An interesting scenario that I'm trying to wrap my mind around.

Let's say you were running a giveaway, and have a probability of 10% for every participant to potentially win a prize. The way you would choose the participant would be using an RNG for each one (1,10) and if the output was 1, they win, otherwise, they don't. Technically not a 10% probability exactly, since the generation is random.

However, it turns out that by error, the raffle was actually conducted using a 7% chance instead.

If re-running the raffle was not an option, but you had to run a second one with the remaining supply of prizes so that you can help rectify the mistake, and restore balance to the force -- would you re-run it again with a 3% chance for each participant of winning (10% - 7%)?

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    $\begingroup$ 1 out of 7 is a ~14 probability so it's too large, not to small. Is this just a hypothetical, and the real problem is that whatever happened reduced the probability of selection? $\endgroup$
    – num_39
    Mar 6, 2022 at 19:02
  • $\begingroup$ Sorry brain fart -- meant to say that it was a 7% chance rather than 10% $\endgroup$ Mar 6, 2022 at 19:21
  • $\begingroup$ @num_39 Yeah, hypothetical. Call it an error in 'programming'. $\endgroup$ Mar 6, 2022 at 19:44
  • $\begingroup$ If in a sequence of draws for these prizes everybody winds up at the end with the same expected number of prizes, then that sequence is demonstrably equitable. One issue is whether you wish to ensure that nobody will win more than one prize (which was true in the original design but might not be true in a sequence of raffles). $\endgroup$
    – whuber
    Mar 6, 2022 at 20:37
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    $\begingroup$ @whuber: Yes, that is my interpretation too (player-wins are independent Bernoulli events, so total number of winners is binomial). $\endgroup$
    – Ben
    Mar 7, 2022 at 2:02

1 Answer 1


My reading of your question is that the win event for each player is independent, so that more than one prize can be won. Under this interpretation we can consider the case for a single player without further complication. If my interpretation is not what you meant in your question then things could get more complicated.

Suppose the desired win-probability was $\pi$ but you accidentally used a lower value $\pi_* < \pi$. The player wins the game if they either win on the first round (with the lower than intended win-probability) or if they lose on the first round but win on the second (your "make-up" round). Consequently, you would need to set the new probability $p$ to attain the overall win-probability:

$$\pi = \pi_* + (1-\pi_*) p,$$

which gives:

$$p = \frac{\pi-\pi_*}{1-\pi_*}.$$

In the present case you stipulate that $\pi = 0.10$ and $\pi_* = 0.07$ so you would require the new win-probability to be $p = 0.03/0.90 = 1/30$.

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    $\begingroup$ I wonder: suppose that despite using a 7% RNG in the first draw, it turned out that 9% of the entrants won a prize. What value would you propose for the second draw: the nominal $\pi_{*}=0.07$ or the observed $\pi_{*}=0.09$? ;-) What if nobody won a prize the first time? Intuitively, we could just rerun the raffle (as if $\pi_{*}=0$) and everyone would be content; but there is an argument for rerunning it with $\pi_{*}=0.07.$ This problem looks like it needs a deeper analysis. $\endgroup$
    – whuber
    Mar 6, 2022 at 20:41
  • $\begingroup$ @whuber: Since I think you and I know the answer to this one, I'll leave your question as a brain-teaser for OP. $\endgroup$
    – Ben
    Mar 6, 2022 at 21:02
  • $\begingroup$ I can think of several somewhat different answers, each taking a different perspective on what was intended to be achieved and what might be consider equitable. I don't claim to know "the" answer. $\endgroup$
    – whuber
    Mar 7, 2022 at 1:04

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