Re: Help in solving this puzzle
- From: matt271829-news@xxxxxxxxxxx
- Date: 31 Mar 2006 16:44:32 -0800
Michael Harrington wrote:
<matt271829-news@xxxxxxxxxxx> wrote in message
news:1143811743.700626.31370@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
I now have 12 tickets which comply with exhaustive testing>>>
john@x wrote:
There is a lottery drawn on 1 through 20 numbers and six distinct balls
are randomly picked by the lottery every day. To play you need to buy a
ticket and select six numbers of your choice. An example ticket might
be [1,20,6,19,5,12] You can buy any number of tickets. If any three of
the numbers in one of your tickets are among the six winning numbers
picked by the lottery, in any order then you win.
What is the minimum number of tickets you have to buy to be certain of
winning with atlease one?
Thanks
John
I think you can do it with 14 tickets, though I do not know if this is
the minimum. This is an example 14-ticket solution (I hope!):
1 2 3 4 5 6
7 8 9 10 11 12
13 14 15 16 17 18
1 2 7 13 19 20
3 4 8 14 19 20
5 6 9 10 15 16
5 6 11 12 17 18
1 3 9 10 17 18
2 4 11 12 15 16
5 6 7 8 13 14
1 2 8 14 19 20
3 4 7 13 19 20
1 2 3 4 7 13
1 2 9 10 19 20
1 2 3 4 5 6
1 2 3 4 7 8
1 2 5 6 7 8
3 4 5 6 7 8
9 10 11 12 13 14
9 10 11 12 15 16
9 10 13 14 15 16
11 12 13 14 15 16
17 18 19 20 1 2
17 18 19 20 3 4
17 18 19 20 5 6
17 18 19 20 7 8
Very good! It's interesting to see how your answer has so many
repetitions - for example, four tickets all with the numbers 17, 18,
19, 20. Intuitively you would think (or I would, anyway) that the best
solutions would mix up the numbers as much as possible. Do you think
that 12 is the best that can be achieved?
Logic:
Divide 20 numbers into 3 blocks ( 2 blocks of 8 numbers & 1 block of 4
numbers)
Defining the blocks as eg (1 to 8 ) & (9 to 16) & (17 to 20)
With 6 numbers drawn, one of these blocks must contain at least 3 of the
drawn numbers or else each block contains exactly 2 drawn numbers.
Every combination of three from eight numbers can be covered with 4
tickets.>>
1 2 3 4 5 6
1 2 3 4 7 8
1 2 5 6 7 8
3 4 5 6 7 8
And
9 10 11 12 13 14
9 10 11 12 15 16
9 10 13 14 15 16
11 12 13 14 15 16
And
17 18 19 20 ( plus any other two numbers) covers every combination of three
from the block of four numbers.
The alternative to one of these blocks containing at least 3 numbers is that
each
block has exactly 2 numbers. This means that the block of four (17 18 19 20)
must
contain two numbers. If we combine this block of four with every number from
one to the blocks of eight this must give at least three numbers in one of
the
resulting tickets.
PS I previously interpreted the puzzle as requiring every combination of 3
to be covered. With six numbers drawn giving 20 combos of 3 numbers the
earlier
solution I posted of 120 tickets( which did contain every combo of 3)
therefore
also contained 20 combos of 3 from any draw.
Regards Richard Spelman.
.
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