I assume the points are ordered 1 to N, we start at point 1, the (N – 2)-nd distinct point is the point we reach such that there are 2 points we have not visited, and the starting point counts as visited.
Let X(t) be a stochastic process such that X(0) = 0 and it either increments or decrements by 1 each time step with equal probability. It is easy to see X(t) is a martingale. Now let Y(t) = X(t) % N + 1 be our position at time t in the circle.
We want to calculate the probability P(p) of reaching a point p without visiting point q, given our current position Y(t). Let dist(p) be the distance we need to travel to reach p and dist(q) be the distance we need to travel in the other direction to reach q. The time taken to reach one of these points is a stopping time with finite expectation. Assuming p is in the positive direction (the probability does not change because of symmetry), the expected value E[dX(t)] = P(p)dist(p) - (1 – P(p))dist(q). Because X(t) is a martingale, this expectation is equal to zero and P(p) = dist(q) / (dist(p) + dist(q)).
Let x, y be points 3 away from each other. Point x is the (N – 2)-nd distinct point iff we reach point y without visiting x or the 2 points in between, and then make a trip around the circle to point x so that the only unvisited points are those 2 in between x and y. For the final trip, we travel N – 3 points without visiting the point 1 away in the other direction. Using the formula above, with dist(p) = N – 3 and dist(q) = 1, we calculate the probability P(x | y) = 1 / (N – 2). This is the probability x is the (N – 2)-nd distinct point given that we reached point y without visiting x or the points in between. Now we need to find the probability of reaching point y without visiting x or the points in between, and starting at point 1. There are 4 cases:
x=r+1,y=r−2:dist(x)=(1+N)−(r+1)=N−rdist(y)=(r−2)−1=r−3P(y)=N−3N−r,r≥3P(y)=0,r<3(x=r+1, y=r+4:\\dist(x)=(r+1)-1=r\\dist(y)=(1+N)-(r+4)=N-r-3\\P(y)=\frac{r}{N-3}, r\leq N-3\\P(y)=0, r>N-3\)
x=r−1,y=r−4:dist(x)=(1+N)−(r−1)=N−r+2dist(y)=(r−4)−1=r−5P(y)=N−3N−r+2,r≥5P(y)=0,r<5
x=r−1,y=r+2:dist(x)=(r−1)−1=r−2dist(y)=(1+N)−(r+2)=N−r−1P(y)=N−3r−2,r≤N−1P(y)=0,r=N
Adding the probabilities of the disjoint cases together, and multiplying by P(x | y) = 1 / (N – 2) yields the solution, the probability that the (N – 2)-nd distinct point is r + 1 or r – 1. For the trivial cases N = 3 or N = 4, the probabilities are always 0 or 1 depending on r. For N > 4:
P(r+1)=(N−2)(N−3)r,r≤2
P(r+1)=(N−2)(N−3)N,2<r≤N−3
P(r+1)=(N−2)(N−3)N−r,r>N−3
P(r−1)=(N−2)(N−3)r−2,r≤4
P(r−1)=(N−2)(N−3)N,4<r≤N−1
P(r−1)=(N−2)(N−3)2,r=N