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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,…,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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