I was at a summer bbq picnic on the weekend and there was a raffle draw. There were different prizes on the table. Everyone had a ticket and they would randomly pick tickets - when a ticket was called, the recipient would come and pick up their prize. Their ticket would then be removed from the raffle and they would keep going. Obviously the "better prizes" would be picked first and the "less desirable prizes" would remain until the end.

At the same time, there were popsicles in another part of the picnic area that were going fast! But if I went to go get a popsicle, I might miss my ticket being called! But if I stayed at the raffle area and my ticket was never called, I would miss out on the popsicles!

I was trying to calculate: based on how many tickets/prizes were remaining in the raffle - would it be worth sticking around for my number to be called?

I tried to think of it this way: Suppose there were 100 tickets (i.e. 100 people) - on average, from the very beginning of the raffle - how many tickets would need to be drawn before mine showed up? Is there a mathematical formula for this?

A wild guess - I say that 50 on average need to be drawn but I dont think this is correct?


  • $\begingroup$ Your ticket is equally likely to be the nth one called, for all n in the range 1 ≤ n ≤ 100. The average of these numbers is 50.5. $\endgroup$
    – Dan Asimov
    Commented May 28, 2023 at 19:01

1 Answer 1


The probability that your ticket has been picked by the 50th draw is

$$ 1 - \frac{99 * 98 * 97 * \dots * 50}{100 * 99 * 98 * \dots * 51} = 1 - \frac{50}{100} = 50 \ \text{percent} $$

where $1$ is the total probability and each numerator-denominator pair in the fraction of falling factorials is the probability that your ticket is not selected on the given draw.

This, moreover, is clearly true for any of the 100 draws. That is, for the $i^{\text{th}}$ draw, the probability that your ticket has already been drawn is $$1 - \frac{99 * 98 * \dots * (99 - i + 1)}{100 * 99 * \dots * (100 - i + 1)} $$

The expected number of draws you'd have to wait is the average of all draws, namely $$ (1 + 2 + \dots + 100) * \frac{1}{100} = \frac{\frac{n(n + 1)}{2}}{100} = 50.5 $$

So your initial guess was quite close.


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