Assume that after breaking the stick into two, the left piece is of length h. If h < 1/2, the probability of P falling into the left pink triangle is h/(1-h) which should be integrated from 0 to 1/2. If h > 1/2, the probability of P falling into the right pink triangle is (1-h)/h which should be integrated from 1/2 to 1. The two integrals are equal (and in fact reduce to each other after a substitution h = 1-h). The sought probability is double their common value of ln(2) - 1/2.
After the first random break, we can (without loss of generality) skip the step of randomly selecting one of the pieces, since the distribution of the left and right pieces will be identical anyway. This simplifies the problem a bit.
Say we always choose the leftmost piece; call its length x.
At this stage we have 2 pieces: [0,x] and [x,1].
Now we want to break [0,x] "at a random point", which amounts to choosing (independent of x) a number y at random in the interval [0,1], where y represents the fraction of x where the break will occur.
So, we end up with 3 pieces: [0,xy], [xy, x], and [x,1]. Their lengths are clearly xy, x-xy = x(1-y), and 1-x.
The distribution on x and y is uniform on the unit square [0,1]×[0,1].
And the region where the 3 pieces form a triangle corresponds to precisely the condition that all 3 sides are of length < 1/2.
The subset of the square, then, where xy < 1/2, x(1-y) < 1/2, and 1-x < 1/2 (i.e. x > 1/2) is the thorn-shaped region bounded on the left by the line x = 1/2 and on the right by the 2 hyperbolas xy = 1/2 and x(1-y) = 1/2.
Calculus then gives the area of this region -- which must be the probability we are seeking -- as Prob(triangle) = ln(2) - 1/2 = .1931471805599453094172321....
