superprismatic Posted July 18, 2009 Report Share Posted July 18, 2009 The country of Ozendia has a weekly lottery drawing. A set of 6 numbers are drawn without replacement from the set of numbers {1,2,3,...,40} randomly. A play of this lottery consists of buying a ticket with a buyer-chosen guess as to what these 6 numbers will be. The drawing order of the numbers does not matter. Each play costs $1. The payoff is usually in the multi-million dollar range and it is a percentage of the ticket sales since the last winner. Ozendians love to gamble, so there is usually a small number of winning tickets each week. If this number is greater than one, the jackpot is shared equally among the winning tickets. So far, this is much like many lotteries familiar to most people. Ozendians are social gamblers. That is, very often a group of people will buy a bunch of tickets with any winnigs divided up equitably to the group members. Ozendian lottery officials want to encourage this behavior as they believe it increases ticket sales. So they developed a scheme to allow groups to buy lots of sets of numbers on a single ticket. They have what they call a 7-play ticket. You get to choose 7 numbers on a 7-play and the ticket wins if any subset of 6 of these 7 numbers is picked. Since this is equivalent to buying 7 different normal tickets, the cost is $7 for this kind of ticket. Similarly, they have an 8-play, which is equivalent to 28 standard tickets and so, costs $28. In fact, they have N-play tickets right up to N=15 which costs a whopping $5005! They, in fact have all N-plays for N=6,7,8,9,10,11,12,13,14,15 (I included the standard game as a 6-play). Well (wouldn't you know it?), there's a group of 143 mathematicians in the Ozendian Defence Department who wish to buy the equivalent of a 16-play costing $8008 ($56 for each mathematician). They have decided on their 16 favorite numbers. They wish to cover all subsets of 6 out of their 16 numbers using standard N-play tickets (N=6 to 15) at minimal cost. After all, they are mathematicians, so they can afford to pay a bit more than $56 each! How many of each type of N-play tickets (include the standatd game as 6-play) are necessary to achieve minimal cost? What is the cost per play? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 19, 2009 Report Share Posted July 19, 2009 (edited) After all, they are mathematicians, so they can afford to pay a bit more than $56 each! How many of each type of N-play tickets (include the standatd game as 6-play) are necessary to achieve minimal cost? What is the cost per play? They could buy 8008 (16C6) standard tickets (6-play) for $1 each. So, the minimum amount cannot be more than $56 each for the 143 mathematicians. And you wrote that they could afford to pay a bit more than $56. Am I missing something or is that statement in the question redundant?? Edited July 19, 2009 by DeeGee Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted July 19, 2009 Author Report Share Posted July 19, 2009 They could buy 8008 (16C6) standard tickets (6-play) for $1 each. So, the minimum amount cannot be more than $56 each for the 143 mathematicians. And you wrote that they could afford to pay a bit more than $56. Am I missing something or is that statement in the question redundant?? Sorry, I meant to ask: What is the minimum number of N-play tickets which would achieve the cover of all subsets of 6 out of the 16 numbers? Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted July 19, 2009 Author Report Share Posted July 19, 2009 I would like to restate the problem: The country of Ozendia has a weekly lottery drawing. A set of 6 numbers are drawn without replacement from the set of numbers {1,2,3,...,40} randomly. A play of this lottery consists of buying a ticket with a buyer-chosen guess as to what these 6 numbers will be. The drawing order of the numbers does not matter. Each play costs $1. The payoff is usually in the multi-million dollar range and it is a percentage of the ticket sales since the last winner. Ozendians love to gamble, so there is usually a small number of winning tickets each week. If this number is greater than one, the jackpot is shared equally among the winning tickets. So far, this is much like many lotteries familiar to most people. Ozendians are social gamblers. That is, very often a group of people will buy a bunch of tickets with any winnigs divided up equitably to the group members. Ozendian lottery officials want to encourage this behavior as they believe it increases ticket sales. So they developed a scheme to allow groups to buy lots of sets of numbers on a single ticket. They have what they call a 7-play ticket. You get to choose 7 numbers on a 7-play and the ticket wins if any subset of 6 of these 7 numbers is picked. Since this is equivalent to buying 7 different normal tickets, the cost is $7 for this kind of ticket. Similarly, they have an 8-play, which is equivalent to 28 standard tickets and so, costs $28. In fact, they have N-play tickets right up to N=15 which costs a whopping $5005! They, in fact have all N-plays for N=6,7,8,9,10,11,12,13,14,15 (I included the standard game as a 6-play). Well (wouldn't you know it?), there's a group of 143 mathematicians in the Ozendian Defence Department who wish to buy the equivalent of a 16-play costing $8008 ($56 for each mathematician). They have decided on their 16 favorite numbers. They wish to cover all subsets of 6 out of their 16 numbers using standard N-play tickets (N=6 to 15) with as few N-play tickets as possible (including the standard game as 6-play). After all, they are mathematicians, so they can afford to pay a bit more than $56 each! What is the minimum number of N-play tickets which would achieve the cover of all subsets of 6 out of the 16 numbers? What would they cost? