"Kepler has observed over 156,000 stars simultaneously and near continuously to search for planets that periodically pass in front of their host star (transit)"
From the article:
"The confirmation of Kepler-452b brings the total number of confirmed planets to 1,030." — and I have verified that this refers specifically to planets confirmed by Kepler, not just planets in total.
From Wikipedia[2]:
"four, including Kepler-296f, were less than 2 1/2 the size of Earth and were in habitable zones where surface temperatures are suitable for liquid water."
These figures give us an approximate sense of the number of planets that occult their host star, as a proportion of the number of stars surveyed.
TRIGGER WARNING: SKETCHY BACK-OF-THE-ENVELOPE MATH
259 stars within 30 light years
1030 / 156000 = 0.00660256410256 (fraction of Kepler-surveyed stars possessing Kepler-confirmed planets that transit)
4 / 1030 = 0.00388349514563 (fraction of significantly Earth-like Kepler-surveyed planets)
259 * 0.00660256410256 * 0.00388349514563 = 0.00664102564102 or 0.66% (chance of a transiting Earth-like within 30 light years)
And that's not even factoring in planets that do not transit their host star.
[1] http://ntrs.nasa.gov/search.jsp?R=20100030619 [2] https://en.wikipedia.org/wiki/List_of_exoplanets_discovered_...
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.
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.
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...