The gap is here: the next randomly chosen point from the circle is equally likely to lie on each of the n + 2 arcs.
For any given configuration of points, the quoted statement is almost certainly false, since the arcs almost certainly have different lengths. To close the gap, you need to show that the statement becomes true when averaging over all configurations consistent with the observed h out of n Heads.
I’m not conditioning on any configuration of points. I agree it’s false for a given configuration of points, but that’s not relevant here. Instead, I’m saying: number the intervals clockwise from 1 to n + 2, starting with the interval clockwise of Z. Since the n + 2 points were chosen uniformly at random, the interval numbered k1 is just as likely to have the new point as the interval numbered k2, for any k1 and k2. This is a probability over the entire space of outcomes, not for any fixed configuration.
(Or, as you put it, the average probability over all configurations of the probability of landing in a given interval is the same for all intervals. But that’s needlessly complicated.)
Nice idea, but it’s not a proof.
The gap is here: the next randomly chosen point from the circle is equally likely to lie on each of the n + 2 arcs.
For any given configuration of points, the quoted statement is almost certainly false, since the arcs almost certainly have different lengths. To close the gap, you need to show that the statement becomes true when averaging over all configurations consistent with the observed h out of n Heads.
I’m not conditioning on any configuration of points. I agree it’s false for a given configuration of points, but that’s not relevant here. Instead, I’m saying: number the intervals clockwise from 1 to n + 2, starting with the interval clockwise of Z. Since the n + 2 points were chosen uniformly at random, the interval numbered k1 is just as likely to have the new point as the interval numbered k2, for any k1 and k2. This is a probability over the entire space of outcomes, not for any fixed configuration.
(Or, as you put it, the average probability over all configurations of the probability of landing in a given interval is the same for all intervals. But that’s needlessly complicated.)
I see it now. Good call.