Date: Fri, 1 Jul 2005 01:36:05 +0000
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Q: There are n tickets in the lottery, k winners and m allowing you to
pick another ticket. The problem is to determine the probability of
winning the lottery when you start by picking 1 (one) ticket.

A lottery has N balls in all, and you as a player can choose m numbers
on each card, and the lottery authorities then choose n balls, define
L(N,n,m,k) as the minimum number of cards you must purchase to ensure
that at least one of your cards will have at least k numbers in common
with the balls chosen in the lottery.

S: This relates to the problem of rolling a random number
from 1 to 17 given a 20 sided die.  You simply mark the numbers 18,
19, and 20 as "roll again".

The probability of winning is, of course, k/(n-m) as for every k cases
in which you get x "draw again"'s before winning, you get n-m-k similar
cases where you get x "draw again"'s before losing.  The example using
dice makes it obvious, however.

L(N,k,n,k) >= Ceiling((N-choose-k)/(n-choose-k)*
                   (n-1-choose-k-1)/(N-1-choose-k-1)*L(N-1,k-1,n-1,k-1))
            = Ceiling(N/n*L(N-1,k-1,n-1,k-1))

The correct way to see this is as follows:  Suppose you have an
(N,k,n,k) system of cards.  Look at all of the cards that contain the
number 1.  They cover all ball sets that contain 1, and therefore these
cards become an (N-1,k-1,n-1,k-1) covering upon deletion of the number
1.  Therefore the number 1 must appear at least L(N-1,k-1,n-1,k-1).
The same is true of all of the other numbers.  There are N of them but
n appear on each card.  Thus we obtain the bound.