Probability of at least one ace in a hand without using complementary event

Question

While going through the textbook, Discrete Math and its Applications, I found a problem, "what is the probability that a 5-card poker hand contains at least one ace?" It uses a standard deck of 52 cards with 4 aces.

The solution given and that I have found on here uses a complement, which I understand.

But my initial answer, which leads to a wrong value, was $$\frac{\binom{4}{1}\cdot\binom{51}{4}}{\binom{52}{5}}$$ I still can't figure out why this is wrong, which suggests I have a misunderstanding of combinatorics. I'll write my reasoning and would appreciate if someone can point out how it's flawed.

I thought the size of the event in which our hand has at least one ace would be equal to the number of ways we can choose one ace, which is $\binom{4}{1}$ times the amount of ways we can choose 4 other cards from the remaining cards in the deck, which is $\binom{51}{4}$. And then this goes over the sample space of $\binom{52}{5}$.

I would greatly appreciate an explanation of why this reasoning is flawed or resources that might help me understand this better.

Answer 1

Your answer counts some hands more than once. For example, [A♡ A♠ 3♢ 4♢ 5♢] is counted both as selecting A♡ and then the four cards A♠ 3♢ 4♢ 5♢, but also as selecting A♠ and then the four cards A♡ 3♢ 4♢ 5♢.

To avoid using the complement event, you could instead add together the counts of hands with

• Exactly one ace: ${4 \choose 1} {48 \choose 4}$
• Exactly two aces: ${4 \choose 2} {48 \choose 3}$
• Exactly three aces: ${4 \choose 3} {48 \choose 2}$
• All four aces: ${4 \choose 4} {48 \choose 1}$

Answer 2

Alternative approach:

With the constraint that complementary events are to be avoided, I feel that the method used in the answer of aschepler is best. An alternative method is Inclusion-Exclusion. See this article for an introduction to Inclusion-Exclusion. Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula.

Let :

• $~S_1~$ denote the set of all 5 card poker hands that contain the Ace of Spades.

• $~S_2~$ denote the set of all 5 card poker hands that contain the Ace of Hearts.

• $~S_3~$ denote the set of all 5 card poker hands that contain the Ace of Diamonds.

• $~S_4~$ denote the set of all 5 card poker hands that contain the Ace of Clubs.

Note that in each of the four sets above, some of the elements in that set will represent a poker hand that contains more than one Ace.

The probability may be expressed as

$$\frac{N}{D} ~: ~~~D = \binom{52}{5}, ~~~N = | ~S_1 \cup S_2 \cup S_3 \cup S_4 ~|. \tag1 $$

So, the problem reduces to computing $~N,~$ as expressed on the RHS of (1) above.

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Let $~T_1~$ denote $~|S_1| + |S_2| + |S_3| + |S_4|.$

For $~r \in \{2,3,4\},~$ let $~T_r~$ denote
$\displaystyle \sum_{1 \leq i_1 < i_2 < \cdots < i_r \leq 4} | ~S_{i_1} \cap S_{i_2} \cap \cdots \cap S_{i_r} ~|.$
That is, $~T_r~$ represents the sum of $~\displaystyle \binom{4}{r}~$ terms.

Then, in accordance with Inclusion-Exclusion theory,

$$N = | ~S_1 \cup S_2 \cup S_3 \cup S_4 ~| = \sum_{r=1}^4 (-1)^{r+1} T_r. \tag2 $$

So, the entire problem has been reduced to computing $~T_r ~: ~r \in \{1,2,3,4\}.$

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$\underline{\text{Computation of} ~T_1}$

$~S_1~$ represents all poker hands that contain the Ace of Spades.

Therefore, $~\displaystyle |S_1| = \binom{51}{4}.$

By considerations of symmetry,
$~|S_1| = |S_2| = |S_3| = |S_4|.$

Therefore,

$$T_1 = \binom{4}{1} \times \binom{51}{4}.$$

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$\underline{\text{Computation of} ~T_2}$

$~S_1 \cap S_2~$ represents all poker hands that contain both the Ace of Spades and the Ace of Hearts.

Therefore, $~\displaystyle |S_1 \cap S_2| = \binom{50}{3}.$

By considerations of symmetry,
for $~1 \leq i_1 < i_2 \leq 4,$
$~|S_{i_1} \cap S_{i_2}| = |S_1 \cap S_2|.$

Therefore,

$$T_2 = \binom{4}{2} \times \binom{50}{3}.$$

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$\underline{\text{Computation of} ~T_3}$

$~S_1 \cap S_2 \cap S_3~$ represents all poker hands that contain the Ace of Spades, the Ace of Hearts, and the Ace of Diamonds.

Therefore, $~\displaystyle |S_1 \cap S_2 \cap S_3| = \binom{49}{2}.$

By considerations of symmetry,
for $~1 \leq i_1 < i_2 < i_3 \leq 4,$
$~|S_{i_1} \cap S_{i_2} \cap S_{i_3}| = |S_1 \cap S_2 \cap S_3|.$

Therefore,

$$T_3 = \binom{4}{3} \times \binom{49}{2}.$$

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$\underline{\text{Computation of} ~T_4}$

$~S_1 \cap S_2 \cap S_3 \cap S_4~$ represents all poker hands that contain all four Aces.

Therefore,

$$T_4 = |S_1 \cap S_2 \cap S_3 \cap S_4| = \binom{48}{1}.$$

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$\underline{\text{Final Computation}}$

The probability is

$$\frac{N}{D} ~: ~D = \binom{52}{5}.$$

$$N = \sum_{r=1}^4 (-1)^{r+1} T_r = \sum_{r=1}^4 (-1)^{r+1} \binom{4}{r} \binom{52-r}{5-r}.$$