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 20, 2009 Report Share Posted July 20, 2009 (edited) This seems too easy again! May be I am still missing something!! A minimum of 16 tickets would be needed (16C15) Buy 16 tickets of 15-Play leaving out 16the number in first, 15th number in second, 14th in the third... and so on The cost would be 16 * 5005 = 80080 and cost per mathematician = 80080/143 = $560 Edited July 20, 2009 by DeeGee Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 20, 2009 Report Share Posted July 20, 2009 The minimum number of tickets is 10, costing $80 per mathematician Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 20, 2009 Report Share Posted July 20, 2009 'Unless I'm mistaken (and I frequently am)...' Such as now...and buy only 4 tickets, costing $68.01 per mathematician Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted July 21, 2009 Author Report Share Posted July 21, 2009 This seems too easy again! May be I am still missing something!! A minimum of 16 tickets would be needed (16C15) Buy 16 tickets of 15-Play leaving out 16the number in first, 15th number in second, 14th in the third... and so on The cost would be 16 * 5005 = 80080 and cost per mathematician = 80080/143 = $560 No, you're not missing something. You have a convincing argument that you have found an upper bound. Can you prove that it can't be done with fewer tickets? Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted July 21, 2009 Author Report Share Posted July 21, 2009 (edited) Such as now...and buy only 4 tickets, costing $68.01 per mathematician Would you post precisely which tickets they would be? Just assume that the mathematicians want to cover all subsets of 6 from the numbers {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16}. I can't seem to find a set of tickets as you describe. (Webmaster, There seems to be a spoiler problem) Edited July 21, 2009 by superprismatic Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 22, 2009 Report Share Posted July 22, 2009 I can't seem to find a set of tickets as you describe. Oops... neither can I. 4 tickets is definitely wrong, and I can't remember anymore how I came up with 10. Quote Link to comment Share on other sites More sharing options...
Question
superprismatic
The country of Ozendia has a weekly lottery drawing.
A set of 6 numbers are drawn without replacement from
the set of numbers {1,2,3,...,40} randomly. A play
of this lottery consists of buying a ticket with a
buyer-chosen guess as to what these 6 numbers will be.
The drawing order of the numbers does not matter.
Each play costs $1. The payoff is usually in the
multi-million dollar range and it is a percentage of
the ticket sales since the last winner. Ozendians
love to gamble, so there is usually a small number
of winning tickets each week. If this number is
greater than one, the jackpot is shared equally among
the winning tickets. So far, this is much like many
lotteries familiar to most people.
Ozendians are social gamblers. That is, very often
a group of people will buy a bunch of tickets with
any winnigs divided up equitably to the group
members. Ozendian lottery officials want to encourage
this behavior as they believe it increases ticket
sales. So they developed a scheme to allow groups
to buy lots of sets of numbers on a single ticket.
They have what they call a 7-play ticket. You get
to choose 7 numbers on a 7-play and the ticket wins
if any subset of 6 of these 7 numbers is picked.
Since this is equivalent to buying 7 different
normal tickets, the cost is $7 for this kind of ticket.
Similarly, they have an 8-play, which is equivalent
to 28 standard tickets and so, costs $28. In fact,
they have N-play tickets right up to N=15 which
costs a whopping $5005! They, in fact have all
N-plays for N=6,7,8,9,10,11,12,13,14,15 (I included
the standard game as a 6-play).
Well (wouldn't you know it?), there's a group of 143
mathematicians in the Ozendian Defence Department who
wish to buy the equivalent of a 16-play costing $8008
($56 for each mathematician). They have decided on
their 16 favorite numbers. They wish to cover all
subsets of 6 out of their 16 numbers using standard
N-play tickets (N=6 to 15) at minimal cost. After all,
they are mathematicians, so they can afford to pay a
bit more than $56 each!
How many of each type of N-play tickets (include the
standatd game as 6-play) are necessary to achieve
minimal cost? What is the cost per play?
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