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Roll the Die Higher (Posted on 2005-04-07) Difficulty: 3 of 5
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?

  Submitted by Charlie    
Rating: 3.8571 (7 votes)
Solution: (Hide)
For an n-sided die, the probability the first player wins is 1 - (1-1/n)^n. When n is 6 this is 31031/46656 = .665102023319616... (See http://www.physics.harvard.edu/probweek/sol54.pdf).

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionPuzzle SolutionK Sengupta2008-12-03 11:57:58
AnswerK Sengupta2008-11-26 00:00:23
No SubjectSean2005-10-26 14:52:05
AnswerGovind2005-04-26 08:19:52
re(2): General ResultSteve Herman2005-04-10 23:24:19
re: General ResultRichard2005-04-08 19:11:26
SolutionGeneral ResultSteve Herman2005-04-08 15:39:50
re(3): Solution w/o expl. and a questionRichard2005-04-07 21:01:45
re(2): Solution w/o expl. and a questionFederico Kereki2005-04-07 20:54:37
Some Thoughtsre: Solution w/o expl. and a questionFederico Kereki2005-04-07 20:47:49
re: Solution w/o expl. and a questionRichard2005-04-07 20:45:51
Solution w/o expl. and a questionJer2005-04-07 20:18:35
Some ThoughtsTheory (maybe solution)Federico Kereki2005-04-07 19:00:47
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