ML Interview Q Series: Combinatorics for Ace vs. Ten Probability Calculations in Card Hands.
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A hand of 13 cards is dealt from a standard deck of 52 cards. What is the probability that it contains more aces than tens? How does this probability change when you have the information that the hand contains at least one ace?
Short Compact solution
The probability that the hand has the same number of aces as tens is
By a symmetry argument, the probability that the hand has more aces than tens is half of the complement of 0.3162, giving 0.3419. Then using the conditional probability formula P(A|B) = P(A and B) / P(B), the updated probability of having more aces than tens given at least one ace is
Comprehensive Explanation
One way to see why the probability that the hand has more aces than tens is 0.3419 relies on comparing the counts of aces and tens in a 13-card hand. There are four aces and four tens in a standard deck. A direct combinatorial approach is to calculate the likelihood that the number of aces equals the number of tens, then use symmetry to find the probability that aces exceed tens.
The expression for the probability that the hand contains exactly k aces and k tens, for k = 0 to 4, multiplies three combinatorial terms: choose k of the 4 available aces, choose k of the 4 available tens, and choose the remaining 13 - 2k cards from the other 44 non-ace, non-ten cards in the deck. Summing over k from 0 up to 4 and dividing by the total number of ways to draw 13 cards out of 52 gives 0.3162. This is the probability that the number of aces equals the number of tens.
Symmetry arises because the deck has the same number of aces and tens. Consequently, the probability that there are more aces than tens in a 13-card hand must be the same as the probability that there are more tens than aces, provided we exclude the cases where they are equal. Since the probability that aces and tens counts are equal is 0.3162, the remaining probability mass of 1 - 0.3162 = 0.6838 must be split equally between the event that there are more aces than tens and the event that there are more tens than aces. Therefore the probability that there are more aces than tens is (0.6838)/2 = 0.3419.
When we know that there is at least one ace in the hand, we condition on the event that the hand is not ace-free. This modifies the sample space by excluding those hands that contain zero aces. Mathematically, we compute P(A|B) = P(A and B) / P(B). Here B is the event "at least one ace" and A is the event "more aces than tens". The relevant probability P(A and B) corresponds to "having more aces than tens and having at least one ace", which is effectively the same as "more aces than tens" but excluding zero-ace hands (those do not satisfy either condition). P(B) is simply the probability of having at least one ace. The formula in the short solution shows how summing the appropriate combinatorial expressions and dividing by the total ways yields approximately 0.4911. The denominator is the probability of having at least one ace, and the numerator is the probability of having more aces than tens while having at least one ace.
This shows that once you know at least one ace is present, it becomes more likely that there are more aces than tens, shifting the probability upward from 0.3419 to about 0.4911. Intuitively, removing all the zero-ace hands from consideration increases the chance of having multiple aces relative to tens.
Real-world insight often comes from realizing that conditional probabilities can shift the likelihood of outcomes significantly once you remove cases that are incompatible with newly acquired information. In this specific card problem, the knowledge that there is at least one ace removes exactly those outcomes that had zero aces, thus boosting the relative number of scenarios in which aces can be in majority.
Follow-up Question: Why does symmetry apply between the number of aces and the number of tens?
Symmetry applies because there are exactly four aces and four tens in a 52-card deck, giving them identical positions in the deck's composition. If we imagine any particular 13-card hand, swapping all aces for tens and all tens for aces would produce another valid 13-card hand with the opposite relationship between the count of aces and tens. No other ranks in the deck affect that one-to-one pairing of hands, so the probabilities that there are more aces than tens or more tens than aces must match as long as we exclude the possibility that the counts are equal.
Follow-up Question: How would these probabilities change if the deck had different numbers of aces and tens?
The symmetry argument would break if the number of aces differed from the number of tens. In such a scenario, the probability that the hand contains more aces than tens would not be simply half of the remaining probability mass after excluding the cases where they match. Instead, one would need to compute it directly using combinatorial counts for each possible pair of counts of aces and tens, or seek a more nuanced argument if some symmetry still exists. For instance, if the deck had six tens and four aces, the event “more tens than aces” would naturally have a higher probability than “more aces than tens.”
Follow-up Question: Could we simulate this probability instead of calculating it combinatorially?
Simulation is possible and often used when analytical formulas become cumbersome or the deck conditions are modified in ways that break straightforward combinatorial symmetries. A Monte Carlo simulation would repeatedly draw 13-card hands from a virtual deck and track how often the counts of aces exceed the counts of tens. Repeating the experiment many times and averaging the results would approximate the probabilities in question. Care must be taken to use enough draws to ensure the estimate is statistically reliable and to properly randomize card selection. Simulation is not exact but is very practical for more complex setups or larger decks where enumerating the entire sample space is infeasible.
Follow-up Question: Could we extend this approach to compute probabilities related to other ranks simultaneously?
Yes, one can extend the same combinatorial approach or simulations to scenarios where multiple ranks are considered simultaneously. For instance, you might want to compute the probability that a hand has more face cards (Jack, Queen, King) than spot cards or more black suits than red suits. Each scenario can be tackled by counting configurations that meet the given criteria or by adapting a simulation strategy. The main difference is that the counting might become more involved and the direct combinatorial formulas can grow in complexity. If direct counting is complicated, simulation methods might be simpler to implement while still producing good numerical approximations.
Below are additional follow-up questions
If we consider the event "strictly more aces than tens" vs. "at least as many aces as tens," how would that affect the probability calculations?
