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.
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))
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