Two players alternate throwing a six-sided die. The first player who fails to roll a higher number than the preceding roll loses. What is the probability that the first player wins?
I did the case for n=2, n=3 and n=4, and the chances of the second
player's win were (1/2)^2, (2/3)^3 and (3/4)^4, which leads me to
suppose the answers to this puzzle are (5/6)^6 and ((n-1)/n)^n that in
the limit is 1/e.
Theory (maybe solution)
