# What's the probability of getting a pair of king and ace with the same suit?

First off, the answer that was given is $$\frac{{4\choose1}{2\choose2}{50\choose11}-{4\choose2}{4\choose4}{48\choose9}+{4\choose3}{6\choose6}{46\choose7}-{4\choose4}{8\choose8}{44\choose5}}{52\choose13} = 0.21978$$

It's easy to tell that the answer here is using the inclusion-exclusion principle.

However, my problem with this solution is that: If we consider the 4 sets, $S_1$,$S_2$,$S_3$,$S_4$, where $S_i$ stands for all the possible combinations of getting $i$ pairs of kings and aces with the same suit. Then isn't $S_4$ a subset of $S_3$, $S_3$ a subset of $S_2$,$S_2$ a subset of $S_1$? If so, then finding all possible combinations of getting only a pair is equal to $|S_1|-|S_2|$. In short, why is the latter two terms in the numerator of the solution needed?And what's wrong with my reasoning above?

I interpret your $S_1$ as the set of all combinations of cards that have at least one same suit Ace King pair. Similarly, I interpret your $S_2$ as the set of all combinations that have at least two same suit Ace King pair.

Then indeed the number of combinations with exactly one same suit Ace King pair is $|S_1|-|S_2|$.

However, the first term $\binom{4}{1}\binom{2}{2}\binom{50}{11}$ does not count $S_1$. In fact, it does not count anything.

Note that $\binom{2}{2}\binom{50}{11}$ counts the hands that have (at least) the Ace and King of $\spadesuit$. We get the same count for the hands that have at least the Ace and King of $\heartsuit$, and so on.

But adding these counts, which is equivalent to multiplying by $\binom{4}{1}$, double-counts each hand that has exactly two same suit Ace King pairs. It triple-counts the hands that have exactly three same suit, and quadruple counts the hands that have exactly four. In particular, $\binom{4}{1}\binom{2}{2}\binom{50}{11}$ overcounts $S_1$.

So, to repeat for emphasis, $\binom{4}{1}\binom{2}{2}\binom{49}{11}$ does not count $S_1$. And similarly, $\binom{4}{2}\binom{4}{4}\binom{48}{9}$ does not count $S_2$. The Inclusion/Exclusion strategy does not attempt to find our number by finding $|S_1|-|S_2|$.

The issue is that the $S_2$ that you define in words does have ${4 \choose 2}{4 \choose 4}{48 \choose 9}$ card hands. There is an over counting here that must be dealt with.

To see this, let's break down these binomial coefficients.

• First, we choose the two suits for our pairs. [${4 \choose 2}$ ways to do this]
• Second, once we've chosen the suits, there are four cards with these suits and we pick all of them. [${4 \choose 4} =1$ ways to do this]
• Third, we choose the remaining 9 cards out of all possible 52-4=48 remaining cards. [${48 \choose 9}$ ways to do this]

To see the overcounting, let us choose $\spadesuit, \clubsuit$ in first step, so after second step we have the cards $A \spadesuit, K \spadesuit, A \clubsuit, K \clubsuit$. In the third step, we pick $A \heartsuit, K \heartsuit$ and 7 other cards. Conversely, let us choose $\heartsuit, \clubsuit$ in first step, so after second step we have the cards $A \heartsuit, K \heartsuit, A \clubsuit, K \clubsuit$. In the third step, we pick $A\spadesuit, K \spadesuit$ and the same 7 other cards. !!!

So this final hand is counted (at least) twice!