If I have x number of balls in a lottery, how many times will I have to draw a ball at random and replace it before I draw a ball that has already been drawn once before?

 

I developed an algorithm to solve this, I think it works, not entirely sure. But I was wondering how to solve it algebraicly, here is my algorithm:

 

t = number of balls in lottery
x = 1
n = 0

while (x>0.5) {
   n = n + 1
   x = x * (t - n + 1) / t
}

Answer = n

the question is badly worded. and the answer is x+1 if we assume you mean "have to draw a ball that has been drawn before". obviously, by the pigeon hole principle in x+1 draws there is at elast one repetition, and in x draws it is possible to get each balle xactly once.

 

however, you might mean "draw with probability freater than 1/2 a balll that has been drawn before, and that is a simple exercise in probability (geometric i think).

Yes, sorry matt. You are right, drawn with a probabilitiy greater than half. Would you care to show me how you would solve this?

what is the probablitly that after n draws you have n different things?

[math]\frac{n!}{n^n}[/math]

But how does that help?

 

EDIT: OK, perhaps this is better, but I how would I then solve for n?

 

[math]\frac{t!}{t^n.(t-n)!}<0.5[/math]

that is a simple exercise in probability (geometric i think).

 

Is it harder than you realised or have you moved on?