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

Solution w/o expl. and a question by Jer)

This solution agrees with mine. To see why, summing the fractions gets
(6^6+ 5^1.6^5+ 5^2.6^4+ 5^3.6^3+ 5^4-6^2+ 5^5.6^1)/6^6. The numerator
equals (6^6-5^5), as can be seen by remembering the formula
(x^n-y^n)=(x-y)(x^n-1+ x^(n-2).y^1+ x^(n-3).y^2 + ... + y^(n-1)). For
n=56, his result is approximately 1-1/e.