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Gambler's triangulation (Posted on 2010-03-18) Difficulty: 3 of 5
I simultaneously toss three standard dice.
I get 3 numbers, 1 to 6, not necessarily distinct. Evaluate the probability that these numbers can represent sides of a triangle.

  Submitted by Ady TZIDON    
Rating: 3.5000 (2 votes)
Solution: (Hide)
ANSWER: 37/72
SOLUTION: Denote this probability by p, and let the three numbers that come up be x, y, and z.
We will calculate 1-p instead: 1-p is the probability that x >=(y+z); y >=(z+x), or z>=( x+y). Since these events are mutually exclusive, 1-p is just 3 times the probability that x >=(y+z).
This happens with probability (0+1+3+6+10+15)/216 = 35/216, so the answer is (1-P)=3*35/216 =35/72.

Therefore P= 37/72.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Including degeneratesJer2010-03-19 14:26:56
re(5): Solution/zero area triangleAdy TZIDON2010-03-19 13:28:03
re(4): SolutionCharlie2010-03-19 10:59:55
Hints/Tipsre(3): SolutionAdy TZIDON2010-03-19 04:26:27
re(2): SolutionDej Mar2010-03-19 04:07:19
re: SolutionAdy TZIDON2010-03-18 19:53:34
SolutionSolutionDej Mar2010-03-18 17:57:47
SolutionsolutionCharlie2010-03-18 15:33:46
SolutionGeneral geometric solutionJer2010-03-18 14:18:10
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