Probability of drawing exactly two aces, at least two hearts and at most three kings when drawing $10$ cards from given $32$ card deck

Question

In a deck of $32$ cards ({7, 8, 9, 10, Jack, Queen, King, Ace} × {♣, ♠, ♥, ♦})

,

$10 $cards are drawn simultaneously. Give the probability of having exactly $2$ aces, at least $2$ hearts and at most $3$ kings.

It's a difficult question. There are a lot of cases to deal with.

Addition 1 It's ok, i have an answer from my friend , this probability is 23,2 %

Addition 2 I give the answer from my french friend in original language (now translated)

For a "hand" of $10$ cards having the required properties, we define the integers $x,y,z$ by: $x:= \text{ number of kings present in the hand: }\: x\in \{0,1,2,3\}.$ $y: = \text{ number of kings of hearts present in the hand: }\: y \in \{0,1\}.$ $z:= \text{ number of aces of hearts present in the hand: } \: z\in \{0,1\}.$ $$\begin{array} {|c|c|}\hline (x,y,z) &\text{nombre de mains de type }(x,y,z)\\ \hline (0,0,0)& 3\times \left[ \binom {24}8 -\binom {18} 8 -6 \binom {18}7 \right]\\ \hline (0,0,1) & 3\times\left[ \binom{24}8 - \binom{18}8\right]\\ \hline (1,0,0) &9 \times\left[ \binom{24}7 - \binom{18}7 -6\binom{18}6\right] \\\hline (1,0,1)& 9 \times \left[\binom{24}7 - \binom{18} 7 \right]\\ \hline (1,1,0) & 3\times \left[ \binom {24}7 - \binom {18} 7 \right]\\\hline (1,1,1)& 3 \times \binom {24 }7 \\ \hline (2,0,0)& 9\times \left[ \binom {24}6 - \binom {18}6 -6\binom {18}5\right] \\ \hline(2,0,1)& 9 \times \left[ \binom{24}6 -\binom {18}6 \right] \\ \hline ( (2,1,0) & 9\times \left[ \binom{24}6 - \binom {18}6 \right] \\ \hline\hline (2,1,1)& 9 \times \binom {24} 6 \\ \hline (3,0,0)& 3 \times \left[\binom{24}5 - \binom{18}5 - 6 \binom{18}4\right] \\ \hline (3,0,1)& 3 \times\left[ \binom{24}5 - \binom {18}5 \right] \\ \hline (3,1,0) & 9 \times \left[ \binom{24}5 - \binom{18}5 \right] \\ \hline (3,1,1) & 9 \times \binom {24}5 \\ \hline\end{array}$$ Let $A$ be the event whose probability is to be determined. $\mathbb P(A) = \dfrac {\text{Card } A}{\text{Card }\Omega}, \:\text{Card }\Omega = \binom {32}{10},\:\text{Card }A =6\binom {24}8 +24 \binom {24}7 + 36 \binom {24}6 +24 \binom {24}5 -6 \binom {18}8 - 39 \binom {18}7 -81 \binom {18}6 - 69 \binom {18}5 -18 \binom {18}4.$ $$ \mathbb P(A) = \dfrac {14931234}{64542240} = 0.2314480.\dots .$$

NB we can simplify with a much simpler count: $$ \text{Card } A= a-b$$ where $a$ is the number of hands of $10$ cards with exactly $2$ aces and at least $2$ hearts and $b$ the number of these hands having the $4$ kings. $a= 3\times \left [\Big(\binom {28}8 -\binom {21}8 -7\binom {21}7 \Big )+ \Big( \binom {28}8 - \binom{21}8 \Big) \right] \qquad b =6\binom {24}4 - 3\binom {18}4, \qquad \text{Card } A= 6\binom{28}8 +3\binom{18}4 -6\binom {21}8 -21 \binom{21} 7 - 6 \binom {24}4 =14931234.$

ADDITION 3 in the answer of leonbloy I rather find

$$ P(A=2,H=0) = \frac{1}{\binom{32}{10}} \binom{3}{2} \binom{21}{8}$$

$$ P(A=2,H=1) = \frac{1}{\binom{32}{10}} \binom{3}{2} \binom{7}{1} \binom{21}{7}+\frac{1}{\binom{32}{10}} \binom{1}{1} \binom{3}{1} \binom{21}{8}$$

$$ P(A=2,K=4) = \frac{1}{\binom{32}{10}} \binom{4}{2} \binom{4}{4} \binom{24}{4}$$

$$ P(A=2,H=0, K=4) = 0$$

$$ P(A=2,H=1, K=4) =\frac{1}{\binom{32}{10}} \binom{3}{2} \binom{4}{4} \binom{18}{4}$$

Answer 1

Alternative approach:

The probability will be

$$\frac{N}{\binom{32}{10}}, \tag1 $$

so the problem reduces to enumerating $N$.

