When drawing 14 cards from a set of 52 cards, is it more likely to have a full house or two consecutive pairs?

Question

This is a problem that has been bothering me for quite some time now:

When drawing 14 cards from a set of 52 cards (standard poker deck), is it more likely to have at least one full house or at least two consecutive pairs?

Both questions seem to resist my attempts to telescope all possible hands into one expression, so here is, where I am:

My sample space in both cases is $52\choose14$. For the 2 consecutive pairs, there are $12\choose1$ different consecutive pairs to consider. It might be that more cards of the same rank like these consecutive pairs are amongst the drawn cards, so I distinguish between the cases:

• case 1: no other cards of the ranks of the pairs are drawn ($4\choose2$$4\choose2$ combinations)
• case 2: of one rank 3 cards were drawn, of the other one 2 ($4\choose2$$4\choose3$ combinations)
• case 3: of one rank 4 cards were drawn, of the other one 2 ($4\choose4$$4\choose2$ combinations)
• case 4: of one rank 4 cards were drawn, of the other one 3 ($4\choose4$$4\choose3$ combinations)
• case 5: of both rank 3 cards were drawn ($4\choose3$$4\choose3$ combinations)
• case 6: of both rank 4 cards were drawn ($4\choose4$$4\choose4$ combinations)

Now I also have to somehow avoid "overlapping", when counting the possible combinations of the remaining cards. I cannot for example count the combinations for case 1 with the expression $12\choose1$$4\choose2$$4\choose2$$44\choose9$, because among the $44\choose9$ combinations of the remaining cards are also many combinations including consecutive pairs, which I count more than once like this.

I also tried to approach the problem via the complement event, but it seems to expand even worse.

The problem, I am facing with the full house, is similar to this.

Maybe I just miss a much simpler way of looking at this. If anybody can give me some advice, it would be much appreciated.

Answer 1

This problem is begging you to apply inclusion–exclusion to it. I’m not sure whether that qualifies as what you had in mind as “a much simpler way of looking at this”, but it’s quite a systematic way, and that’s rather necessary for your endeavour.

I’ll do the calculation for two consecutive pairs and then see whether you feel like doing the one for the full house yourself after being equipped with the necessary tools, or whether you’d like me to do that as well.

We can do this in two stages. First I’ll determine the number $a_k$ of ways to draw at least two cards of each of $k$ particular ranks; and then I’ll use those numbers to compute the probabilities that you want using inclusion–exclusion.

The calculation of the $a_k$ proceeds pretty much like you described it for two consecutive pairs, except that for the inclusion–exclusion calculation we’ll need all the $a_k$ up to $k=7$, not just $a_2$. (Beyond $k=7$, they’re zero because it’ obviously impossible to have at least two cards in each of more than $7$ ranks if you draw $14$ cards.)

The following fact comes in very handy for us at this point: A binomial coefficient is zero if the upper index is positive and the lower index is negative. That means that we can write

$$ a_k=\sum_{i_1=2}^4\cdots\sum_{i_k=2}^4\binom4{i_1}\cdots\binom4{i_k}\binom{52-4k}{14-i_1-\cdots-i_k} $$

without worrying about the fact that it’s not possible e.g. to draw $4$ cards of each of $4$ ranks when drawing $14$ cards – the right-most coefficient has negative lower index in such cases and is thus zero.

This is of course a computation that we wouldn’t want to carry out by hand, but our electronic friends over at Sage don’t mind doing it for us.

Here’s the Sage code (essentially Python) for the computation of the $a_k$:

i,j,k,l,m,n,r = var ('i,j,k,l,m,n,r')
a = [0] * 15
a [2] = sum (binomial (4,j) * sum (binomial (4,i) * binomial (44,14-i-j),i,2,4),j,2,4)
a [3] = sum (binomial (4,k) * sum (binomial (4,j) * sum (binomial (4,i) * binomial (40,14-i-j-k),i,2,4),j,2,4),k,2,4)
a [4] = sum (binomial (4,l) * sum (binomial (4,k) * sum (binomial (4,j) * sum (binomial (4,i) * binomial (36,14-i-j-k-l),i,2,4),j,2,4),k,2,4),l,2,4)
a [5] = sum (binomial (4,m) * sum (binomial (4,l) * sum (binomial (4,k) * sum (binomial (4,j) * sum (binomial (4,i) * binomial (32,14-i-j-k-l-m),i,2,4),j,2,4),k,2,4),l,2,4),m,2,4)
a [6] = sum (binomial (4,n) * sum (binomial (4,m) * sum (binomial (4,l) * sum (binomial (4,k) * sum (binomial (4,j) * sum (binomial (4,i) * binomial (28,14-i-j-k-l-m-n),i,2,4),j,2,4),k,2,4),l,2,4),m,2,4),n,2,4)
a [7] = sum (binomial (4,r) * sum (binomial (4,n) * sum (binomial (4,m) * sum (binomial (4,l) * sum (binomial (4,k) * sum (binomial (4,j) * sum (binomial (4,i) * binomial (24,14-i-j-k-l-m-n-r),i,2,4),j,2,4),k,2,4),l,2,4),m,2,4),n,2,4),r,2,4)
print(a)

And here are the results: \begin{array}{r|r} k&a_k\\\hline 2&128630045544\\ 3&26328445104\\ 4&4106040168\\ 5&429861360\\ 6&23219136\\ 7&279936 \end{array}

We can check them by noting that we should have $a_7=\binom42^7=6^7$, which checks out.

For the inclusion–exclusion calculation, we have $12$ conditions of having at least two cards in two particular consecutive ranks (since there are $12$ pairs of consecutive ranks), and we want to count the hands that fulfill at least one of these conditions.

So consider how to choose $j$ of these conditions. They can form $m$ overlapping runs (with $1\le m\le j$), and then they cover $j+m$ different ranks, which can be chosen in $\binom{j-1}{m-1}\binom{13-j}m$ ways (since we can distribute the $j$ conditions over the $m$ non-empty runs in $\binom{j-1}{m-1}$ ways (see stars and bars) and then we choose $m$ positions for the runs among the $m$ runs and the $13-(j+m)$ remaining ranks). So our count is

$$ \sum_{j=1}^6(-1)^{j+1}\sum_{m=1}^j\binom{j-1}{m-1}\binom{13-j}ma_{j+m}\;. $$

Here’s the corresponding Sage code:

    print(sum((-1)**(j+1) * binomial (j-1,m-1) * binomial (13-j,m) * a [j + m] for j in range(1,7) for m in range(1,j+1)))

The result is $1104417845112$. Here’s Java code that confirms the result by enumeration. Thus, the probability for a $14$-card hand drawn from a standard $52$-card deck to contain at least one pair of consecutive pairs is

$$ \frac{1104417845112}{\binom{52}{14}}=\frac{46017410213}{73706931025}\approx62.433\%\;. $$