"Two Cards are picked from a deck with replacement. Let X= number of aces, and Y= number of kings.

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

I'm trying to show whether or not X an Y are independent in the case that the sampling is done with replacement. I first assumed that X and Y are independent simply because replacement was involved but my calculations showed otherwise.

edit: Confirmed in the comments that X and Y are not independent. Can someone please provide a proof that X and Y are dependent and explain the error in the following attempt a proof that they are independent:

$P(X,Y) = P(X \cap Y) = P(X|Y)\times P(Y) $

but $P(X|Y) = P(X)$ because the sampling is done with replacement

so $P(X,Y) = P(X)\times P(Y)$ thus X and Y are independent.

In particular, can anyone explain why this particular scenario of drawing two cards with the random variables as they are defined makes X and Y dependent even though they are sampled with replacement? I understand now that how many X are selected changes the probability of how many Y are selected and I can sense that this has to do with the constraint of how many cards are selected in the event, i.e. the more of one type that are selected changes not the probability of selecting the second type but instead, it changes the amount of the second type that can be selected.

And since the random variable is defined as the amount of the card selected, its clear that this is the heart of the issue. However, I'm having trouble articulating or reasoning further about why this makes sampling with replacement or without replacement irrelevant to the question of dependence/independence of the variables. If anyone could shed light from this angle, I would greatly appreciate it.

Reference: another post asks about the joint probability function of this situation without replacement.

  • 1
    $\begingroup$ $0=Pr(X=2,Y=2)\neq Pr(X=2)Pr(Y=2) > 0$ so $X$ and $Y$ are not independent, no calculations necessary beyond recognizing that it is impossible to get two kings and two aces simultaneously when drawing two cards, but it is possible (occurs with nonzero probability, that is all we care at the moment) to get two kings by themselves as well as it is possible to get two aces by themselves. $\endgroup$
    – JMoravitz
    Nov 3 '18 at 15:31
  • $\begingroup$ @JMoravitz Yes, something along these lines dawned me as I was re-reading my post. $ Pr(Y=1|X=2) = 0 $ but $Pr(Y=1)>0 $ and $Pr(X=2)>0)$. $\endgroup$
    – Elfen Dew
    Nov 3 '18 at 15:37
  • $\begingroup$ So, can you confirm what your actual question is? Is your question asking for confirmation that $X$ and $Y$ are in fact dependent variables? They are as showed by our first comments. Are you asking why your friend is wrong about them being independent? His line "$Pr(X\mid Y)=Pr(X)$" is incorrect and unjustified, as shown again by our earlier comments. Is your question what the joint pdf is? That was answered in the linked question in your post. What else might your question be? $\endgroup$
    – JMoravitz
    Nov 3 '18 at 15:43
  • $\begingroup$ $Pr(X\mid Y)$ would have been equal to $Pr(X)$ and $X$ and $Y$ would have been independent since sampling was done with replacement in the case that $X$ had to do only with the first draw and $Y$ had to do only with the second draw. That is not the case here. $\endgroup$
    – JMoravitz
    Nov 3 '18 at 15:44
  • $\begingroup$ Ah, yes sorry I haven't been sleeping well and am not thinking clearly. I will edit my post to make my question clear and state it here: I am asking for a demonstration/explanation of the independence or dependence of X and Y and what is the error in the "proof" of independence. $\endgroup$
    – Elfen Dew
    Nov 3 '18 at 15:52

The lack of independence is intuitively clear. If you got two aces in your two picks then you have a lot of information about the number of kings. So the ace count and the king count are not independent.

For me this is a convincing "proof" of dependence. It may not be enough for you or your instructor. I leave it to others to find the error in your formal argument.

  • $\begingroup$ @kimchilover Yes if course, Fixed thanks (you could have.) $\endgroup$ Nov 3 '18 at 17:29
  • $\begingroup$ Thanks for answering. $\endgroup$
    – Elfen Dew
    Nov 6 '18 at 6:50

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