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 (n-1)st tetrahedral number they are the (n-2)nd.

The remaining volume
n^3 - 3*(n-2)(n-1)(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 2010-03-19 14:26:56