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Red-Dog's Last Theorem
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Topic: Red-Dog's Last Theorem (Read 11905 times)
thetank
Hero Member
Offline
Posts: 19278
Re: Red-Dog's Last Theorem
«
Reply #45 on:
May 02, 2009, 07:19:28 PM »
Quote from: Cf on May 02, 2009, 05:12:39 PM
Quote from: thetank on May 02, 2009, 04:58:01 PM
There will be k riffles with stacks of n chips
That's about as far as I got before lack of education inhibited my progress.
Love the picture, but you need to use Z+. Your set allows a negative number of riffles
I was just gonna use Z, but go on to specify that k must fall into the range
1<k<(2n)!
(With < denoting less than or equal to but cba going to paint again to use the proper symbol
)
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For super fun to exist, well defined parameters must exist for the super fun to exist within.
Grier78
www.AllInOnADraw.com
Hero Member
Offline
Posts: 1136
www.AllInOnADraw.com
Re: Red-Dog's Last Theorem
«
Reply #46 on:
May 02, 2009, 07:45:46 PM »
Quote from: thetank on May 02, 2009, 06:27:49 PM
Quote from: kinboshi on May 02, 2009, 12:09:00 PM
Chips
Riffles
2
2
3
3
4
3
5
5
6
6
7
4
8
4
9
9
10
6
I'd like to question this data.
I simulated the effect of a regimented riffling system on two stacks of chips under laboratory conditions and oserved the following
2 chips = 2 riffles
3 chips = 4 riffles
4 chips = 3 riffles
5 chips = 6 riffles
6 chips = 10 riffles
7 chips = 12 riffles
8 chips = 4 riffles
I can corroborate your data
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UpTheMariners
Hero Member
Offline
Posts: 1888
We Only Sing When We're Fishing
Re: Red-Dog's Last Theorem
«
Reply #47 on:
May 02, 2009, 08:48:39 PM »
the answer is 3.5 bags of sugar
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ACE2M
Hero Member
Offline
Posts: 7832
Re: Red-Dog's Last Theorem
«
Reply #48 on:
May 02, 2009, 10:47:00 PM »
this battered my brain a few years a go, way beyond us.
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Jon MW
Hero Member
Offline
Posts: 6202
Re: Red-Dog's Last Theorem
«
Reply #49 on:
May 02, 2009, 11:04:18 PM »
Quote from: thetank on May 02, 2009, 07:19:28 PM
Quote from: Cf on May 02, 2009, 05:12:39 PM
Quote from: thetank on May 02, 2009, 04:58:01 PM
There will be k riffles with stacks of n chips
That's about as far as I got before lack of education inhibited my progress.
Love the picture, but you need to use Z+. Your set allows a negative number of riffles
I was just gonna use Z, but go on to specify that k must fall into the range
1<k<(2n)!
(With < denoting less than or equal to but cba going to paint again to use the proper symbol
)
ffs I was trying to keep out of it, wtf is Z+ ?
Do you mean
?
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Jon "the British cowboy" Woodfield
2011 blonde MTT League August Champion
2011 UK Team Championships: Black Belt Poker Team Captain - - runners up - -
5 Star HORSE Classic - 2007 Razz Champion
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thetank
Hero Member
Offline
Posts: 19278
Re: Red-Dog's Last Theorem
«
Reply #50 on:
May 03, 2009, 01:16:44 AM »
Oooh, is that for Natural numbers.
Wil use that then
for kay is less than or equal to (two times enn) factorial
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For super fun to exist, well defined parameters must exist for the super fun to exist within.
StuartHopkin
Hero Member
Offline
Posts: 8145
Ocho cinco
Re: Red-Dog's Last Theorem
«
Reply #51 on:
May 03, 2009, 06:17:36 AM »
Quote from: Grier78 on May 02, 2009, 07:45:46 PM
Quote from: thetank on May 02, 2009, 06:27:49 PM
Quote from: kinboshi on May 02, 2009, 12:09:00 PM
Chips
Riffles
2
2
3
3
4
3
5
5
6
6
7
4
8
4
9
9
10
6
I'd like to question this data.
