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Say I'm doing a raffle, there's 5 prizes and 50 entrants. Everyone gets one entry. Once a person has won a prize they cannot win another one.

I'm wondering what the probability differences are between the following ways of drawing winners:

  1. Random number generator between 1 and 50. If a number that's won before comes up again, they don't win again instead another random number is chosen again, from the same pool of 1 - 50.
  2. Similar to the above, but winning numbers are removed from the pool so that duplicate wins cannot occur.

In my mind, method two seems to give better odds to people who haven't won yet since winners are removed from the ability to be drawn again. I read this question/answer and think it may apply so below is my math; there's probably mistakes.

Method 1: is it something like 1 - (49/50)^5 => 0.096079203 => 9.6% chance of a winning draw? i don't think this factors in re-drawing if a previous winner has won though. each draw still has 1/50 chance of winning, but if a duplicate occurs, another 1/50 chance of winning occurs.

Method 2: is it like 1 - (49 choose 5)/(50 choose 5) => 10% chance of winning? or something like 1/50 + 1/49 + 1/48 + 1/47 + 1/46 => ~10.426% chance of a winning draw?

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1 Answer 1

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In Method 1, you say another drawing from among 1-50 is done if the second draw is the same as the first (say, 43). If you modify Method 1 so that you keep on drawing again if the 43 should come up again, etc., then the two methods are the same. A subsequent draw counts only if it comes up 43. In the long term any number but 43 has probability $1/49$ of being the second winner,

Method 2, is sampling without replacement. It's a lot simpler because the first winning number is simply removed so that other numbers have probability $1/49$ of being chosen on the second draw.

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  • $\begingroup$ That makes sense, thank you very much! $\endgroup$
    – a-person
    Jan 22, 2021 at 14:05

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