随机概率选择(Random Probability Selection)

 假设我有10个奖品给100个人。 每个人一次射击一次。 因此，如果第一个人未能赢得奖品，则概率会上升，99分中有10分，所以一个......所有10个奖项必须去。 
 
 写这篇文章的最佳方式是，如果还剩下奖品，那么这个人将有一个机会获得奖金...... 
 
 我这样想的是： 
 
int playersLeft = 100
int winners = 0

while (winners < 10)
    winners += (random.Next(playersLeft--)<(10-winners)) ? 1 : 0;
 
 我想知道是否有更好或更直接的方法来做到这一点。 我知道它看似简单，但这个简单的任务是应用程序非常重要的一部分，它必须是正确的。 
 
 澄清：为什么我想做这样的事情： 
 
 实际上有无限数量的玩家，每个玩家有一个X在Y的概率下赢，比如说10/100 = 10％。 但是如果我把它留给随机数发生器，那么在100个玩家中，只有9个会赢，或者最差11个。在我的应用程序中，我必须保证每100个玩家不会再少于10个玩家赢得。 

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.

原文：https://stackoverflow.com/questions/6949221