Author 
Message 
Proginoskes science forum Guru
Joined: 29 Apr 2005
Posts: 2593

Posted: Mon Apr 03, 2006 6:08 am Post subject:
Re: Help in solving this puzzle



Michael Harrington wrote:
Quote:  "James Waldby" <jwaldby@pat7.com> wrote in message
news:442CDE7C.89CBBFB4@pat7.com...
Michael Harrington wrote:
"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?
...
Now for your stated 20 numbers we have 10 blocks of 2
{1,2} {3,4}.............{17,18}{19,20}
This will equate to 10C3 = 120 tickets.
This is unproven as the minimum, but it is some partial solution.
Would be interested to see any improvement on this.
This is not the simplistic problem others are making it out to be.
You need to cover at least 17 of the numbers to force a win,
not 20 of them. Rather than 10C3 tickets your scheme should
need somewhere between 8C3 and 9C3, ie, between 56 and 84.
Return on investment might be better or worse with your 120
ticket scheme vs. an 84ticket one, depending on how many of
the 20 (6C3) winning triples occur on distinct tickets.
Having multiple winning triples on some tickets will decrease
the total payoff if each winning ticket pays only once.
jiw
Good point on only covering 17 numbers. Would be good to
know if there is a mathematical method of finding the minimum
number of covering tickets.

The problem clearly belongs in the field of combinatorics. If you want
every triple to be in exactly K tickets, then you're in the area of
design theory/tdesigns.
 Christopher Heckman 

Back to top 


BDH science forum beginner
Joined: 25 Oct 2005
Posts: 11

Posted: Thu Jun 29, 2006 10:06 am Post subject:
Re: Combinations satisfying linear equations?



Exactly. 

Back to top 


David DeLaney science forum beginner
Joined: 08 May 2005
Posts: 20

Posted: Sat Jul 08, 2006 4:28 pm Post subject:
Re: arithmetic Robak 0,(9)+{1+}0=1 Time Theory



ksrobak <ksrobak@o2.pl> wrote:
Quote:  Rich Holmes <rsholmes+usenet@mailbox.syr.edu
And why, WHY, has this information been withheld from a.r.k until now?
[Rich Holmes]

Manley Hubbel had a kid?
Quote:  ~~~~~~~~~~~~~~~~~~~~~~~~
X + oo = oo <== religion
X + oo = X + oo <== LOGIC
~~~~~~~~~~~~~~~~~~~~~~~~

X + oo = oo ; oo + X = oo + X <== ordinal transfinite arithmetic
* <== PERTH
Dave "time=inertia?" DeLaney

\/David DeLaney posting from dbd@vic.com "It's not the pot that grows the flower
It's not the clock that slows the hour The definition's plain for anyone to see
Love is all it takes to make a family"  R&P. VISUALIZE HAPPYNET VRbeable<BLINK>
http://www.vic.com/~dbd/  net.legends FAQ & Magic / I WUV you in all CAPS! K. 

Back to top 


Rich Holme science forum beginner
Joined: 06 Jun 2005
Posts: 45

Posted: Mon Jul 10, 2006 2:24 pm Post subject:
Re: arithmetic Robak 0,(9)+{1+}0=1 Time Theory



dbd@gatekeeper.vic.com (David DeLaney) writes:
Quote:  ksrobak <ksrobak@o2.pl> wrote:
Rich Holmes <rsholmes+usenet@mailbox.syr.edu
And why, WHY, has this information been withheld from a.r.k until now?
[Rich Holmes]
Manley Hubbel had a kid?
~~~~~~~~~~~~~~~~~~~~~~~~
X + oo = oo <== religion
X + oo = X + oo <== LOGIC
~~~~~~~~~~~~~~~~~~~~~~~~
X + oo = oo ; oo + X = oo + X <== ordinal transfinite arithmetic
* <== PERTH
A=A
Dave "time=inertia?" DeLaney

Ah yes, I remember that post. Good times, good times.

 Doctroid Doctroid Holmes <http://www.richholmes.net/doctroid/>
Ancient use of incendiary pigs as an antielephant measure is
disqualified on grounds of pigs not being cows, even when on fire.
 John D Salt 

Back to top 


Proginoskes science forum Guru
Joined: 29 Apr 2005
Posts: 2593

Posted: Sun Jul 16, 2006 4:34 am Post subject:
Re: Combinatorics



Thomas Mautsch wrote:
Quote:  In news:<1152934283.850075.249680@b28g2000cwb.googlegroups.com
schrieb Proginoskes <CCHeckman@gmail.com>:
Whatever5k@web.de wrote:
Suppose you have n fields, where n is at least 10. Each field is to be
filled with a ball. There are infinitely many balls, but they can be
distincted by their colour. In total there are 10 different colours. So
now, you want to find out how many distinct possibilities there are to
fill the n fields with those balls. There are, however, two conditions:
a) at the end, each colour has to appear at least once
b) the first field cannot contain the colour blue
Any idea how to tackle this?
I am an absolute beginner in combinatorics.
Please help.
The answer should be "infinitely many possibilities". You should check
the wording of the problem.
From what can be gathered from the OP's postings in de.sci.mathematik,
the problem is to determine the number of ndigit decimal numbers
with first digit different from zero
that contain each of the ten digits 0,1,2,3,4,5,6,7,8,9 at least once.

I don't see this at all. _IF_ there was a condition that stated each
field had to have at most one ball, then that would be the case. But as
I read the question, the following are all solutions to the problem,
for any nonnegative integer K: (where 0 = "blue")
Field 1: Infinitely many balls of colors 1 ... 9,
Field 2: Infinitely many balls of color 0, K balls of color 1,
Fields 3n: Empty.
This satisfies conditions (a) and (b). If fields aren't allowed to be
empty, put an infinite number of balls of color 1 in fields 3n.
That was what I was asking clarification about.
 Christopher Heckman 

Back to top 


Google


Back to top 



The time now is Fri Oct 20, 2017 7:04 am  All times are GMT

