## probability about the ball drawing game

Hi All,

Here is a question about probability calculation in the ball drawing game needed your help.

In the game, there are n balls provided, including m black balls and n-m white balls. We are required to draw one ball after another till getting a black one. The problem is to calculate the probability of we winning the game at the k-th drawing (i.e., just drew white balls from the 1st to the (k-1)-th time, while drew a black one at the k-th time).

Attach my understanding first as the follows:
Let Pr(n,m,k) represent the probability that we draw the first black ball at the k-th time, given m black balls and n-m white balls.
1) for k=1, certainly Pr(n,m,k)=m/n;
2) for k>n-m, certainly Pr(n,m,k)=1, because there are up to n-m white balls and there must be at least 1 black balls included if we draw more than n-m balls;
3) **Just where I got confused**
for 1<k<=n-m:
at the k-th time, only n-k+1 balls left, then the probability of drawing a black one is m/(n-k+1);
however, this is on the condition that at the (k-1)-th time we didn't draw a black with probability of 1-Pr(n,m,k-1);
so, I consider it as a conditional probability as Pr(n,m,k)=m/(n-k+1)/(1-Pr(n,m,k-1)).
Using n=5, m=2 for test, just found that Pr(n,m,3)>1, which means that the expression must go wrong.

Anyone can help figure it out?