Re: [OT] Math

Here we go.

Granted that my initial answer of p=(1/6)^4=7.716*10^(-4) only takes 
into account four hits and totally disregards the other two dice.

The chance of the other two dice missing is p=(5/6)^2.

Thus EXACTLY 4 dice hitting is p=(1/6)^4*(5/6)^2=5.358*10^(-4)

...or .05358%