First of all, thanks to all for the replies.

Ok. So YellowJack and AaronBrown are saying that the non-six showing die has 5 non-six faces which then is used to multiply my original combination computation of 4 to get 20 plus the 1 event of all 4 dice showing six. (What a long sentence :-)

So I end up with 21 over 1296 or about 3.5 times my previous computation. So for 5 dice the math would be C(5,3)=10 x 5 x5=250+1=251 then divided by 7776.

As for where this started; I was playing Zonk against someone and was using a strategy of not taking all my dice/points on the first roll. That way I left more dice for my next roll with a higher chance of getting a triple and a higher score than if I took all of my dice/points and then did or did not roll again.

When pressed by my opponent, I was unable to do the math to convince him, and that sparked a trip to the Internet, library and book store to brush up on probability. Usually that does it, but not this time--I think probability is Latin for "makes one cry." No, I don't plan to confront my opponent with this; he wouldn't understand it anyways. I'll just keep on winning.