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.
Since it was mentioned.
My geometric solution of a cube with 3 corners cut off needs only a slight adjustment.
More possibilities are now included. These basically just make the corners cut off smaller by one. So instead of the (n1)st tetrahedral number they are the (n2)nd.
The remaining volume
n^3  3*(n2)(n1)(n)/6 = n^3/2 + 3n^2/2  n
And so the probability is to divide by n^3
1/2 + 3/(2n)  1/n^2
For n=6 this gives 13/18 = .722222 which is very close to Charlie
It still converges to 1/2 in the limit.

Posted by Jer
on 20100319 14:26:56 