It isn't clear to me whether the number of hearts drawn is (necessarily) independent of either the number of hearts drawn from the Aces, the number of Kings drawn, or the number of hearts drawn from the Kings. Therefore, I will consider the following $16$ cases individually:

• Two Aces with or without a heart.

• $0,1,2,~$ or $3$ Kings, with or without a heart.

In each of the $16$ cases, I will deduct the enumeration where the remaining cards (non-Aces, non-Kings) did not bring the total number of hearts drawn up to $(2)$.

---

Let $a$ denote the number of hearts drawn from the Aces.
So, $a \in \{0,1\}.$

Let $n$ denote the number of Kings drawn.
So, $n \in \{0,1,2,3\}.$

Let $b$ denote the number of hearts drawn from the Kings.
So, $b \in \{0,1\}.$

Let $f(a)$ denote the number of ways of drawing $a$ hearts from the Aces.

Let $g(n,b)$ denote the number of ways of drawing $n$ Kings, $b$ of which are hearts.

Let $h(a,n,b)$ denote the number of ways of drawing a sufficient number of hearts from the remaining $(24)$ cards.

Then,

$$\text{Let} ~N = \sum_{a=0}^1 f(a) \times \left\{ ~\left[ ~\sum_{n=0}^3 ~\left( ~\sum_{b=0}^1 g(n,b) \times h(a,n,b) ~\right) ~\right] ~\right\}. \tag2 $$

---

$f(a)$ equals:

• $\displaystyle \binom{3}{2} = 3 ~: ~a = 0.$
• $\displaystyle \binom{3}{1} = 3 ~: ~a = 1.$

Therefore, $f(a)$ is always equal to $(3)$.

---

$g(n,b)$ equals:

• $\displaystyle \binom{3}{n} ~: ~b = 0.$

• $0 ~: n = 0, b = 1.$

• $\displaystyle \binom{3}{n-1} ~: n > 0, b = 1.$

---

When $(n)$ cards are drawn from the Kings,
then $[10 - (n+2)] = (8 - n)$ other cards must be drawn from the $(24)$ cards that are not Aces or Kings.

This can be done in $~\displaystyle \binom{24}{8-n}~$ ways.

In order to compute $h(a,n,b)$ you have to deduct the number of ways of drawing an insufficient number of hearts.

The number of ways of drawing an insufficient number of hearts is :

• $(0) ~: ~(a + b) = 2.$

• $\displaystyle \binom{18}{8-n} ~: ~(a + b) = 1.$

• $\displaystyle \binom{18}{8-n} + \left[\binom{18}{7-n} \times \binom{6}{1}\right] ~: ~(a + b) = 0.$

Therefore,

$h(a,n,b)$ equals:

• $\displaystyle \binom{24}{8-n} ~: ~(a + b) = 2.$

• $\displaystyle \binom{24}{8-n} - \binom{18}{8-n} ~: ~(a + b) = 1.$

• $\displaystyle \binom{24}{8-n} - \left\{ ~\binom{18}{8-n} + \left[\binom{18}{7-n} \times \binom{6}{1}\right] ~\right\} ~: ~(a + b) = 0.$

---

At this point, the challenge is to apply the above formulas to the expression in (2) above, as elegantly as possible.

I will divide the enumerations into $(4)$ cases, depending on the possible values of $a$ and $b$. For $k \in \{1,2,3,4\}$, I will let $N_k$ denote the enumeration for Case $k$.

Then $N = N_1 + N_2 + N_3 + N_4.$

Some Shortcuts:

• $~\displaystyle \binom{24}{5} = 42504.$

• $~\displaystyle \binom{24}{6} = \binom{24}{5} \times \frac{19}{6} = 134596.$

• $~\displaystyle \binom{24}{7} = \binom{24}{6} \times \frac{18}{7} = 346104.$

• $~\displaystyle \binom{24}{8} = \binom{24}{7} \times \frac{17}{8} = 735471.$

• $~\displaystyle \binom{18}{4} = 3060.$

• $~\displaystyle \binom{18}{5} = \binom{18}{4} \times \frac{14}{5} = 8568.$

• $~\displaystyle \binom{18}{6} = \binom{18}{5} \times \frac{13}{6} = 18564.$

• $~\displaystyle \binom{18}{7} = \binom{18}{6} \times \frac{12}{7} = 31824.$

• $~\displaystyle \binom{18}{8} = \binom{18}{7} \times \frac{11}{8} = 43758.$

---

$\underline{\text{Case 1:} ~a=0, ~b=0}$.

