ML Interview Q Series: Calculating Card Pair Probability with Complementary Counting and Combinations.
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You draw N cards from a standard deck of 52 unique cards (without putting any card back). Given 4 suits and 13 ranks, determine the probability that your hand of N cards contains at least one pair of identical rank.
Comprehensive Explanation
One effective approach to derive this probability is to use the idea of complementary counting. Instead of directly calculating the probability of getting at least one pair, you find the probability that all N cards in the hand have different ranks, and then subtract that from 1.
To count how many ways to get N cards all of distinct ranks:
First choose which N distinct ranks out of the 13 available ranks.
For each chosen rank, you can select 1 card from its 4 possible suits.
The total number of ways to choose any N cards (regardless of rank) from the 52 cards is the combination of 52 cards taken N at a time.
Hence, the probability of getting no pairs (i.e., all ranks distinct) is: C(13, N) * 4^N / C(52, N) where C(a, b) denotes the binomial coefficient "a choose b".
Therefore, the probability of having at least one pair is:
In this expression:
P(pair) represents the probability that there is at least one pair among your N cards.
C(13, N) is the number of ways to choose N distinct ranks from the 13 possible ranks.
4^N is the number of ways to pick one suit from 4 possibilities for each of those N ranks.
C(52, N) is the total number of ways to pick any N cards out of 52.
When N exceeds 13, it is impossible to choose all cards with distinct ranks, so the probability of getting at least one pair will be 1.
Below is a Python snippet illustrating how one might compute this probability:
import math
def combinations(n, k):
return math.comb(n, k)
def probability_of_pair(N):
# total combinations of N cards from 52
total_ways = combinations(52, N)
# ways to select N different ranks (from 13) and then pick 1 suit for each
distinct_rank_ways = combinations(13, N) * (4**N)
# probability of no pairs
prob_no_pairs = distinct_rank_ways / total_ways
# probability of at least one pair
return 1 - prob_no_pairs
# Example usage:
for n in [2, 5, 10, 13, 14]:
print(f"N={n}, Probability of at least one pair = {probability_of_pair(n)}")
Potential Follow-up Questions
How is the formula derived from first principles?
The complementary counting technique starts by looking at the scenario where every card has a distinct rank. To make sure no two cards share the same rank, you first select which ranks to use, then pick exactly one card from each of those ranks. The total number of ways to choose N cards with no duplicates is the number of ways to pick N distinct ranks (C(13, N)) multiplied by the number of ways to select exactly 1 suit for each of those ranks (4^N). Dividing that by the total ways to pick any N cards from 52 (C(52, N)) yields the probability of no pairs. Subtracting from 1 gives the probability that at least one pair exists.
A critical assumption is that the cards are drawn without replacement, which means each draw reduces the total pool of cards by one. This sets up the combinatorial structure neatly without needing complicated inclusion-exclusion for the "no pairs" event.
What happens if N is larger than 13?
If N is 14 or above, it's impossible for all cards to be of distinct ranks, because there are only 13 possible ranks in total. By the pigeonhole principle, you must have at least one pair of the same rank. So for N > 13, the probability of getting a pair is 1.
Could we adapt this approach to compute the probability of exactly one pair?
Yes, but in that case you would not simply use a single subtraction from 1. You would count:
The number of ways to choose 1 rank for the pair and 2 distinct cards (suits) from that rank.
The number of ways to choose the remaining N-2 cards from the remaining 12 ranks (ensuring those are all distinct), multiplied by 4^(N-2) for each of their suits. Then divide that product by C(52, N). You would have to be careful not to include scenarios with multiple pairs or three-of-a-kind, etc. A direct enumeration or use of the inclusion-exclusion principle helps in such expansions.
How would we compute this probability if the draws were with replacement?