When we say "strictly more aces than tens," we exclude the scenario where the number of aces and tens is equal. On the other hand, "at least as many aces as tens" allows the counts to be equal or higher for aces. These two definitions produce different events:
Strictly more aces than tens: Probability = 0.3419 (as given by the symmetry argument).
At least as many aces as tens: Probability = Probability of strictly more aces than tens + Probability of equal numbers of aces and tens.
Because the latter event includes the equality case, its probability would be 0.3419 + 0.3162 = 0.6581. A potential pitfall occurs if one confuses these two events and inadvertently calculates or interprets the probability of one event while meaning the other. Ensuring that “strictly greater” vs. “greater or equal” is carefully distinguished in combinatorial sums or simulations is crucial.
What if the deck is missing some cards or contains extra jokers?
In a real-world scenario, you might not always have a complete 52-card deck. Missing cards or extra cards (like jokers) can break the symmetry between aces and tens or alter the total count of cards and the rank distribution. For instance, if a joker is treated as a wild card (capable of substituting for any rank), the probability of having more aces than tens is no longer purely about counting actual aces. Instead, each joker could be an ace or a ten or another rank, depending on the game rules.
A pitfall here is assuming that adding wild jokers keeps the deck structure simple. In truth, you would have to carefully define how jokers can be used. If a joker can act as an ace, it effectively raises the number of possible aces, changing the probability calculations substantially. One would then handle this scenario by enumerating or simulating all possible ways the jokers could be assigned to different ranks.
How does knowledge of other players' cards (partial or full) influence these probabilities?
If you are playing in a multi-player setting and you learn something about other players’ cards (e.g., they hold some tens or some aces), that knowledge changes the composition of the deck for your own draw or your own hand. For instance, if you know another player already has two aces in their hand, then the probability of your hand having more aces than tens is reduced because fewer aces remain in the deck.
A subtle point is that knowing someone else has “some aces” is not the same as knowing exactly how many. The more precise the information, the more accurately you can update your probability. However, partial knowledge introduces Bayesian updates that are less straightforward than simple combinatorial counts. One pitfall is to overestimate or underestimate the impact of partial information if you do not carefully model the probabilities for each possible distribution of remaining aces and tens in the deck.
How can we use Bayesian analysis to refine probabilities with partial data about the deck?
Bayesian analysis allows you to start with a prior distribution about how many aces might be left in the deck and update that belief as you gain information (for example, seeing a card dealt face up). If you see a single card turned over and it’s a ten, you adjust your prior to reflect that only 3 tens remain in the 51 unknown cards. The probability of your next hand having more aces than tens would then be recalculated based on that updated deck distribution.
A pitfall arises if one tries to apply only the unconditional probabilities despite having conditional information about the number of remaining aces or tens. Failing to update the count of unseen cards properly, or forgetting that each new piece of information alters the sample space, would yield incorrect results. Ensuring that each observation is factored into the posterior distribution is essential for accurate Bayesian updates.
What if we deal multiple 13-card hands in sequence without replacement from the same deck?
When multiple 13-card hands are dealt in sequence (to multiple people, for example), the events are not independent. If the first 13-card hand contains a certain number of aces, those cards are removed from the deck, affecting the second hand’s probability of having more aces than tens.
To handle this scenario, one would either:
Enumerate the possible outcomes for each sequential deal, which can be very large combinatorially.
Use conditional probabilities repeatedly or employ Bayesian updating as cards are removed from the deck.
Use simulation to draw hands sequentially and track the distribution of aces and tens across multiple hands.
A common pitfall is to treat the second hand’s probability as if it came from a fresh 52-card deck. This would overlook the depletion of aces (and tens) from the first hand. Correctly accounting for the fact that cards are not replaced is key for an accurate probability assessment.
How does the hypergeometric distribution appear in these calculations?
The hypergeometric distribution is the distribution of the number of “successes” (in this context, aces or tens) drawn from a finite population without replacement. For instance, the probability that a hand of 13 cards contains exactly k aces can be modeled by a hypergeometric formula:
number_of_ways_to_choose_k_aces_from_4 multiplied by the number_of_ways_to_choose_(13-k)_non-aces_from_48, divided by the total number_of_ways_to_choose_13_cards_from_52.
In formulas, this is usually represented as:
(#ways of choosing k from 4) * (#ways of choosing 13-k from 48) / (#ways of choosing 13 from 52).
A subtlety arises when combining two or more hypergeometric distributions (for aces and tens simultaneously). You can’t simply multiply their probabilities independently because the events of how many aces you draw is correlated with how many tens you can still draw. Hence, you need joint combinatorial calculations, leading to expressions like binomial coefficients over the 4 aces, 4 tens, and the other 44 cards. Missing that correlation is a pitfall, as it would lead to double counting or ignoring overlapping constraints in the deck.
Could out-of-order combinatorial summations lead to confusion?
Yes, if you are not systematic about summing over all valid ways to distribute aces and tens, you might accidentally omit or double-count certain configurations. A common pitfall is to compute the number of ways to choose aces and tens separately but then fail to properly account for how many cards remain to be chosen from the rest of the deck. Another source of confusion occurs when you switch between events like “exactly k aces” vs. “at least k aces” vs. “more aces than tens.” Mixing up these conditions in the summation limits is easy if not approached carefully.
An effective approach is to write the exact combinatorial expressions for each scenario and confirm that each term is counting disjoint events. If you use a computer program to assist, verifying that your summation iterates over the correct ranges of aces and tens is crucial.