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If there are N participants, k winners total, and y'all really do choose tickets randomly (proven false, but let's go with your assumption), your chances of win
by swolchok 17y ago
If there are N participants, k winners total, and y'all really do choose tickets randomly (proven false, but let's go with your assumption), your chances of winning are k/N. If you know that one of the winners is not you, your chances of winning are (k-1)/N.
EDIT: above is assuming that you can win more than once if k > 1. If not, the math is more complicated than I can be bothered to try to format in an HN post, but the principle should hold.
- smanek 17y agoI don't think there are a fixed number of winners in advance ... My expected return may go down if the person in front of my wins (I have to split with more people, so each winner gets less) - but my chances of winning don't change.
- brent 17y agoThat is an interesting form of lottery. Do you have any references to lotteries run in this manner (and proof that the tickets are not randomly generated)? I think the form of lottery I (and others) are referring to is the one where there are N unique tickets and you have a 1/N chance of matching those N numbers. Therefore, there is no dependence between different tickets and the likelihood of matching the winning numbers.
- swolchok 17y agoWell, the tickets where y'all fill in the bubbles with pencil to choose numbers and the form can be read by an optical scanner are surely not randomly generated. People will draw pictures, play birthdays, and attempt to choose layouts that "look random", none of which result in a uniform ticket distribution, which is what most people seem to mean when they say that something is "random". The takeaway is that "distributed uniformly at random" is not a great assumption at all when inputs are generated by a process. It's one of the relatively few things I did learn in my graduate algorithms class.