If the deck were reshuffled and the card put back every time, each draw becomes an independent event with identical probabilities. In that scenario, you would calculate the probability of never drawing two cards of the same rank via a different approach:
Probability that each new draw is a distinct rank from all previously drawn cards. But because each draw is independent, you would handle the probability differently (e.g., for large N, it increasingly becomes likely to draw a repeated rank, because each draw can produce any of 13 ranks with probability 1/13, repeatedly).
What if we only cared about suits, not ranks?
Then you would adapt the counting to focus on suits. For instance, to find the probability that you get at least two cards of the same suit, you would do a parallel counting method:
Probability of all distinct suits in a hand of N (though distinct suits would mean you are picking from among 4 suits, and if N>4 you must have repeats of suits). That is simpler than focusing on ranks, because there are only 4 suits, and so for any N>4, you definitely have at least one repeated suit.
How might floating-point precision issues appear in such calculations?
When using combinations for large N, the numbers can be huge. In Python, the built-in math.comb function handles exact integer arithmetic, but once you divide, you get floating-point numbers that can lead to precision issues if N is large. One way to circumvent this is to rely on logarithms of combinations or to use rational arithmetic libraries. However, for moderate N, straightforward floating-point arithmetic is typically sufficient in Python.
Are there any real-world pitfalls to consider?
One pitfall is incorrectly assuming independence of events when using combinatorial reasoning, especially if you try to break down the problem in a naive probability product approach. Another subtlety is off-by-one errors when considering boundary cases (such as N=13 or N=14). Ensuring that the combinatorial formulas do not inadvertently overcount or undercount certain scenarios is important, especially for more complicated events like exactly two pairs or full house probabilities.
When computing, care must also be taken with large N close to 52, because the ratio of large combinations can stress floating-point calculations. Testing with known boundary conditions (e.g., N=1 or N=52) helps validate correctness.
Below are additional follow-up questions
How would the probability change if we were looking for at least one three-of-a-kind (or more) instead of a pair?
When searching for at least one three-of-a-kind, the direct approach (counting all hands that have at least one triplet) can get complicated because the hand could contain multiple triplets or even a full house. A more systematic approach involves the principle of inclusion-exclusion:
First compute the number of ways to choose exactly one particular rank to form a triplet.
Then choose the suits from that rank (C(4, 3) ways), and pick the remaining N-3 cards from the other 48 cards.
Subtract out overlaps where multiple triplets occur.
Continue with inclusion-exclusion for each possible number of triplets.
Once you get the count of all possible hands with at least one three-of-a-kind, you divide by the total C(52, N). This is more tedious compared to the pair problem, because overlapping events (multiple triplets) must be carefully handled.
Potential pitfalls and real-world issues:
Failing to account correctly for multiple possible triplets or overlap with higher-order combinations like four-of-a-kind can lead to overcounting or undercounting.
Inclusion-exclusion is error-prone if the sets for each event (e.g., "three-of-a-kind in rank 1," "three-of-a-kind in rank 2," etc.) are not carefully defined.
Large N complicates the combinatorial expansions, and floating-point arithmetic might introduce inaccuracies unless you handle large numbers with care.
How do partial knowledge or visible cards affect the probability calculation?
If you already know that certain cards are missing from the deck (for example, if someone else’s hand is revealed, or some cards have been discarded), you must adjust the sample space. Instead of 52 unknown cards, you have fewer unknown cards:
Reduce the effective deck size to 52 minus however many visible or known-removed cards.
If some of those known-removed cards share ranks or suits you care about, reflect that knowledge in the counting logic.
Compute new combinations from the updated pool.
Potential pitfalls:
Not properly accounting for the ranks or suits of the known cards, leading to incorrect combinatorial denominators.
Mixing partial knowledge with unconditional probabilities, which can confuse the scenario and produce incorrect results.
How would the formula adapt to a non-standard deck or multiple decks?
When the deck’s size or composition changes, the main idea remains the same but you must alter the numbers accordingly:
If you have multiple decks combined, say 2 decks for a total of 104 cards, then each rank appears 8 times instead of 4. You then recast your combinatorial arguments with 8 suits per rank (for 2 standard decks) instead of 4, and a total of 104 cards.
