Playing with 16 cards. 4 A, 4 K, 4 Q, and 4 J;

Deal 8 2 card hands. Notice all cards are dealt.

Number of 2 card hands $\binom{16}{2} = 120$
Of the 120 there are 6 paired Q and 24 unpaired
For every rank there are 6 paired and 24 unpaired 4(6 + 24) = 120

What is the chance of

  • 2 hands with a pair of queens each
  • 1 hands with a pair of queens and 2 hands with 1 queen each
  • 4 hands with a single queen

2 Answers 2


One way to solve this is to create $n=8\times 2=16$ places to hold the cards, where the places are grouped into $m=8$ sets of $k=2$ cards (the "hands"). Mark the $q=4$ queens in the deck and deal them randomly into those $n$ places.

Let's say that a "pattern" is a vector $p=(p_1, p_2, \ldots, p_k)$ in which $p_i$ counts how many of the groups contain exactly $i$ queens. For example:

  • Two hands with two queens gives the pattern $p=(0,2)$.

  • One hand with a pair of queens and two single-queen hands gives the pattern $p=(2,1)$.

  • Four single-queen hands gives $p=(4,0)$.

Notice that each $p_i$ is between $0$ and $m$, the sum of the $p_i$ cannot exceed $m$, and $1p_1 + 2p_2 + \cdots + kp_k = q$ because it counts the total number of queens.

We're practically done:

  1. There are $\binom{n}{q}$ possible subsets of the $n$ places that will correspond to the locations of the $q$ queens.

  2. Count the number of ways to create a pattern $p$ by starting with the largest possible counts of queens and working down:

    • There are $\binom{m}{p_k}$ ways of choosing the $p_k$ hands that will contain $k$ queens. All are equally probable.

    • After those hands are selected, there remain $m-p_k$ hands and therefore, conditional on selecting the hands with $k$ queens, there are $\binom{m-p_k}{p_{k-1}}$ ways of selecting the hands with $k-1$ queens. Within each of those hands, independently, there are $\binom{k}{k-1}$ ways of selecting which cards are the queens. Thus, $$\binom{m-p_k}{p_{k-1}}\binom{k}{k-1}^{p_{k-1}}$$ counts the possible arrangements.

    • Proceed in the same manner for indexes $i$ from $k$ down through $1$. The total number of possible ways to distribute the queens to create pattern $p$ is the product of the individual counts, equal to $$\binom{m}{p_k}\cdot\binom{m-p_k}{p_{k-1}}\binom{k}{k-1}^{p_{k-1}}\cdots \binom{m-p_k-\cdots-p_2}{p_1}\binom{k}{1}^{p_{1}}.$$

Dividing (2) by (1) gives the desired probabilities.

In the example of the question,

  • Two hands with two queens, $p=(0,2)$, occur with probability $$\frac{\binom{8}{2}}{\binom{16}{4}} = \frac{1}{65} \approx 0.01538462.$$

  • $p=(2,1)$ occurs with probability $$\frac{\binom{8}{1}\binom{7}{2}\binom{2}{1}^2}{\binom{16}{4}} = \frac{24}{65} \approx 0.36923077.$$

  • $p=(4,0)$ occurs with probability $$\frac{\binom{8}{4}\binom{2}{1}^4}{\binom{16}{4}} = \frac{40}{65} \approx 0.61538462.$$

Let's check with a simulation. Here are the results of 10,000 computer-generated deals as output by the following R program:

               0,2    2,1    4,0
simulation 0.01600 0.3731 0.6109
theory     0.01538 0.3692 0.6154

The agreement is excellent.

# Specify the problem.
m <- 8
k <- 2
q <- 4
# Create a data structure.
n <- m*k
Deck <- c(rep(1,q), rep(0, n-q))
# Perform a simulation.
p <- replicate(1e4, {
  paste(tabulate(colSums(matrix(sample(Deck, n), k)), k), collapse=",")
# Theory.
f <- function(p, m, k){
  p <- rev(p[1:k])
  n <- m*k
  q <- sum(k:1 * p)
  a <- m - c(0, cumsum(p[-k]))
  exp(sum(lchoose(a, p) + lchoose(k, 0:(k-1))*p) - lchoose(n,q))
# Summarize the simulation.
simulation <- table(p) / length(p)
theory <- sapply(names(simulation), function(s) {
  eval(parse(text=paste0("f(c(", s, "), ", m, ", ", k, ")")))
# Compare to theory.
signif(rbind(simulation, theory), 4)
  • $\begingroup$ Could you give more explanation on the $\binom{k}{k-1}^{p_{k-1}}$ $\endgroup$
    – paparazzo
    Apr 26, 2017 at 17:01
  • 1
    $\begingroup$ The general term is $\binom{k}{k-i}^{p_{k-i}}$. The Binomial coefficient counts the number of ways of distributing $k-i$ queens among the $k$ cards in a group. These ways have to be multiplied together, with one for each group that has exactly $k-i$ queens: there are $p_{k_i}$ such groups. You might want to work it out by brute force in a small example, such as $m=3$ and $k=2$. The full formula is implemented in the last two lines of the function f in the R code: lchoose is the logarithm of the binomial coefficient. $\endgroup$
    – whuber
    Apr 26, 2017 at 17:36

This is probably wrong but a naive attempt

    a       b        c       d           e  f   combin(e,f)          net    fraction
QQ  6   QQ  1                   6       90  6   622614630     3735687780    0.004664
QQ  6   Q   24  Q   23          3312    90  5    43949268   145559975616    0.181739
Q   24  Q   23  Q   22  Q   21  255024  90  4     2555190   651634774560    0.813597

First line: a 6 ways to make QQ, b only 1 QQ left, e 90 other cards, f with 6 possible positions

Second line: a 6 ways to make QQ, b 24 Q left, c 23 Q left, f with 5 possible positions

The first fraction is pretty close to chance of QQQQ = 1/190 which kind of makes sense

  • $\begingroup$ The correct answers are in the ratio $1:24:40$. $\endgroup$
    – whuber
    Apr 26, 2017 at 13:30
  • $\begingroup$ @whuber Could you post an answer how you got that? $\endgroup$
    – paparazzo
    Apr 26, 2017 at 13:42

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