At my department's tea yesterday, the following problem came up: Given three random points on a circle, what is the probability that they lie in the same semicircle?
(Incidentally, why do we insist on calling it "tea" when most of the people there drink coffee? I suspect British influence at some point.)
First, by "random" I mean "uniformly at random, with respect to angle or arc length"; a classic problem in this sort of thing is to determine the probability that the length of a random chord of a circle is larger than the length of a side of an equilateral triangle inscribed in the circle. This is Bertrand's paradox. A chord is uniquely determined by its midpoint, and the result is different if you choose a chord by picking its midpoint uniformly at random from the interior of the circle than by choosing its endpoints uniformly at random from the boundary. I feel this is "more ambiguous" in the two-point case than in the three-point case, because there's a nice way to associate a single point in the interior with two points on the boundary.
The answer to my problem is 3/4.
Consider the circle as the interval [0, 1] with the endpoints identified. Then the problem is one about choosing triplets of real numbers (x1, x2, x3) in that interval, where these numbers are chosen uniformly at random and independently. Replacing these with (0, y2, y3) = (0, x2-x1, x3 - x1) corresponds to rotating the circle by x1 and therefore doesn't change the probability of our event. Finally, if either of y2 and y3 are between 1/2 and 1, subtract 1 from it; now we're just viewing the circle as the interval [-1/2, 1/2] (centered on the first point) instead of [0, 1].
So now our event is just that |y3 - y2| ≤ 1/2. If y3 and y2 have the same sign, this is obviously true, and we can take the semicircle [0, 1/2] or [-1/2, 0]. If y3 and y2 have different signs, but |y3 - y2| ≤ 1/2, then some interval of size 1/2 containing y2 and y3. But if |y2 - y3| ≤ 1/2, -- for example, if y2 = 0.3, y3 = -0.3 -- then we lose. Any interval of width one-half containing both of those -- say the union of [0.25, 0.5] and [-0.5, -0.25], recalling that 0.5 and -0.5 are identified -- won't contain zero.
And the probability that two random points in a unit interval are within distance 1/2 of each other is just 3/4.
Alternatively, if we're trying to find this interval, one of the points is at the "counterclockwise" end of the interval. So fix which point is counterclockwisemost in the interval, there are three ways to do that. Then the other two points must be within 180 degrees clockwise from that point; the probability of that is 1/4. The "most counterclockwise" point is unique, so we can just add up the three probabilities to get 3/4. In general, if "semicircle" is replaced with some section of the circle making up a proportion c of the whole, where c ≤ 1/2, the probability that three randomly chosen points lie in an interval of that size is 3c2.