@user31264 The problem isn't the OP's simulation. The problem is that the OP's intuition is wrong that a greater likelihood of hitting a node sooner implies a greater likelihood of hitting the node last. To reach a node $X$ last, you need to reach a neighbor of $X$ then come entirely around the ring to reach $X$ from the other direction. At all times, this is equally likely for any node not on the boundary of the interval visited. And initially, it is equally likely for all nodes. These explanations are excellent: math.stackexchange.com/questions/116446/random-walk-on-n-cycle

You say that my situation was discussed earlier here. But, that analysis does shifts focus away from the paradox I address. Thanks for pointing out that I have no code bugs. Now, it is time to realize that I do not have any false assumptions either. My code confirms all the assumptions and theoretical results and confirms the intuitive paradox I had. Just think about it: you hit A, the the left and right elements of the table sooner than reach B, the center of the table. However, getting later to A is as likely as later B. How is it possible? Why do not you see neither my question nor paradox?

@LittleAlien What's wrong with your intuition is that getting RIGHT next to a node doesn't increase the probability at all! If you're right next to it, there's a 50% probability you hit it and then it can NEVER be the last node. The only way the probability for a node $X_i$ goes up is that: (1) you happen to have gotten next to it and (2) you get lucky and move away. I advise reading the math.stackexchange.com link and going through those two answers very carefully.

@MatthewGunn I do not analyze "getting next to". I am comparing the probabilities of getting into: getting into A vs. getting into B. As you say, if we run into A sooner then we reduce its chances to be the last visited. Yet, we see that the chances are not reduced. That puzzles me. But, I think that it is less paradoxical after your explanation. You basically say that it is irrelevant if we visit A first or after B in average.

Once a node has been visited, it cannot be the last visited node.

@MatthewGunn That what makes me puzzling: you visit A first (in 2/3 of cases). This must make it less likely to be the last visited. BUT IT IS NOT!

Let's say you have n nodes.

Maybe, let's make it concrete, and say you have 11 nodes

each node (other than 5) has a 1/10 chance of being hte last node.

50% of the time, your first move will be to node 4

50% of the time your first move will be to node 6

if you move to node 4, it goes to 0% chance of being the last node

and node 6 goes to a 2/10 chance of being the last node

if you move to node 3, node 6 goes to 3/10

the probability of each node being the last node is 1/10

EXCEPT for the nodes at the edge of the interval you have visited

let's say k is hte length of the interval you have visited