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 Gambler's triangulation (Posted on 2010-03-18)
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.

 See The Solution Submitted by Ady TZIDON Rating: 3.5000 (2 votes)

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 solution | Comment 2 of 9 |

If all 3 dice show the same value (prob. 1/36) an equilateral triangle is possible.

If two are the same and one different (prob. (1/6)*(5/6)*3 = 15/36 = 5/12), the two equal values must add to more than the differing one. There are 6*5 = 30 such sets of two match and one different. When the matches are 2, then 1 or 3 will work for the odd-man-out; when the matches are 3, then any of 1,2,4 or 5 will work; matches above 3 will always work. That's a total of 2 + 4 + 3*5 = 21 cases out of the 30 that work. So the overall contribution of one pair matching is (5/12)*(21/30) = (5/12)*(7/10) = 7/24.

If all are different (prob. (5/6)*(4/6) = 20/36 = 5/9) the smaller two must add to more than the largest. Those that work: 2,3,4; 2,4,5; 2,5,6; 3,4,5; 3,4,6; 3,5,6; 4,5,6. That's 7 combinations out of the C(6,3)=20, so the overall contribution of "all different" is (5/9)*(7/20) = 7/36.

Therefore the total probability is 1/36 + 7/24 + 7/36 = 37/72.

Simulation confirmation:

DEFDBL A-Z
FOR trial = 1 TO 1000000
a = INT(RND(1) * 6 + 1)
b = INT(RND(1) * 6 + 1)
c = INT(RND(1) * 6 + 1)
IF b < a THEN SWAP a, b
IF c < b THEN SWAP b, c
IF b < a THEN SWAP a, b
IF a + b > c THEN hit = hit + 1
tot = tot + 1
PRINT hit; tot, hit / tot, 72 * hit / tot
NEXT

produces 514,689 out of 1,000,000 trials. that's 37.057608/72

 Posted by Charlie on 2010-03-18 15:33:46

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