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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?

See The Solution Submitted by Charlie    
Rating: 3.8571 (7 votes)

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Some Thoughts Theory (maybe solution) | Comment 1 of 13
I did the case for n=2, n=3 and n=4, and the chances of the second player's win were (1/2)^2, (2/3)^3 and (3/4)^4, which leads me to suppose the answers to this puzzle are (5/6)^6 and ((n-1)/n)^n that in the limit is 1/e.
  Posted by Federico Kereki on 2005-04-07 19:00:47
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