Re: easy dice problem, please quickly help me.
The answer is a summation of the chances of N=0 to 49 times that a 5 or a 6 is rolled.
The formula for any N between 0 and 200 is F(N)=[200!/(N!*(200-N)!)](1/3)^N*(2/3)^(200-N)]
So use this formula to determine F(N) for N=0 to 49 and then add them up and voila you are done. This isn't a hard problem just very tedious.
|