Probabilitical Players

Since $S_{1}$ has an equal chance of being paired with any of the other $15$ players, there is a probability of $\frac{1}{15}$ that $S_{1}$ and $S_{2}$ will be paired together and a probability of $\frac{14}{15}$ that they will be in different pairings.

If they are paired together then it is certain that exactly one of them will be among the eight winners.

If they are not paired together, then there is a probability of $\frac{1}{2}$ that exactly one of them will win their pairing.

Thus the desired probability is $(\frac{1}{15})(1) + (\frac{14}{15})(\frac{1}{2}) = \frac{8}{15}$ ,

and so $a + b = 8 + 15 = \boxed{23}$ .

$Can\quad think\quad of\quad two\quad cases\\ (1)\quad S1\quad and\quad S2\quad are\quad in\quad pair\quad and\quad (2)\quad it's\quad complement\\ P(S1\quad and\quad S2\quad are\quad in\quad pair)\quad =\quad \frac { 1 }{ 15 } \\ P(S1\quad and\quad S2\quad are\quad not\quad in\quad pair)\quad =\quad \frac { 14 }{ 15 } \\ P(Exact\quad one\quad from\quad S1\quad and\quad S2\quad in\quad final\quad 8)\\ =\quad P(S1\quad and\quad S2\quad are\quad in\quad pair)\quad P(Exact\quad 1\quad winner\quad from\quad S1\quad and\quad S2)\quad +\quad P(S1\quad and\quad S2\quad are\quad not\quad in\quad pair)\quad P(Exact\quad 1\quad winner\quad from\quad S1\quad and\quad S2)\\ =\quad \frac { 1 }{ 15 } \times 1+\frac { 14 }{ 15 } (1-\frac { 1 }{ 4 } -\frac { 1 }{ 4 } )\quad =\quad \frac { 1 }{ 15 } +\frac { 7 }{ 15 } =\frac { 8 }{ 15 } .\\ 8+15\quad =\quad 23$