Hard to create a regular polygon

Clearly the only regular polygons we can create are of degree $3, 4, 6, 12$ .

For degree $3$ , the possible sets of points are: $\{1,5,9\}, \{2,6,10\}, \{3,7,11\}$

The chance of getting one of theses is $1 \cdot \frac{2}{11} \cdot \frac{1}{10} = \frac{1}{55}$ because any initial point can be part of the polygon, but once that point is fixed there are only 2 of the remaining 11 points that can form the polygon, and after choosing one of those, then only 1 more of the remaining 10 works.

Similarly, the chances of getting a square are $1 \cdot \frac{3}{11} \cdot \frac{2}{10} \cdot \frac{1}{9} = \frac{1}{165}$ , and all of these are independent to the triangle cases.

For the regular hexagon, it is not independent of the triangle selection (but is independent of the square). We just have to make sure we exclude the cases where the first three points chosen formed a regular triangle.

The two sets of points are $\{1,3,5,7,9,11\}, \{2,4,6,8,10,12\}$ After our first choice we split into two cases: the second point is part of a regular triangle (but the third point is not); or the second point is not part of a regular triangle. These cases each have probabilities respectively

$1 \cdot {2 \over 11}{3 \over 10}{3 \over 9}{2 \over 8}{1 \over 7} = {1 \over 1540}$ and $1 \cdot {3 \over 11}{4 \over 10}{3 \over 9}{2 \over 8}{1 \over 7} = {1 \over 770}$

The sum of these four cases is $11 \over 420$ . So the total expected number of points is

$E = 3 {1 \over 55} + 4 {1 \over 165} + 6 {1 \over 770} + 6{1 \over 1540} + 12(1 - {11 \over 420})$ so then $12 - E = 12({11 \over 420}) - 3 {1 \over 55} - 4 {1 \over 165} - 6 {1 \over 770} - 6{1 \over 1540}$

which comes out to $\boxed{47 \over 210}$ .

Let us label the twelve points around the circle 1 to 12.

Without loss of generality, we may assume that point 1 is chosen first.

Then we shall consider the 11 possibilities of which point is chosen next.

If 2 or 12 are chosen next, the shape formed must be a dodecagon. So in both of these cases, the expected number of points chosen is 12.

If 3 or 11 are chosen next, we can either form a hexagon or a dodecagon. For a hexagon, we need points 5, 7, 9 and whichever of 3 or 11 was not chosen first. The probability of this is $\frac{4}{10} \times \frac{3}{9} \times \frac{2}{8} \times \frac{1}{7}=\frac{1}{210}$ . The probability of a dodecagon is thus $1-\frac{1}{210}=\frac{209}{210}$ . We multiply the number of points of the polygon formed by the probability of it happening, so the expected number of points chosen if 3 or 11 are chosen next is $12\times(1-\frac{1}{210})+6\times\frac{1}{210}=12-\frac{1}{35}$

If 4 or 10 are chosen next, we can either form a square or a dodecagon. For a square, we need point 7 and whichever of 4 or 10 was not chosen first. The probability of this is $\frac{2}{10}\times\frac{1}{9}=\frac{1}{45}$ . The probability of a dodecagon is thus $1-\frac{1}{45}=\frac{44}{45}$ . So the expected number of points chosen if 4 or 10 are chosen next is $12\times(1-\frac{1}{45})+4\times\frac{1}{45}=12-\frac{8}{45}$

If 5 or 9 are chosen next, we can either form a triangle, a hexagon or a dodecagon. For a triangle, we only need choose whichever of 5 or 9 was not chosen first. The probability of this is $\frac{1}{10}$ . For a hexagon, we need to choose points 3, 7, 11, and whichever of 5 or 9 was not chosen first. However, we need to be careful because the first point to be chosen cannot be whichever of 5 or 9 was not chosen first, because then that completes the {1,5,9} equilateral triangle. So the probability of a hexagon forming is $\frac{3}{10}\times\frac{3}{9}\times\frac{2}{8}\times{1}{7}=\frac{1}{280}$ . The probability of a dodecagon is thus $1-\frac{1}{10}-\frac{1}{280}$ . And so, the expected number of points chosen if 5 or 9 are chosen next is $12\times(1-\frac{1}{10}-\frac{1}{280})+3\times\frac{1}{10}+6\times\frac{1}{280}=12-\frac{129}{140}$

If 6 or 8 are chosen next, the shape formed must be a dodecagon. So in both of these cases, the expected number of points chosen is 12.

If 7 is chosen next, we can either form a square, a hexagon or a dodecagon. For a square, we need points 4 or 10 to be chosen next. The probability of this is $\frac{2}{10}\times\frac{1}{9}=\frac{1}{45}$ . For a hexagon, we need points 3, 5, 9, and 11 to be chosen next. The probability of this is $\frac{4}{10} \times \frac{3}{9} \times \frac{2}{8} \times \frac{1}{7}=\frac{1}{210}$ . The probability of a dodecagon is thus $1-\frac{1}{45}-\frac{1}{210}$ . The expected number of points chosen if 7 is chosen next is $12\times(1-\frac{1}{45}-\frac{1}{210})+4\times\frac{1}{45}+6\times\frac{1}{210}=12-\frac{8}{45}-\frac{1}{35}$

Since the answer is in the format $12-\frac{a}{b}$ , we need only work out the part that is $\frac{a}{b}$ . Since our answers have been written in the form $12-\frac{x}{y}$ , we can work this out by adding up all the fractional parts $\frac{x}{y}$ of our answers, and then dividing by eleven, since there are eleven possibilities in total for which point is chosen next after 1.

So the answer is $\huge\frac{3\times\frac{1}{210}+3\times\frac{8}{45}+2\times\frac{129}{140}}{11}=\frac{47}{210}$ .

$\large a+b=47+210=257$