Your lucky day ......

How do you shuffle a deck of cards? Here's one way: Place the four 7s down on the table, with some space between. Now there are 3 spaces between the 7s and 2 spaces on either end. Place each of the remaining cards randomly in 1 of the 5 spaces. Then collect up the deck. The number of cards after the last 7 is equal to the number of cards in the last pile, and the expected value is 48*(1/5).

So S=52-48/5=212/5

This scenario is equivalent to finding the expected position of the last $7$ in a string of $x$ 's and $7$ 's, where the $x$ 's represent any denomination other than a $7$ .

Now there are $\binom{52}{4}$ ways the four $7$ 's can be positioned in the deck. There is one way that the string can have the last $7$ in the $4$ th position, $\binom{4}{3}$ ways in which the last $7$ shows up in the $5$ th position, and in general, $\binom{n}{3}$ ways in which the last $7$ can show up in the $(n + 1)$ st position for $3 \le n \le 51.$

The expected value is then the weighted average of these individual values, i.e.,

$\dfrac{1}{\binom{52}{4}}*\displaystyle\sum_{n=3}^{51} (\dbinom{n}{3}*(n + 1)) = \dfrac{4}{\binom{52}{4}}*\sum_{n=4}^{52} \dbinom{n}{4} = \dfrac{4}{\binom{52}{4}}*\dbinom{53}{5}$

where this last step made use of the Hockey-Stick Identity .

Simplifying, this last value becomes $\dfrac{\frac{4*53!}{48!*5!}}{\frac{52!}{48!*4!}} = \dfrac{4*53}{5} = \dfrac{212}{5}.$

Thus $a + b = 212 + 5 = \boxed{217}.$