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 Roll the Die Higher (Posted on 2005-04-07)
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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 re: Solution w/o expl. and a question | Comment 4 of 13 |
(In reply to Solution w/o expl. and a question by Jer)

This solution agrees with mine. To see why, summing the fractions gets (6^6+ 5^1.6^5+ 5^2.6^4+ 5^3.6^3+ 5^4-6^2+ 5^5.6^1)/6^6. The numerator equals (6^6-5^5), as can be seen by remembering the formula (x^n-y^n)=(x-y)(x^n-1+ x^(n-2).y^1+ x^(n-3).y^2 + ... + y^(n-1)). For n=56, his result is approximately 1-1/e.
 Posted by Federico Kereki on 2005-04-07 20:47:49

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