Sorry that I couldn't type in Chinese.

Maybe an old probability problem:

We are tossing a coin. The probability of the coin to face up is p (0=n>=1)?

It is not (m-n+1)*p**n, because otherwise we are counting some situation multiple times.

Let the answer be f(m,n), then I can get:
1. f(n,n) = p**n ;
2. f(n,1) = 1-(1-p)**n ;
3. f(l,n) = 0 if l4. f(m,n) = f(m-1, n) + p**n*(1-p)*f(m-n-1,n).

But I couldn't get the general ????ula from it, except that f(m,n)=p**n*(m-n+1-mp+np) if(m<=2n).

I think the answer is already there since this is a quite old question.
Anyone can help?

Thanks.