14:44
  • $\begingroup$ Slip of the pen. Repaired. $\endgroup$
    – drhab
    May 25, 2016 at 14:45
  • $\begingroup$ Should all of those add up to 1? @drhab $\endgroup$
    – user300011
    May 25, 2016 at 15:06
  • $\begingroup$ Edit: yep they do! THANKS AGIAN! $\endgroup$
    – user300011
    May 25, 2016 at 15:14

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