The double Monty Hall

There are four different ways to play the game: you could stick both times, stick then swap, swap then stick, or swap both times.

If you stick both times, you have a $\frac{3}{13}$ chance of winning the car, regardless of the revealing and swapping of Monty, because there are $3$ cars out of $13$ doors at the time when you pick in the first round. These odds are equivalent to a $\frac{3}{13} = \frac{48}{208}$ chance

If you swap then stick, there are two paths to winning the car: picking a door with a goat in the first round then swapping to a door with a car in the second round (goat-car), or picking a door with a car in the first round then swapping to a door with a car in the second round (car-car). For the goat-car path, there is a $\frac{10}{13}$ chance of picking the door with the goat in the first round (because there are $10$ goats out of $13$ doors) and then a $\frac{2}{8}$ chance of swapping to a door with a car (because there are $2$ cars out of the $8$ doors available to swap to), for a $\frac{10}{13} \cdot \frac{2}{8} = \frac{20}{104}$ chance. For the car-car path, there is a $\frac{3}{13}$ chance of picking the door with the car in the first round (because there are $3$ cars out of $13$ doors) and then a $\frac{1}{8}$ chance of swapping to a door with a car (because there is only $2 - 1 = 1$ car left out of the $8$ doors available to swap to since you are currently on a door with a car), for a $\frac{3}{13} \cdot \frac{1}{8} = \frac{3}{104}$ chance. These two paths add up to a $\frac{20}{104} + \frac{3}{104} = \frac{23}{104} = \frac{46}{208}$ chance.

If you stick then swap, there are two paths to winning the car: picking a door with a goat in the first round then swapping to a door with a car in the third round (goat-car), or picking a door with a car in the first round then swapping to a door with a car in the third round (car-car). For the goat-car path, there is a $\frac{10}{13}$ chance of picking the door with the goat in the first round (because there are $10$ goats out of $13$ doors) and then a $\frac{1}{2}$ chance of swapping to a door with a car (because there is $1$ car out of the $2$ doors available to swap to), for a $\frac{10}{13} \cdot \frac{1}{2} = \frac{5}{13}$ chance. For the car-car path, there is a $\frac{3}{13}$ chance of picking the door with the car in the first round (because there are $3$ cars out of $13$ doors) and then a $\frac{0}{2}$ chance of swapping to a door with a car (because there no cars available out of the $2$ doors left to swap to since you are currently on a door with a car), for a $0$ chance. These two paths add up to a $\frac{5}{13} + 0 = \frac{5}{13} = \frac{80}{208}$ chance.

If you swap both times, there are four paths to winning the car: picking a door with a goat in the first round then swapping to a door with a goat in the second round then swapping to a door with a car in the third round (goat-goat-car), picking a door with a goat in the first round then swapping to a door with a car in the second round then swapping to a door with a car in the third round (goat-car-car), picking a door with a car in the first round then swapping to a door with a goat in the second round then swapping to a door with a car in the third round (car-goat-car), or picking a door with a car in the first round then swapping to a door with a car in the second round then swapping to a door with a car in the third round (car-car-car). For the goat-goat-car path, there is a $\frac{10}{13}$ chance of picking the door with a goat in the first round (because there are $10$ goats out of $13$ doors) and then a $\frac{6}{8}$ chance of swapping to a door with another goat (because there are $7 - 1 = 6$ goats left out of the $8$ doors available to swap to since you are currently on a door with a goat) and then a $\frac{1}{2}$ chance of swapping to a door with a car (because there is $1$ car out of the $2$ doors available to swap to), for a $\frac{10}{13} \cdot \frac{6}{8} \cdot\frac{1}{2} = \frac{60}{208}$ chance. For the goat-car-car path, there is a $\frac{10}{13}$ chance of picking the door with a goat in the first round (because there are $10$ goats out of $13$ doors) and then a $\frac{2}{8}$ chance of swapping to a door with a car (because there are $2$ cars out of the $8$ doors available to swap to) and then a $\frac{0}{2}$ chance of swapping to a door with a car (because there no cars available out of the $2$ doors left to swap to since you are currently on a door with a car), for a $0$ chance. For the car-goat-car path, there is a $\frac{3}{13}$ chance of picking the door with a car in the first round (because there are $3$ cars out of $13$ doors) and then a $\frac{7}{8}$ chance of swapping to a door with another goat (because there are $7$ goats out of the $8$ doors available to swap to) and then a $\frac{1}{2}$ chance of swapping to a door with a car (because there is $1$ car out of the $2$ doors available to swap to), for a $\frac{3}{13} \cdot \frac{7}{8} \cdot\frac{1}{2} = \frac{21}{208}$ chance. For the car-car-car path, there is a $\frac{3}{13}$ chance of picking the door with a car in the first round (because there are $3$ cars out of $13$ doors) and then a $\frac{1}{8}$ chance of swapping to a door with a car (because there are $2 - 1 = 1$ cars left out of the $8$ doors available to swap to since you are currently on a door with a car) and then a $\frac{0}{2}$ chance of swapping to a door with a car (because there no cars available out of the $2$ doors left to swap to since you are currently on a door with a car), for a $0$ chance. These four paths add up to a $\frac{60}{208} + 0 + \frac{21}{208} + 0 = \frac{81}{208}$ chance.

Therefore, the optimal play would be to swap both times for a $\frac{81}{208}$ chance of winning. This means that $a = 81$ and $b = 208$ , and $a + b = \boxed{289}$ .