Real world raffle occurrence - this really happened! A friend of mine and I joined a raffle the other week. There were about 80 participants and 40 prizes, which were actually items donated by 40 of the participants.
Just over 200 tickets were sold (let's assume 240).
The prizes were given away to holders of winning numbers in random order by someone who had no idea who had brought which prize (and winning numbers were discarded).
My friend bought 2 tickets and won once. He was handed the prize he had donated!
I had bought 4 tickets and won once. I was also handed the item that I had donated as a prize..!
In addition, there were 10 tables of 8 and my friend I were sitting side by side. No one else won the same item they had donated.
After a lot of laughter, we started wondering the odds of what just happened but couldn't manage. Can someone help please us?
Cheers!
 A: Very rough calculation for part of the question.
For each of the $40$ people the probability that they get their contribution as a prize is $1/80$. The probability that they don't is thus $79/80$. So the probability that no one gets what he or she brought is 
$$
\left( \frac{79}{80} \right) ^ {40} \approx 0.60
$$
so about $60\%$. 
A: For a simpler calculation we'll assume that everyone bought 3 tickets each.
Probability of having exactly one winning ticket out of 3 is
40/240 * 200/239 * 200/238  +  200/240 * 39/239 * 200/238  +  200/240 * 200/239 * 38/238 
which is approximately 49.4% Probability that this is the item you brought is 1/40. So probability of winning exactly the same item you brought is
49.4/40 = 0.12341338 
approximately 12 out of 10.000
Probability that 2 specific people winning the same item they brought is
C(80,2) p^2 (1-p)^78 = (80*79)/(2*1) * 0.0012341338^2 * (1.0 - 0.0012341338)^78 
which is 6.81488535036258 E-21 which is practically 0.
In this case, we will not even calculate for the probability of the 2 sitting side by side. 
