Chance of a rotten hand

The set of possible denomination pairings that satisfy the 1st and 3rd conditions is

$\{(2,7), (2,8), (2,9), (2,10), (3,8), (3,9), (3,10), (4,9), (4,10), (5,10)\}$ .

For each of these $10$ pairings we can then choose the first element in $4$ ways and the second in $3$ ways, (in order to satisfy the 2nd condition). So there are a total of $10 \times 4 \times 3 = 120$ "rotten" hands, leading to a probability of

$\dfrac{120}{\dbinom{52}{2}} = \dfrac{120}{26 \times 51} = \dfrac{20}{221} \Longrightarrow a + b = 20 + 221 = \boxed{241}$ .