Draw 21 cards from a standard deck, what are the odds of 6 or more being Q or higher

Question

Given a standard 52 deck of cards with [12/52 being an Ace, King, or Queen], if you draw 21, what are the odds that 6 (or more) of the 21 cards are either (A,K,Q)?

How do you go about calculating that and how could I do the same for exactly 6, or 5+, 5 etc.

Answer 1

In general, we want to use the hypergeometric distribution to find the probability of a given number of successes when drawing from without replacement from a finite pool with a given number of success objects in the inital population. The PMF is given by

$$ \text{Pr}(X = k) = \frac{{K\choose k}{{N -K}\choose{n-k}}}{N\choose n} $$ where $N$ is the population size, $K$ is the number of successes in the population, $n$ is the number of draws, and $k$ is the number of observed successes.

So your answer is given by evaluating the above for $N = 52$, $K = 12$, and $n = 21$. Then, you want to sum the answers for all $k$ such that $12 \geq k \geq 6$. You can obviously choose different $k$ values to get the answers to your other questions.

The formula looks non-trivial but intuitive; ask if you want to see how to derive this formula.

Answer 2

Use the hypergeometric distribution, which is analagous to the binomial distribution but the draws are without replacement.

• success is defined by being (A, K, Q)
• the number of draws $n$ is 21
• the population size $N$ is 52
• there are $K=12$ objects with the success criteria

If $X$ is distributed $\text{Hypergeometric}(N,K,n)$, then

• the probability that you get $k$ successes, is given by its PDF evaluated at $k$ on the Wikipedia page.
• the probability you get less or equal to $k$ successes is given by the CDF evaluated at $k$.
• if you want the probability you get $k+1$ or more successes, you can use $1-\text{CDF}$ evaluated at $k$.