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Five cards are randomly chosen from a standard deck, one at a time with replacement. Let $X$, $Y$, $Z$ be the numbers of chosen queens, kings, and other cards.

(a) Find the joint PMF of X, Y, Z.

(b) Find the joint PMF of X and Y .

a) The joint PMF of $X$, $Y$ and $Z$ $$ P(X=x,Y=y,Z=z) = \left(\frac{1}{13}\right)^{x+y} \left(\frac{11}{13}\right)^z, $$ where the support is $\{(x,y,z) \in \mathbb{Z} \mid x+y+z = 5 \text{ and } x,y,z \geq 0 \}$.

b) To find the joint PMF of $X$ and $Y$ alone, I marginalize out $Z$.

$$ P(X=x,Y=y) = \left(\frac{1}{13}\right)^{x+y} \sum_{z=0}^{5-x-y} \left(\frac{11}{13}\right)^z = \left(\frac{1}{13}\right)^{x+y} \frac{1-\left(\frac{11}{13}\right)^{6-x-y}}{1-\frac{11}{13}} $$

Is this correct?

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1 Answer 1

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You are dealing here with multinomial distribution and 'forgot' the coefficients.

Under $x+y+z=5$ and $x,y,z\in\{0,1,\dots\}$:

$$P\left(X=x,Y=y,Z=z\right)=\frac{5!}{x!y!z!}\left(\frac{1}{13}\right)^{x}\left(\frac{1}{13}\right)^{y}\left(\frac{11}{13}\right)^{z}=\frac{5!}{x!y!z!}\frac{11^{z}}{13^{5}}$$

Under $x+y\leq5$ and $x,y\in\{0,1,\dots\}$:

$$P\left(X=x,Y=y\right)=P\left(X=x,Y=y,Z=5-x-y\right)=\frac{5!}{x!y!\left(5-x-y\right)!}\frac{11^{5-x-y}}{13^{5}}$$

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  • $\begingroup$ Thank you for your answer. Could you explain why plugging in the constraint $5-x-y$ for $Z$ is 'marginalizing' and why not do that already in the joint PMF, as it must be always true? $\endgroup$
    – NoBackingDown
    Commented Dec 30, 2014 at 12:57
  • $\begingroup$ You are asked in a) for the PMF of $X$, $Y$ and $Z$ which is exactly finding an expression for $P(X=x,Y=y,Z=z)$ in the variables $x$, $y$ and $z$. This is achieved. You can indeed substitute $z=5-x-y$ if you like, but what do you win by that? The formula will only loose some of its elegance. Nothing is gained. In b) an expression for $P(X=x,Y=y)$ must be found in the variables $x$, $y$. Now you are somehow forced to take that step after all. This because variable $z$ is not allowed on the RHS here. $\endgroup$
    – drhab
    Commented Dec 30, 2014 at 13:12
  • $\begingroup$ Thank's for your explanation. I now have a clearer picture. $\endgroup$
    – NoBackingDown
    Commented Dec 30, 2014 at 13:22

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