reaverprog,i dont think you mean vertices, e.g., when n=1, there are only 2 steps, there would be no path for t=1, because the last step always ends at 0, one has P(2,2)=P(1,2)=1/2 and P(0,2)=0.the question would make sense if one considers instead the number of steps on the positive side (2t) and negative side (2n-2t). the answer is an equidistribution, i.e., P(2t, 2n)=1/(n+1) for t=0,1,...,n.this is startling if you happen to know the arcsine law in the discrete setup. when one removes the constraint that the final step return to zero, the probability instead is P'(2t,2n)=(2t choose t)*(2n-2t choose n-t)/2^(2n). fixing x=t/n then taking the limit of n->infty would result the arcsine distribution (by using Stirling formula). analogous results hold respectively for the position of maxima as well. a good reference for these results is Feller's textbook vol 1 chapter III.