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)

Your expression is (1/6)*[sum from k=0 to 6-1 of (5/6)^k]. The geometric series sums to [1-(5/6)^6]/(1-5/6), so the result is just

               1-(5/6)^6

and  presumably this becomes

               1-[(n-1)/n]^n=1-(1-1/n)^n

for general n and goes to 1-1/e as n gets large. Your answer appears then to give the probability of the opposite event to that for which the probability was calculated in the first comment below.

Edited on April 7, 2005, 8:58 pm

Posted by Richard on 2005-04-07 20:45:51