Assume the plane needs to be within 1 degree from the line joining the star and kepler. I would guess it probably less. It would be a function of size of the planet, size of the star, radius of the orbit and distance of star from kepler.
That would give about 179 planes that we cannot see, which is two orders of magnitude, but the probability calculation won't be that simple.
Assuming you mean 1 degree in either direction, or 2/360 = 1/180, this is off by about a factor of pi/2 ~= 1.57. The correct answer is sin(pi/360) ~= pi/360 ~= 1/115. Still two orders of magnitude, but slightly better. The reason is that the degrees are not equally likely (small degrees, that you want, are more likely, and large degrees less likely).
Here's the calculation: Consider the unit vector normal to the equator of the orbit, using say the right-hand rule. This vector is uniformly distributed over the unit sphere (surface area 4pi). You want the vector to lie within +-1 degree of the rim of the sphere. Call the region of such vectors R; the probability of a good planet is P = area(R)/(4pi).
R is an annulus going around the equator and up/down by 1 degree. In area, this is close to a rectangle with width 2pi (circumference of sphere) and height 2pi/360 (i.e., 2 degrees), so P ~= (pi^2/90) / (4pi) = pi/360.
The area of R (an equatorial annulus) can be computed exactly using calculus---or looked up on Wikipedia [1]: area(R) = 4pi - 2(2pi(1-sin(pi/360))) = 4pi sin(pi/360). Thus, P = sin(pi/360).
[1] https://en.wikipedia.org/wiki/Spherical_cap#Volume_and_surfa...
A naive calculation for Earth would be Theta ~= 2 * (Earth diameter) / (Mean distance to Sun), which is ~2*10^-4. Since most planets found by kepler were much larger than earth, it seems 10^-2 is a reasonable guesstimate.