The puzzle may have run its course. Probability is ratio of areas, and the triangle area is an hellacious multiple integral. The problem is known as Sylvester's four point problem.
I did find the number, as RG gave it, from simulation, not calculation. Uniform point-picking, necessary to form uniformly distributed triangles was an issue. Equally likely radius and angle gives preference to central points.
What was interesting to me when I simulated was that I was off by a factor of four that the four points would not be convex. That is, and should be 4 time the probability the 4th dart lands inside the triangle. There are three other areas the 4th dart can land that make a non-convex configuration, and those areas, on average, are equal to the triangle size. Not an intuitive result until you realize any of the darts can be taken to be the 4th dart.
So my last question now does not need to be asked, namely, to compare the probabilities of the two cases in the OP. also, that question has been answered in this thread.
Also, the mean triangle area was (to me) surprisingly small.
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