A pair or two ....

A quick, cryptic solution ....

For any given card, the probability of it being in a pair is $\dfrac{3}{51}$ . Thus the expected number of pairs will be $26 * \dfrac{3}{51} = \dfrac{26}{17}$ . This means that $a = 26, b = 17$ and $a + b = \boxed{43}$ .

There are $\frac{\binom{52}{2} \binom{50}{2} \cdots \binom{4}{2}}{26!}$ ways to divide the deck into piles. There are $78$ pairs. Each pair appears in $\frac{\binom{50}{2} \binom{48}{2} \cdots \binom{4}{2}}{25!}$ of the divisions. The answer is the product of the last two numbers divided by the first one. This is $\frac{78 \cdot 26}{\binom{52}{2}} = \frac{26}{17},$ so the answer is $\fbox{43}$ .

The number of cards in the deck in $n=52$ . Let $I_i$ the indicator variable such that

$I_i = \begin{cases} 1, & \text{if the couple} \ i \ \text{has a matching pair} \\ 0, & \text{otherwise} \end{cases}$

for $i \in [1, n/2]$ . The number of matching pairs $M$ is

$\displaystyle M = 13\binom{4}{2}$

given that there are cards from $1$ to $13$ (aces to kings) and $4$ suites. The total number of possible pairs $T$ is

$\displaystyle T = \binom{n}{2}$ .

Hence, the probability of getting a matching pair in

$\displaystyle \mathbb{P}(I_i=1) = \frac{M}{T} = 13\binom{4}{2}\binom{n}{2}^{-1}$

The number of total matching pairs $C$ is

$\displaystyle C = \sum_{i=1}^{n/2} I_i$ ,

so that the expected number of matching pairs is

$\displaystyle \mathbb{E}[C] = \mathbb{E}\bigg[\sum_{i=1}^{n/2} I_i\bigg] = \sum_{i=1}^{n/2} \mathbb{E}[I_i] = \sum_{i=1}^{n/2} [1\cdot \mathbb{P}(I_i=1) + 0 \cdot \mathbb{P}(I_i=0)] = \sum_{i=1}^{n/2} \mathbb{P}(I_i=1) = \frac{n}{2} \cdot 13\binom{4}{2}\binom{n}{2}^{-1} =\frac{26}{17}$

The sum of the numerator and denominator in $\displaystyle 26 + 17 = \boxed{43}$