[Racer26]

Originally Posted by BHS_STopping

Just wanting to correct you slightly on your math here.

If you throw three dice, the probability of at least one of them coming up with a 1 is not simply 3 * 1/6. Using this logic we could then assume that if we throw 6 dice then the number 1 is going to appear every single time (which is false, the actual probability in this case is about 66.5%). When throwing three dice, the probability of throwing at least one 1 is equal to 1 - (5/6)^3. This number turns out to be about 42.1%.

The probability of throwing exactly two 1's is a little bit trickier, but it's not too difficult. There are 216 possible dice rolls for three dice, and 15 of those rolls have exactly two 1's in them. 15/216 is roughly 6.9%. If we include the 1, 1, 1 case (that is, all situations where at least two 1's come up) then our probability is 16/216, or 7.4%.

The probability that all three dice show 1's is 1/216, or .46%, so you were right about that one.

And this is where the house edge on the casino game of Sic Bo comes from. Yay math.