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2886RE: Changes to the Hardrock qualifications and lottery for 2013

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  • Wood, Blake P
    Jun 24, 2012
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      No Matt, we mean 2^N. So first time applicants get 2^0 = 1 ticket. Those with one DNS get 2^1 = 2 tickets. And so on to those three unfortunate souls who will have 6 DNSs without a DNF or finish, and will get 2^6 = 64 tickets, should they apply again.

      - Blake

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      From: Matt Mahoney [mattmahoneyfl@...]

      > Modeling suggests that giving applicants 2^N tickets, where N is the number of previous DNSs,

      I think you mean (N)x(N+1)/2 (N tries with N tickets on the N'th try).
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