Random Probability Selection

Question

Say I have 10 prizes to give to 100 people. Each person gets a shot, one at a time. So if the first person fails to win a prize, the probability goes up, 10 in 99, and so one... Also all 10 prizes MUST go.

What would be the best way to write this in such a way that by the end if there is still a prize left, that person would have a 1 in 1 chance to get a prize...

What I was thinking like this:

int playersLeft = 100
int winners = 0

while (winners < 10)
    winners += (random.Next(playersLeft--)<(10-winners)) ? 1 : 0;

I wanted to know if there was a better or more straight forward way to do it. I know it seems simple but this simple task is part of a very important aspect of the app and it must be right.

TO CLARIFY: Why I want to do something like this:

In reality there is an unlimited number of players, each with an X in Y probability to win, say 10/100 = 10%. However if I leave it to the random number generator, there is a chance that in 100 players, only 9 would win, or worst, 11. In my app, I must assure that no more and no less than 10 players for every 100 will win.

Answer 1

I have thought about this some more and have come up with the following. We can give the first guy a fair shot at winning and then if the rest of the rewards are distributed fairly among the rest of the people (no matter if he wins or loses) the whole thing will be fair. Of course that's far from formal proof, so feel free to correct me. The following should give a fair system:

int prizes = 10;
for(int i = 100; i >= 1; i++)
{
  var result = random.Next(people);
  if(result < prizes)
  {
     Console.WriteLine("{0} won", i);
     prizes--;
  }
}

Edit: Proof this works:

1. The first person trivially has n/k chance of winning (n being the number of prizes, k being the number of people.
2. Let's assume we distribute the remaining prizes fairly among the rest of the people. In that case they will have with probability n/k, n-1 prizes distributed between them and with probability (k-n)/k, n prizes. That adds up to (n*(n-1))/k + (n*(k-n))/k = n*(k-1)/k on average which is their fair share of the prizes.
3. We use the same method to either distribute n-1 or n prizes among the k-1 people. Q.E.D.