Probability of drawing exactly two aces, two hearts and two kings when drawing 8 cards from given 52 card deck

Question

I want a simple aproch to find the Probability of drawing exactly two aces, two hearts and two kings when drawing 8 cards from given 52 card deck

My approch

Let $A,H,K$ be the amount of aces, hearts and kings. Define $E=(A=2)\cap (K=2)\cap (H=2)$

We want $p=P(E)$. I define also $F=(\text{Ace of heart})$ and $G=(\text{King of heart})$ then

$p=P(E\cap \Omega)=P\big((E\cap F\cap G)\cup (E\cap F\cap \bar G)\cup (E\cap \bar F\cap G)\cup (E\cap \bar F\cap \bar G)\big)\\= P(E\cap F\cap G)+P (E\cap F\cap \bar G)+P (E\cap \bar F\cap G)+ P(E\cap \bar F\cap \bar G)$

$=\frac {\binom{1}{1}\binom{3}{1}\binom{1}{1}\binom{3}{1}\binom{11}{0}\binom{33}{4}}{\binom{52}{8}}+\frac {\binom{1}{1}\binom{3}{1}\binom{3}{2}\binom{11}{1}\binom{33}{3}}{\binom{52}{8}}+\frac {\binom{3}{2}\binom{1}{1}\binom{3}{1}\binom{11}{1}\binom{33}{3}}{\binom{52}{8}}$ + $\frac {\binom{3}{2}\binom{3}{2}\binom{11}{2}\binom{33}{2}}{\binom{52}{8}}$

Answer 1

For me, the simplest approach is that taken in my answer to the parallel problem.

In that parallel problem, I was impressed with the elegant alternative approach taken by the OP (i.e. original poster) in the middle of the OP's (now edited) posting. However, I will stick with the simple approach that the probability is

$$\frac{N}{\binom{52}{8}}. \tag1 $$

Let $a,b$ denote the number of hearts drawn from the Aces and Kings, respectively. Then $a,b \in \{0,1\}$, so there are $(4)$ cases to examine.

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Let $f(a)$ denote the number of ways of drawing two Aces, given that $(a)$ of the Aces are hearts. Then, $f(a)$ equals:

• $\displaystyle \binom{3}{2} = 3 ~: ~a = 0.$

• $\displaystyle \binom{3}{1} = 3 ~: ~a = 1.$

Therefore, in all cases, $f(a) = 3.$

Let $g(b)$ denote the number of ways of drawing two Kings, given that $(b)$ of the Kings are hearts. Then, by symmetrical considerations, in all cases, $g(b) = 3.$

Let $h(a,b)$ denote the number of ways of drawing exactly $4$ (non-Ace, non-King) cards, of which $[2 - (a+b)]$ of the $(4)$ cards are hearts.

Then, overall,

$$N = \left\{ ~\sum_{a=0}^1 f(a) \times ~\left[ ~\sum_{b=0}^1 g(b) \times h(a,b) ~\right] ~\right\}. \tag2 $$

Since $f(a)$ and $g(b)$ are both constants, the problem reduces to calculating $h(a,b).$ Then, $h(a,b)$ denotes the number of ways of drawing $(4)$ cards from the $(44)$ cards (non-Aces, non-Kings) in the deck such that $[2 - (a+b)]$ of the cards drawn are hearts.

In the $(44)$ card deck, there are $(11)$ hearts.

Therefore,

$$h(a,b) = \binom{11}{2 - [a+b]} \times \binom{33}{2 + [a+b]}.$$

That is, in the $(44)$ card deck, when $(a+b)$ hearts are drawn from the Aces and Kings combined, then $(2 - [a+b])$ hearts and $(2 + [a+b])$ non-hearts must be drawn.

In the $(4)$ cases represented in (2) above, 1 of them will have $(a+b) = 0$, 2 of them will have $(a+b) = 1$, and 1 of them will have $(a+b) = 2.$

Therefore,

$$N = (3)^2 \times $$

$$\left\{ ~\left[ ~1 \times \binom{11}{2} \times \binom{33}{2} ~\right] + ~\left[ ~2 \times \binom{11}{1} \times \binom{33}{3} ~\right] + ~\left[ ~1 \times \binom{11}{0} \times \binom{33}{4} ~\right] ~\right\}. $$

Therefore,

$$\frac{N}{\binom{52}{8}} = \frac{1709928}{752538150} \approx 0.00227.$$