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MathGroup Archive 2002

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RE: Need a algorithm

  • To: mathgroup at smc.vnet.net
  • Subject: [mg33890] RE: [mg33857] Need a algorithm
  • From: "DrBob" <majort at cox-internet.com>
  • Date: Sun, 21 Apr 2002 06:14:28 -0400 (EDT)
  • Reply-to: <drbob at bigfoot.com>
  • Sender: owner-wri-mathgroup at wolfram.com

If I understand your problem properly, it takes only two lottery
tickets, not twelve, to match 3 or more out of 7 numbers drawn from a
set of size 12.

For instance, let w1 be the first 7 elements of q and let w2 be the last
7.  There are 771 lottery draws that share 3 or more numbers with w1,
and only 21 lottery draws that do NOT.  Those 21 draws each include two
elements of w1 and ALL the last 5 elements of q, so they share at least
5 elements (up to 7) with the other ticket, w2.

If we're working on k=4, observe that if w1 includes 3 or less winning
number, then the last 5 numbers of q include 4 or more of them, so again
w1 and w2 is enough.

If k=5, it gets harder.  w1 and w2 are not enough, since a draw could
consist of the first 4 and last 3 elements of q.  I believe four tickets
are required.

That doesn't give you the general algorithm, but it's a start.

Bobby Treat

-----Original Message-----
From: Juan [mailto:erfa11 at hotmail.com] 
To: mathgroup at smc.vnet.net
Subject: [mg33890] [mg33857] Need a algorithm

Hi,
I ask for help to find an algorithm to resolve this question:

There is 36 numbers T={1,2,3,&#8230;,36}, and we play 7 numbers,  for 
example : w={3,9,11,19,22,27,33}, then we compare w with the price , for

example : p={2,7,11,22,30,31,33}, and in this case we have 3 goods: 
{11,22,33}.

One way to play is to buy for example 1000 w (7 random numbers from T,
1000 
times), and to compare each with p, to see if you have been lucky.

Another way if to choose 12 numbers from T , for example: 
q={8,11,13,14,16,22,23,28,31,32,34,35}, and to play for all
combinations: 
Binomial[12,7]=792.

If you have q, and you want to get at least k rights, assuming that the 
numbers in the prize p are in q, then you don&#8217;t need to buy all
the 
792 w.
I have bought that for k=3 and I got only 12 w.
W= {{8,11,13,14,16,22,23},{8,11,13,14,28,31,32},{8,11,16,22,28,31,34},  
{8,11,23,31,32,34,35},{8,13,14,22,32,34,35},{8,13,16,23,28,32,34},{8,14,
16,23,28,31,35},{11,13,14,16,31,34,35}, 
        
{11,13,22,23,28,34,35},{11,14,16,22,28,32,35},{13,16,22,23,31,32,35},{14
,22,23,28,31,32,34}}
(This is one of the solutions, and you get at least 3 rights).

  If you want at least k=5 rights then you have to select more
combinations 
of the all 792.

Knowing T=Range[36], q=(n numbers from T) and k<7. How do we get W?

Thanks. Juan






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