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?
I got 21031/46656 witch is 45.0767318244.
I I listed all combinations that ended with six and if it would be the highest number before they lost. then added up all the all the ones with odd combinatins, but first multiplied the # by the chance it would take to hit that #.
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Posted by Sean
on 2005-10-26 14:52:05 |