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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.

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 General geometric solution | Comment 1 of 9
The space of all triplets (a,b,c) can be graphed forming a 6x6x6 cube [in general nxnxn]

The numbers cannot form a triangle if one side is greater than or equal to the sum of the other two.  In other words the largest side cannot be too big.

Say c is the largest side.  If c=2 then a and b cannot both be 1.
if c=3 then a and b cannot sum to 2 or 3.  (rules out (1,1) (2,1) (1,2))
etc.
This amounts to shaving a corner off of the cube.

If a or b are the largest we shave off two corners of the cube.

These shaven corners are actually the fifth tetrahedral number [in general n-1] because it is the sum of 5 [n-1] triangular numbers.
The 5th tetrahedral number is 35.  So there are 35*3=105 rolls that do not work and 216-105=111 that do work.
The solution then is 111/216 = .51389

[The nth tetrahedral number is n(n+1)(n+2)/6 so 3 times the (n-1)st is (n-1)(n)(n+1)/2 = (n^3-n)/2 rolls that don't work and n^3 - (n^3-n)/2 = (n^3+n)/2 that do work.
Dividing by n^3 gives
1/2 + 1/(2n^2)

Interestingly then as the number of sides on the dice increases the probability converges to 1/2.]

 Posted by Jer on 2010-03-18 14:18:10

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