Probability of a Void Suit in a Card Hand
You are dealt $13$ cards from a standard deck of $52$ cards (four suits of $13$ cards each). A hand is void in a suit if it contains zero cards from that suit.
What is the probability that your hand is void in at least one suit? Round your answer to the nearest thousandth.
Hints
- Think about what "at least one" suggests -- what counting principle handles unions of events?
- By symmetry all four suits are identical. Let $A_k$ be the event of having no cards from suit $k$. Apply inclusion-exclusion to $P(A_1 \cup A_2 \cup A_3 \cup A_4)$ and notice that each term depends only on how many suits you are avoiding.
- To compute $P(A_1)$, count how many 13-card hands can be formed from the 39 cards outside suit 1: $P(A_1) = \binom{39}{13} / \binom{52}{13}$. Similarly, $P(A_1 \cap A_2) = \binom{26}{13} / \binom{52}{13}$. Plug these into the inclusion-exclusion formula.
Worked Solution
How to Think About It: "At least one" screams inclusion-exclusion. Label the four suits $1, 2, 3, 4$ and let $A_k$ be the event that your hand has no cards from suit $k$. You want $P(A_1 \cup A_2 \cup A_3 \cup A_4)$. By symmetry every suit looks the same, so $P(A_1) = P(A_2) = \cdots$, and similarly for intersections. That makes the inclusion-exclusion formula very clean. Gut feeling: being void in a single suit means all 13 cards come from 39 cards -- that should be a small probability, so the union will be dominated by the first-order term $4 \cdot P(A_1)$.
Quick Estimate: $P(A_1)$ is the chance all 13 cards come from the 39 non-suit-1 cards. Think of it sequentially: the first card avoids suit 1 with probability $39/52 = 3/4$, the second with $38/51$, and so on. Roughly $(3/4)^{13} \approx 0.75^{13}$. Since $0.75^4 \approx 0.316$ and $0.75^{13} \approx 0.75^{12} \cdot 0.75 \approx (0.316)^3 \cdot 0.75 \approx 0.032 \cdot 0.75 \approx 0.024$. That is a rough upper bound (the actual hypergeometric probability is smaller because we draw without replacement from a finite deck). The exact value is about $0.0128$, and $4 \times 0.0128 \approx 0.051$. The higher-order corrections ($P(A_1 \cap A_2)$, etc.) are tiny by comparison, so the answer is approximately $0.051$.
Approach: Apply inclusion-exclusion with four symmetric events, computing each intersection probability via hypergeometric counting.
Formal Solution:
By inclusion-exclusion:
$$P\!\left(\bigcup_{k=1}^{4} A_k\right) = \binom{4}{1}P(A_1) - \binom{4}{2}P(A_1 \cap A_2) + \binom{4}{3}P(A_1 \cap A_2 \cap A_3) - \binom{4}{4}P(A_1 \cap A_2 \cap A_3 \cap A_4)$$
Compute each probability by counting favorable hands over total hands $\binom{52}{13}$:
- $P(A_1)$: all 13 cards come from the 39 cards not in suit 1.
$$P(A_1) = \frac{\binom{39}{13}}{\binom{52}{13}}$$
- $P(A_1 \cap A_2)$: all 13 cards come from the 26 cards in suits 3 and 4.
$$P(A_1 \cap A_2) = \frac{\binom{26}{13}}{\binom{52}{13}}$$
- $P(A_1 \cap A_2 \cap A_3)$: all 13 cards come from suit 4 alone -- there are exactly 13 such cards, so there is exactly one such hand.
$$P(A_1 \cap A_2 \cap A_3) = \frac{\binom{13}{13}}{\binom{52}{13}} = \frac{1}{\binom{52}{13}}$$
- $P(A_1 \cap A_2 \cap A_3 \cap A_4) = 0$, since you cannot avoid all four suits with 13 cards.
Substituting:
$$P = 4 \cdot \frac{\binom{39}{13}}{\binom{52}{13}} - 6 \cdot \frac{\binom{26}{13}}{\binom{52}{13}} + 4 \cdot \frac{1}{\binom{52}{13}}$$
Numerically: - $4 \cdot P(A_1) \approx 0.05116$ - $6 \cdot P(A_1 \cap A_2) \approx 0.0000983$ - $4 \cdot P(A_1 \cap A_2 \cap A_3) \approx 6.3 \times 10^{-12}$ (negligible)
$$P \approx 0.05116 - 0.0000983 + 0 \approx 0.05107$$
Answer: The probability of being void in at least one suit is approximately $\boxed{0.051}$.
Intuition
This problem is a textbook application of inclusion-exclusion, and it illustrates why that principle is so useful: the events "void in suit 1," "void in suit 2," etc. overlap in complicated ways, so you cannot just add $P(A_1) + P(A_2) + \cdots$. Inclusion-exclusion systematically corrects for double-counting. The symmetry across suits is a huge simplifier -- instead of computing $2^4 - 1 = 15$ different intersection probabilities, you only need three distinct values.
The practical lesson is that the first-order term almost always dominates. Here $4 \cdot P(A_1) \approx 0.0512$ while the correction from double-intersections is only $\sim 10^{-4}$ and the triple-intersection is negligible. In quant interviews, recognizing when higher-order corrections are tiny lets you give a fast, confident estimate before grinding through the exact algebra. This pattern -- "at least one" over symmetric events, dominated by the singles term -- shows up constantly in problems about coincidences, collisions, and birthday-type arguments.