4
$\begingroup$

There are $100$ people who bought a ticket for a raffle with $5$ prizes being offered. They randomly draw a name for the $5^{th}$ place prize, then they randomly draw a name for the $4^{th}$ place prize, etc., all the way until they randomnly draw a name for the $1^{st}$ place prize. We are interested in the probability of a single individual who bought a ticket winning a prize.

My initial thoughts were that in general, for $n$ people who bought a ticket for a raffle where $p$ prizes are being offered, that the probability of somebody winning a prize was $\Pi_{k = 1}^{p}\frac{1}{n-p+1}$ because each draw is an independent event but then I remembered that for problems where you are sampling without replacement and have binary outcomes such as win/lose, pass/fail, etc., that the Hypergeometric distribution exists for these kinds of problems.

I consulted the wikipedia entry on this probability distribution (https://en.wikipedia.org/wiki/Hypergeometric_distribution) and plugged into what I believe are the correct values for the constants in what they list as the probability mass function:

$ K = 5\\ k = 1\\ N = 100\\ n = 5. $

Thus, the probability of interest is:

$\\ P(X = k) = \frac{{K\choose k}{N-K\choose n-k}}{N\choose n} = \frac{{5\choose 1}{95\choose 4}}{100\choose 5} \approx 21 \%$

I always just get confused with this distribution because each article and textbook describes the set up of this distribution and parameters slightly differently and it just gets confusing trying to know how to assign what parameters to what values (for instance, trying to figure out this probability using the phyper command in R was confusing because I got $0$ as an answer back).

If you're truly interested in the background surrounding this problem, my parents are trying to set up a raffle for $100$ people and are offering up $5$ really nice bottles of bourbon/whiskey and want to make sure it's a fair raffle and are interested in the probability of a single participant winning a bottle. I did this hypergeometric probability calculation and my dad isn't convinced, and now I'm not either.

Thanks!

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
5
$\begingroup$

The above is a close answer. The hypergeometric distribution, for an individual raffle participant, will have $𝐾=1$, as from their perspective only one ticket is a success. So both the $k$ and $K$ are one. This gives a probability of .05.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ Thanks! I didn't think about the problem set up in that I should be thinking about it from the participants perspective. Tricky! Turns out there are a few other ways to solve it. You can do it with this kind of telescoping series-esque design. The hypergeometric approach was most intuitive to me though, so I appreciate you answering my question from this perspective. $\endgroup$
    – BonnieKlein
    Jan 22, 2021 at 22:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.