Need a algorithm
- To: mathgroup at smc.vnet.net
- Subject: [mg33857] Need a algorithm
- From: "Juan" <erfa11 at hotmail.com>
- Date: Sat, 20 Apr 2002 02:49:47 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Hi,
I ask for help to find an algorithm to resolve this question:
There is 36 numbers T={1,2,3,…,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’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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