If the deck is non-standard in other ways (e.g., 54 cards including jokers), define whether jokers can form a pair with a rank or are treated as wild cards. This changes how you count possible combinations for distinct or matching ranks.
Potential pitfalls:
Overlooking that multiple decks multiply not just total cards but also the count of each rank.
Miscounting jokers if they can serve as wild substitutes for any rank.
What if some ranks are duplicated more than four times and others less, as in custom or faulty decks?
You can think of each rank r having S_r “copies” (i.e., suits or sub-cards) in the deck. If you are trying to avoid pairs, you cannot pick more than one card from the same rank:
The number of ways to pick N cards of all distinct ranks is the product of S_r for each chosen rank, summed over all ways of choosing those ranks.
If the deck has unusual distribution (like rank 1 having 3 cards, rank 2 having 5 cards, etc.), you must carefully handle each S_r.
Potential pitfalls:
Failing to treat each rank’s available suits distinctly can lead to undercounting or overcounting.
Summations become necessary if S_r varies per rank, complicating the formula.
How would you handle a requirement that the final hand must contain a certain minimum or maximum number of suits?
If you impose suit constraints (for instance, “I want no more than two clubs,” or “I want at least one card from each suit”), the probability of having a pair can still be computed, but the sample space is constrained:
First, calculate how many ways to pick N cards that satisfy the suit restriction.
Among those valid draws, count how many contain at least one pair.
Divide the latter by the former to get the conditional probability.
Potential pitfalls:
Suit constraints can invalidate large swaths of the original sample space, so you must be sure to only count from the reduced pool.
Overlaps with multiple constraints (like suit distribution plus at least one pair) may need a more intricate inclusion-exclusion method.
Could we interpret the pair event using a hypergeometric distribution perspective?
Yes. A hypergeometric distribution typically describes the probability of a certain number of “successes” (e.g., a specific rank) when drawing from a finite population. Although the original formula is more directly combinatorial, you can sometimes break it down into hypergeometric components if you label certain ranks as “success” categories. However, ensuring that the standard hypergeometric framework applies (which usually describes success vs. failure, rather than multiple different ranks) might be tricky.
Potential pitfalls:
If you try to directly apply a hypergeometric distribution to “any pair,” you have to track multiple ranks simultaneously, which is more complex than the typical single-success scenario.
The hypergeometric approach is more straightforward for a single rank or multiple ranks tracked independently, but pairs from “any rank” require combining events across all ranks.
How do we ensure numerical stability when N is large and we compute probabilities close to 1?
When N is large, the probability of getting a pair can be extremely high, potentially leading to numerical underflow or overflow in intermediate steps:
Use a logarithmic approach: compute ln of combinations and use exponent rules to manage large numbers. Then exponentiate only at the end if needed.
Python’s built-in
math.combhandles exact integers for binomial coefficients, but dividing might produce extremely small or large floating-point values.Sometimes, for large N, you might approximate using expansions (e.g., the complement probability ~ e^(-small_value)) to avoid direct computation of enormous factorials.
Potential pitfalls:
Underestimating or overestimating probabilities near the extremes (like 0 or 1).
Relying too heavily on floating-point arithmetic for large combinatorial ratios without verifying the results with smaller test values.
How can sampling simulations provide additional insights or sanity checks?
In practice, especially if you are unsure about your combinatorial logic or for large N:
You can run Monte Carlo simulations, repeatedly draw N cards (digitally) from a standard deck, and empirically count how often a pair appears.
Compare the empirical estimate to the theoretical value to confirm you haven't missed an important detail.
Potential pitfalls:
Simulations require sufficient trials to get a stable estimate, or else random fluctuations might mask the true probability.
Implementation errors in the simulation logic (like inadvertently drawing the same card more than once if the simulation is coded incorrectly) can lead to misleading results.