around the table

The Riddler has a variant on the classical (discrete) random walk around a circle where every state (but the starting point) has the same probability 1/(n-1) to be visited last. Surprising result that stems almost immediately from the property that, leaving from 0, state a is visited couterclockwise before state b>a is visited clockwise is b/a+b. The variant includes (or seems to include) the starting state 0 as counting for the last visit (as a return to the origin). In that case, all n states, including the origin, but the two neighbours of 0, 1, and n-1, have the same probability to be last. This can also be seen on an R code that approximates (inner loop) the probability that a given state is last visited and record how often this probability is largest (outer loop):

w=0*(1:N)#frequency of most likely last
for(t in 1:1e6){
 o=0*w#probabilities of being last
 for(v in 1:1e6)#sample order of visits
   o[i]=o[i<-1+unique(cumsum(sample(c(-1,1),300,rep=T))%%N)[N]]+1
 w[j]=w[j<-order(o)[N]]+1}

However, upon (jogging) reflection, the double loop is a waste of energy and

o=0*(1:N)
for(v in 1:1e8)
   o[i]=o[i<-1+unique(cumsum(sample(c(-1,1),500,rep=T))%%N)[N]]+1

should be enough to check that all n positions but both neighbours have the same probability of being last visited. Removing the remaining loop should be feasible by considering all subchains starting at one of the 0’s, since this is a renewal state, but I cannot fathom how to code it succinctly. A more detailed coverage of the original problem (that is, omitting the starting point) was published the Monday after publication of the riddle on R bloggers, following a blog post by David Robinson on Variance Explained.