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?
for n=6 the solution is
1/6 + 5/6^2 + 5^2/6^3 + 5^3/6^4 + 5^4/6^5 + 5^5/6^6 = 31031/46656 = .6651020233
One can easily see the pattern that extends this to any n-sided die.
My question: what value does this tend to as n increases?
The highest I can go is n=56 solution = .6354298564 which is not very close to anything I recognize.
Posted by Jer
on 2005-04-07 20:18:35