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Not sure if this falls specifically under mathematics, but I didn't see a more suitable location. If it needs to be moved elsewhere please do so.

Say I have a prize drawing with a top prize, a 2nd place prize, and a 3rd place prize. Each participant gets their name dropped into a hat. When it's time to do the drawing, typically the emcee draws the 3rd place name, then the 2nd place name, then finally the 1st place winner.

However, this seems unfair to the 2nd and 3rd place winners, as the 3rd place "winner" now has no chance to get the 2nd or 1st place prize as their name has been removed from eligibility due to winning a lower prize. Correspondingly, the 2nd place winner now has no chance to win the 1st place prize.

Obviously they do this to build suspense and excitement, but it seems unfair that the 2nd and 3rd place winners have no chance at the 1st place prize, which is assumed to be the most desirable.

It would seem to me to make sense that the emcee should draw, but not reveal, the 1st place ticket first, then the 2nd place ticket, and finally the 3rd, then reveal the winners in 3rd, 2nd, 1st order, as typical.

Yes, the 2nd and 3rd place winners are better off than those that don't win anything, but by being picked for a lesser prize, they are no longer eligible for the top prize. This seems unfair, if the goal is for everyone to have an equal chance at the top prize. Is my thinking correct on this?

Milwrdfan
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2 Answers2

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Your thinking is not correct, and this paragraph in your question explains why

It would seem to me to make sense that the emcee should draw, but not reveal, the 1st place ticket first, then the 2nd place ticket, and finally the 3rd, then reveal the winners in 3rd, 2nd, 1st order, as typical.

Whether or not the third and second prize winners are revealed before the first is chosen clearly doesn't change the odds for the second and third prize winners.

There's another question asked and answered today that deals with the a similar issue: Is this lot drawing fair?

Community
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Ethan Bolker
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Your reasoning is, in fact, fallacious. In fact, each person has equal probability of winning the first three top prizes.

Suppose there are $n$ people from which the three winners are announced as you explained in your question.

Then, the probability that a person will be selected for the third prize is: $$ P(3^{rd}) = \frac{1}{n} $$ and the probability that a person will be selected for the second prize is: $$ P(2^{nd}) = \frac{n-1}{n} \cdot \frac{1}{n-1} = \frac{1}{n} $$ since he/she has to not receive the third prize and then receive the second prize.

Similarly, a person will win the first prize with probability: $$P(1^{st}) = \frac{n-1}{n} \cdot \frac{n-2}{n-1} \cdot \frac{1}{n-2} = \frac{1}{n}$$

An intuitive way to see this is to imagine the equivalent situation where the show host first chooses three names at the same time and then randomly chooses from the three names such that the first person chosen wins the third prize, the second the second prize, and the last person the first prize.

The three names first chosen each obviously has the same probability of being chosen as the other two, and assigning $(1,2,3)$ randomly to three people also obviously is fair to all three.

Peter G. Chang
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