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?

What if the die is n-sided?

(In reply to

re: Solution w/o expl. and a question by Richard)

This is wrong: the sum is NOT to infinity, but to n=6. Also, you forgot
the (1-5/6) that divides [1-(5/6)]^n. The answer is (6^6-5^6)/6^6.