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