Monday, November 14, 2011
Hardrock: What are the chances?
Been seeing a lot of chatter on the interwebs lately on how bleak the chances are of getting in to some of the increasingly popular big events for 2012 (namely Hardrock). With the lottery soon approaching (December 1 deadline on the applications), I thought I would crunch some numbers to see just how bleak my chances are for a July 13 date in Silverton with my 2 tickets in the hat this year.
Disclaimer: I am not a professional statistician but just pretend to be now and then. Feel free to comment on any of the BS below.
As of November 5, there were 427 applicants for the 140 spots. Some projections I have read put the final number in the 700 to 800 range. I'll assume 800 applicants on that final list. Of course, some of those applicants are already in with automatic entries. I'll project 30 of those. That leaves 770 applicants for 110 spots, however past history shows that about 30 more will get in from the wait list. Therefore, 770 applicants for 140 spots (110 plus 30 wait-listers).
This problem sets up nicely as a hypergeometric probability distribution, however we have another issue to deal with here. The 770 applicants may have anywhere from 1 to 6 tickets in the hat. I expect the distribution of the number of tickets per applicant to be skewed, with probably at least half of the applicants holding just 1 or 2 tickets. I'll assume a mean of 2 tickets, giving a total of 1540 tickets in the hat. Since a person can only take up one spot in the race, once their name is drawn any remaining tickets they have in the hat are excluded from the analysis. Therefore, after each unsuccessful draw of your name, the next draw will have, on average, 2 less tickets in the hat (increasing your chances slightly with each draw).
If you have 1 ticket
P(not getting drawn in the 1st 140 draws) = (1539/1540)*(1537/1538)*(1535/1536)*...*(1261/1262)
P(getting drawn) = 1 - 0.905 = 0.095 = 9.5% (about a 1 in 10 chance with 1 ticket)
By the same equations,
P(getting drawn with 2 tickets) = 0.182 = 18.2%
P(getting drawn with 3 tickets) = 0.260 = 26.0%
P(getting drawn with 4 tickets) = 0.331 = 33.1%
P(getting drawn with 5 tickets) = 0.395 = 39.5%
P(getting drawn with 6 tickets) = 0.453 = 45.3%
I would analyze the Western States lottery, but I'm not even qualified to throw my name in that hat. Go figure. Qualified for Hardrock, but not Western States?