dice probability

[rja211077]

what is the probability P of throwing EXACTLY x sixes with y standard six-sided dice? I'd like the answer in terms of x and y please. I have got as far as P = z/(6^y) but what is z? This is doing my head in. Help please.

[Dsg]

Is this along the right lines: P=(1/6)^x*(5/6)^(y-x) Being the probability of x sixes and (x-y) non-sixes ?

[Dsg]

Cont - the last bit should read (y-x) sixes - Sorry

[rja211077]

your answer only seems to work for x=0 and x=y (i.e. no sixes and all sixes).

[lisaj]

Assuming you are throwing them all together, and that all the dice are independent, then this should be a binomial distribution. So using the good old formula... yCx * (1/6)^x * (5/6) ^ (y-x) should do the trick. If not, then MSN Messenger me to figure it out : [email protected]

[Hamish]

I Make the answer as y/(6^x) I don't think that binomials have anything to do with it as these usually involve factorials which are used in calculations where the outcome of the first probablility affects the next etc. My formula seems to work for the first few combinations but I must do some work now!! Hamish

[rja211077]

I don't think so Hamish! Your formula would give the probability of throwing 2 sixes with 2 dice as 2/36 instead of 1/36 - and think about when x=0!!

[rja211077]

I'm pretty sure lisaj is right but my A level maths is so rusty I have forgotten what yCx means - can anybody fill me in?