I simulated the effect of a regimented riffling system on two stacks of chips under laboratory conditions and oserved the following
2 chips = 2 riffles
3 chips = 4 riffles
4 chips = 3 riffles
5 chips = 6 riffles
6 chips = 10 riffles
7 chips = 12 riffles
8 chips = 4 riffles
I can corroborate your data
3 chips = 3 riffles
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thetank
Hero Member
Offline
Posts: 19278
Re: Red-Dog's Last Theorem
«
Reply #52 on:
May 03, 2009, 07:54:00 AM »
Quote from: StuartHopkin on May 03, 2009, 06:17:36 AM
3 chips = 3 riffles
after 0...
b y
b y
b y
after 1...
b y
y b
b y
after 2...
y y
y b
b b
after 3... (not quite there yet)
y y
b y
b b
after 4... (now we're home)
b y
b y
b y
«
Last Edit: May 03, 2009, 07:56:53 AM by thetank
»
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For super fun to exist, well defined parameters must exist for the super fun to exist within.
thetank
Hero Member
Offline
Posts: 19278
Re: Red-Dog's Last Theorem
«
Reply #53 on:
May 03, 2009, 08:01:40 AM »
Only way to do it in 3 is if you break from the regimented system and suddenly change the direction on the 3rd riffle.
We won't be able to come up with a formula if people are changing the direction of the riffle at arbitary times.
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For super fun to exist, well defined parameters must exist for the super fun to exist within.
kinboshi
ROMANES EUNT DOMUS
Administrator
Hero Member
Offline
Posts: 44239
We go again.
Re: Red-Dog's Last Theorem
«
Reply #54 on:
May 03, 2009, 09:56:37 AM »
I still get 3 riffles for stacks of 3 chips.
The top three chips always going off to the left and being the first chip at the bottom:
£ $
£ $
£ $
First riffle gives:
$ £
£ $
$ £
2nd:
£ £
$ £
$ $
3rd:
£ $
£ $
£ $
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Cf
Global Moderator
Hero Member
Offline
Posts: 8081
Re: Red-Dog's Last Theorem
«
Reply #55 on:
May 03, 2009, 10:04:45 AM »
Quote from: Jon MW on May 02, 2009, 11:04:18 PM
Quote from: thetank on May 02, 2009, 07:19:28 PM
Quote from: Cf on May 02, 2009, 05:12:39 PM
Quote from: thetank on May 02, 2009, 04:58:01 PM
There will be k riffles with stacks of n chips
That's about as far as I got before lack of education inhibited my progress.
Love the picture, but you need to use Z+. Your set allows a negative number of riffles
I was just gonna use Z, but go on to specify that k must fall into the range
1<k<(2n)!
(With < denoting less than or equal to but cba going to paint again to use the proper symbol
)
ffs I was trying to keep out of it, wtf is Z+ ?
Do you mean
?
Z+ is the set of positive integers, {1,2,3...}
N is the set of natural numbers, {1,2,3...} or {0,1,2,3...}. Using Z+ saves this ambiguity. I believe 0 is commonly said to be a natural number these days.
So if you want to allow 0 riffles in your calculations use N, if not use Z+.
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Rupert
:)
Hero Member
Offline
Posts: 2119
Re: Red-Dog's Last Theorem
«
Reply #56 on:
May 03, 2009, 10:31:52 AM »
Hmm pretty sure someone from poker society at Warwick did a paper on this. Will see if I can get in touch with them.
At Walsall GUKPT this year, Red Dog was
boring
engaging the people at his table with this problem:
http://www.funtrivia.com/askft/Question23938.html
That site also has the solution FWIW, so thought Red Dog might be interested. The problem is you have 12 chips and one of the chips is of a different weight to the others (either heavier or lighter). On a set of balancing scales, determine in 3 weighs which chip it is and whether it is heavier or lighter.
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RichEO
Hero Member
Offline
Posts: 1493
Re: Red-Dog's Last Theorem
«
Reply #57 on:
May 03, 2009, 10:33:43 AM »
If I have 3 chips and cut them to the left every time it takes 4 riffles. If I cut them to the right everytime it takes 3 riffles. It is to do with having the same colour chip on the bottom (for the 2nd riffle) as you started with or a different colour one.
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sharky_uk
Sr. Member
Offline
Posts: 654
Re: Red-Dog's Last Theorem
«
Reply #58 on:
May 03, 2009, 12:05:54 PM »
I only have a 2000 piece chipset so had to stop at n=1000
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david3103
Hero Member
Offline
Posts: 6089
Re: Red-Dog's Last Theorem
«
Reply #59 on:
May 03, 2009, 12:29:21 PM »
Some interesting stuff here...
http://echochamber.me/viewtopic.php?f=17&t=33796
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