Geometric Probabilities: Random Quadrilaterals?

I found manjyomesando1's analysis quite interesting but I don't agree with her numbers. This seems like quite a difficult problem, so to get a rough idea a wrote a program to simulate this. The program is thrown together without great care, so I may have to revise, but tentatively approximate values are:

P(convex) = 0.23

P(concave) = 0.31

P(self-crossed) = 0.46

An accurate value, incidentally, but not used above, for the expected length of each edge of the quadrilateral is 0.521405433*R.

I would be interested to learn if these numbers, for convex, concave, self-crossed, agree more or less with what anyone else has found.

EDIT: My program seems to be OK, it worked like this: for convex check that rotation sense does not change as you move around the quadrilateral. For self-crossing actually check for line segment intersection for edges 23, 14 or 12, 34. For concave all the rest, no check.

The results were consistent with the new program I wrote this morning following Duke's outline in his additional comments. Here are the frequency results calculated from 12 batches (from the new program) each containing 100,000 random quadrilaterals:

convex: 0.23189, with s.d. 0.00117

concave: 0.30513, with s.d. 0.00102

self-crosses: 0.46298, with s.d. 0.00156

The standard deviations are based on the 12 batch frequency numbers.

behaviour of probabilities when R -> infinity,

P(convex) -> 1/6

P(concave) -> 1/2

P(self-crossed) -> 1/3

for any 3 points chosen, the other one will make its shape.

with only 3 points, we only have 2 straight lines. if we connect P1 to P3 and make the 3 lines longer on both sides, we'll have 6 areas surrounding triangle P1P2P3. concave is when P4 to be within P1P2P3 or the 3 areas that do not share sides with the triangle. convex is when P4 inside the area sharing side P1P3 with triangle. self-cross is when P4 inside the other 2 areas, sharing side P1P2 and P2P3 each with the center triangle.

the probability = (sum of areas) / R^2

sorry, i was only "frying" the answer. i don't even know how to measure the areas if it is the way to do it (which is not). the one thing useful is that the 7-areas idea showed P(self-crossed) = 2P(convex), as JB's sims confirmed (thank you, JB!). maybe the answer is more toward 3 : 4 : 6 or 2 : 3 : 4 ratio.

i think of I see the place you made your mistake. you have S - S(a million - p) = a million - (a million - p)^n. I even have S - S(a million - p) = p(a million - (a million - p)^n). right this is my artwork. define S = p((a million-p)^0 + (a million-p)^2 + ... + (a million-p)^n-a million). Then S(a million-p) = p((a million-p)^0 + (a million-p)^a million + ... + (a million-p)^n-a million)(a million-p) = p((a million-p)^a million + (a million-p)^2 + ... + (a million-p)^n). And S - S(a million - p) = p((a million-p)^0 + (a million-p)^a million + ... + (a million-p)^n-a million) - p((a million-p)^a million + (a million-p)^2 + ... + (a million-p)^n) = p[(a million-p)^0 + (a million-p)^a million + ... + (a million-p)^n-a million - {(a million-p)^a million + (a million-p)^2 + ... + (a million-p)^n}] = p[(a million-p)^0 + (a million-p)^a million - (a million-p)^a million + ... + (a million-p)^n-a million - (a million-p)^n-a million + (a million-p)^n] = p[a million - (a million-p)^n] for this reason S - S(a million-p) = p[a million - (a million-p)^n] S[a million-(a million-p)] = p[a million-(a million-p)^n] S[a million-a million+p] = p[a million-(a million-p)^n] S*p = p[a million-(a million-p)^n] S = a million - (a million-p)^n finally, we've an interest in the case while the better cut back of the sum is infinity. If 0 < p < a million, then 0 < a million-p < a million meaning that lim (n---> infinity) S = lim (n---> infinity) [a million-(a million-p)^n] = a million - lim (n---> infinity) (a million-p)^n = a million - 0 = a million.