$$N_1 = \sum_{n=0}^3 \left[ ~3 \times \binom{3}{n} \times h(0,n,0)\right]$$

where

$$h(0,n,0) = \binom{24}{8-n} - \left\{ ~\binom{18}{8-n} + \left[\binom{18}{7-n} \times \binom{6}{1}\right] ~\right\}.$$

$h(0,n,0)~$ equals :

• $500769 ~: ~n = 0.$

• $202896 ~: ~n = 1.$

• $64624 ~: ~n = 2.$

• $15576 ~: ~n = 3.$

Therefore,

$$N_1 = \left[3 \times 1 \times 500769\right] + \left[3 \times 3 \times 202896\right] $$

$$+ \left[3 \times 3 \times 64624\right] + \left[3 \times 1 \times 15576\right] = 3956715. \tag3 $$

---

$\underline{\text{Case 2:} ~a=0, ~b=1}$.

$$N_2 = \sum_{n=1}^3 \left[ ~3 \times \binom{3}{n-1} \times h(0,n,1)\right]$$

where

$$h(0,n,1) = \binom{24}{8-n} - \binom{18}{8-n}.$$

$h(0,n,1)~$ equals :

• $314280 ~: ~n = 1.$

• $116032 ~: ~n = 2.$

• $33936 ~: ~n = 3.$

Therefore,

$$N_2 = \left[3 \times 1 \times 314280\right] + \left[3 \times 3 \times 116032\right] $$

$$+ \left[3 \times 3 \times 33936\right] = 2292552. \tag4 $$

---

$\underline{\text{Case 3:} ~a=1, ~b=0}$.

$$N_3 = \sum_{n=0}^3 \left[ ~3 \times \binom{3}{n} \times h(1,n,0)\right]$$

where

$$h(1,n,0) = \binom{24}{8-n} - \binom{18}{8-n}.$$

$h(1,n,0)~$ equals :

• $691713 ~: ~n = 0.$

• $314280 ~: ~n = 1.$

• $116032 ~: ~n = 2.$

• $33936 ~: ~n = 3.$

Therefore,

$$N_3 = \left[3 \times 1 \times 691713\right] + \left[3 \times 3 \times 314280\right] $$

$$+ \left[3 \times 3 \times 116032\right] + \left[3 \times 1 \times 33936\right]= 6049755. \tag5 $$

---

$\underline{\text{Case 4:} ~a=1, ~b=1}$.

$$N_4 = \sum_{n=1}^3 \left[ ~3 \times \binom{3}{n-1} \times h(1,n,1)\right]$$

where

$$h(1,n,1) = \binom{24}{8-n}.$$

$h(1,n,1)~$ equals :

• $346104 ~: ~n = 1.$

• $134596 ~: ~n = 2.$

• $42504 ~: ~n = 3.$

Therefore,

$$N_2 = \left[3 \times 1 \times 346104\right] + \left[3 \times 3 \times 134596\right] $$

$$+ \left[3 \times 3 \times 42504\right] = 2632212. \tag6 $$

---

$\underline{\text{Final Computations}}$

$$N = N_1 + N_2 + N_3 + N_4 $$

$$= 3956715 + 2292552 + 6049755 + 2632212 = 14931234.$$

$~\displaystyle \binom{32}{10} = 64512240.$

So, the probability equals $~\displaystyle \frac{14931234}{64512240} \approx 0.2314.$

Edit Just confirmed via sanity checking with Java program (assuming no programming error).

Pseudocode:
Used integer variables a thru j.
Let 1 thru 4 represent Aces, 5 thru 8 represent Kings.
Let numbers congruent to 1 mod 4 represent hearts.

Forced a thru j to be in strictly ascending order:
a : 1 thru 3.
b : a+1 thru 4.
This guarantees exactly two aces.

c : 5 thru 25.
d : c+1 thru 26.
e : d+1 thru 27.
f : e+1 thru 28.
g : f+1 thru 29.
h : g+1 thru 30.
i : h+1 thru 31.
j : i+1 thru 32.

if f = 8 : skip, since this represents 4 kings.

add 1 to counter if at least 2 of the 10 variables are congruent to 1 mod 4. this represents that there are at least 2 hearts drawn.