A poor genie's lottery
The Qurqirish Dragon
come up with a better solution than I have:
A man finds an old magic lamp, cleans it up, and a genie appears. The
genie says, "I am not as strong as I once was, but I will try to grant
you any one wish, to the best of my ability."
The man thinks for a moment and says "There is a lottery where six
numbers are drawn randomly from a set of 70. If you match them all,
then you win. I wish to buy one ticket and have it be the winner."
The genie pauses a moment before responding. "I have granted your wish
to the best of my ability. The laws of chance and luck would not let
me force all the numbers you choose to appear, however through a
loophole I have the ability for you to create your OWN luck. On the
next time you buy any tickets- as many as you want- then you will
match at least one number on every ticket you buy, if it is possible
for that to happen. It is up to you to figure out how to win with as
few tickets as possible."
How many tickets minimum must he buy to guarantee that he wins the
lottery? To summarize the rules:
1) this is a "pick 6 out of 70" lottery - each ticket has 6 distinct
numbers between 1 and 70
2) If it is possible, the winning six numbers, drawn after all tickets
are purchased, will be such that at least one of them is a number
picked on each ticket. (so if you buy 7 tickets with no numbers in
common, then some will not have a number match)
I have a solution, but I do not know if it is the best possible. I'll
give info about it if nobody matches or beats it in a few days.
Peter Smyth
"The Qurqirish Dragon" <qurqi...@aol.com> wrote in message
news:f7f27b96-9205-40d2...@27g2000yqr.googlegroups.com...
Buying every combination of numbers between 1 and 11 works (11C6 = 462
tickets). I suspect there is a more efficient way though.
Peter Smyth
Mark Tilford
Spoiler:
I can do it with 13 tickets
1 2 3 4 5 6 X
1 7 8 9 10 11 A
2 12 13 14 15 16 B
3 17 18 19 20 21 C
4 22 23 24 25 26 D
5 27 28 29 30 31 E
6 32 33 34 35 36 F
1 37 38 39 40 41 A'
2 37 38 39 40 41 B'
3 37 38 39 40 41 C'
4 37 38 39 40 41 D'
5 37 38 39 40 41 E'
6 37 38 39 40 41 F'
Tickets A,B,C,D,E,F have no members in common, so force all numbers to
be in 1-36. For at least one number on A' to win, it must be 1.
Similarly, B'-F' force 2-6 to be on the winning ticket. That makes
ticket X a winner.
The Qurqirish Dragon
That's a better solution than I originally had. It has a similar idea,
but executed in a much better way. Since you beat it, I have no
problem giving my original solution now.
My idea used 43 tickets in this way:
1 2 3 4 5 X
6 7 8 9 10 X
11 12 13 14 15 X
16 17 18 19 20 X
21 22 23 24 25 X
26 27 28 29 30 X
31 32 33 34 35 X
These are seven tickets, disjoint in the first five numbers, so if
these tickets are played, X must be one of the drawn numbers. Choose
six different values for X, and play those 42 tickets, and a 43rd
which consists of those six X values. that last ticket must win.
As I said, this is a similar idea to what you had, but yours is much
nicer.
Of course, you wouldn't want 1,2,3,4,5,6 to be the winning combination
(since you'd likely share the prize), but that's a whole different
problem.
dgates
Someone has probably already answered this, but I'll take a swing at
it.
The man has to kind of force the genie's hand by taking advantage of
the fact that the genie will do everything possible to have a winning
number on each ticket.
The man's first two tickets will be:
1,2,3,4,5,6 and 6,7,8,9,10,11.
This will force a 6 to be a winner (at least, once the man has done
this same trick five more times).
Next two tickets:
12,13,14,15,16,17 and 17,18,19,20,21,22.
That will force a 17.
Then:
23 thru 28, and 28 thru 33: will force a 28.
34 thru 39, and 39 thru 44: will force a 39.
45 thru 50, and 50 thru 55: will force a 50.
56 thru 61, and 61 thru 66: will force a 61.
So, for the man's final (thirteenth, and winning) ticket, he'll play
6,17,28,39,50, and 61. For the win!
There may be some clever way for the man to win it with less tickets,
but, if there is, I'll bet the man won't be 100% sure which ticket is
the winner, like he is with this strategy.
Prai Jei
space-time continuum:
> My idea used 43 tickets in this way:
> 1 2 3 4 5 X
> 6 7 8 9 10 X
> 11 12 13 14 15 X
> 16 17 18 19 20 X
> 21 22 23 24 25 X
> 26 27 28 29 30 X
> 31 32 33 34 35 X
>
> These are seven tickets, disjoint in the first five numbers, so if
> these tickets are played, X must be one of the drawn numbers. Choose
> six different values for X, and play those 42 tickets, and a 43rd
> which consists of those six X values. that last ticket must win.
> As I said, this is a similar idea to what you had, but yours is much
> nicer.
> Of course, you wouldn't want 1,2,3,4,5,6 to be the winning combination
> (since you'd likely share the prize), but that's a whole different
> problem.
Your solution would even work with the UK national lottery (6 from 49) as
would Mark's. Dgates' solution would not work because numbers higher than
49 were invoked in the analysis.
--
ξ:) Proud to be curly
Interchange the alphabetic letter groups to reply
The Qurqirish Dragon
That's a second problem I haven't looked at: If there are 40 numbers,
how many tickets are needed? 30 numbers? I can do it using 28 tickets,
and a maximum number of 26 this way:
A) use seven tickets to choose all possibilities of six numbers from 1
to 7. This forces two numbers to be between 1 and 7.
B) use seven more (total 14) to do the same, but with 8 through 14
C) seven more (total 21) does the same with 15 through 21.
D) use six more tickets (total 27) with the numbers 22, 23, 24, 25,
26, X, where X is one of the six numbers you want to be in the winning
combination.
E) the 6 numbers isolated are the winning ticket, so buy it.
This will work on as few as 26 numbers.
question 1) can you do better on a 26-number lottery?
question 2) how many do you need for a 25 (or fewer) number lottery?
Obviously, if there are too few numbers, the only solution is to buy
all possible tickets. For example, if there are 11 numbers total, then
you need to buy all 462 possible tickets, and in no case do you need
to buy more (as with fewer than 11 numbers, there are fewer possible
tickets, and with more than 11 numbers, the 11 solution will work, but
will not necessarily be the best.)
Remysun
common jackpot, keeping with the higher numbers keeps you from sharing
a jackpot with folks who pick birthday numbers to try to win.
The Qurqirish Dragon
That, and no obvious patterns (common differences, for example)