Load the dice
PT
7 is most likely, followed by 6 and 8, etc, where
all arithmetic is performed in fractions of 36.
This group requires no elaboration.
Your mission, Jim, should you decide to accept,
follows: weight a pair of dice, such that the 11
sums are equiprobable. i.e. denote them A and
B, and weight (probabilistically, not gravity) the
6 faces of each, appropriately.
Good luck, Jim -
--
Paul T.
Vincent Zweije
<fb6f3bf6-880d-4b37...@j32g2000prh.googlegroups.com>,
PT <ptane...@consultant.com> wrote:
> Your mission, Jim, should you decide to accept,
> follows: weight a pair of dice, such that the 11
> sums are equiprobable. i.e. denote them A and
> B, and weight (probabilistically, not gravity) the
> 6 faces of each, appropriately.
Assign probabilities a1..a6, b1..b6
Require:
1) a1*b1 = 1/11
2) a1*b2 + a2*b1 = 1/11
3) a1*b3 + a2*b2 + a3*b1 = 1/11
...
10) a5*b6 + a6*b5 = 1/11
11) a6*b6 = 1/11
As well as:
12) a1 + a2 + ... + a6 = 1
13) b1 + b2 + ... + b6 = 1
There are 12 variables, and 13 equalities. There is probably no solution,
unless some of the equations are dependent. I'm too lazy to go symbol
shunting and solve this.
Should there be one or more solutions, there are also 24 inequalities
to satisfy:
1..12) 0 <= ai <= 1
13..24) 0 <= bi <= 1
Ciao. Vincent.
--
Vincent Zweije <zwe...@xs4all.nl> | "If you're flamed in a group you
<http://www.xs4all.nl/~zweije/> | don't read, does anybody get burnt?"
[Xhost should be taken out and shot] | -- Paul Tomblin on a.s.r.
TTman
"PT" <ptane...@consultant.com> wrote in message
news:fb6f3bf6-880d-4b37...@j32g2000prh.googlegroups.com...
So why is 7 more likely than any other combination ?
k...@att.bizzzzzzzzzzzz
# Possible rolls #Possible rolls
2: 1-1 = 1
3: 1-2, 2-1 = 2
4: 1-3, 2-2, 3-1 = 3
5: 1-4, 2-3, 3-2, 4-1 = 4
6: 1-5, 2-4, 3-3, 4-2, 5-1 = 5
7: 1-6, 2-5, 3-4, 4-3, 5-2, 6-1 = 6
8: 2-6, 3-5, 4-4, 5-3, 6-2 = 5
9: 3-6, 4-5, 5-4, 6-3 = 4
10: 4-6, 6-6, 6-4 = 3
11: 5-6, 6-5 = 2
12: 6-6 = 1
+---
36
Robert Israel
I don't accept.
Let P(A=j) = a_j and P(B=j) = b_j for j=1..6. Thus
P(A+B=2) = a_1 b_1 = 1/11 and P(A+B=12) = a_6 b_6 = 1/11.
But P(A+B=7) >= a_1 b_6 + a_6 b_1
= (1/11) (a_1/a_6 + a_6/a_1)
>= 2/11
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
John O'Flaherty
Assign them all zero. All combinations equiprobable.
--
John
A N Niel
Try this...
http://mathoverflow.net/questions/41310/
Now, PT, your assignment is to answer honestly: Was this your homework
problem?
Don Lancaster
Back when I first ran this dice project <
http://www.tinaja.com/glib/eldicepe.pdf >, a bunch of readers really got
bent out of shape when they did not realize that shooting two six sided
dice at once was exactly the same as shooting one thirty six sided dice.
"But they are locked together!"
Golly Gee mister science.
--
Many thanks,
Don Lancaster voice phone: (928)428-4073
Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552
rss: http://www.tinaja.com/whtnu.xml email: d...@tinaja.com
Please visit my GURU's LAIR web site at http://www.tinaja.com
m II
> Back when I first ran this dice project <
> http://www.tinaja.com/glib/eldicepe.pdf >, a bunch of readers really got
> bent out of shape when they did not realize that shooting two six sided
> dice at once was exactly the same as shooting one thirty six sided dice.
>
> "But they are locked together!"
>
> Golly Gee mister science.
Would that 36 sided dice have a '2' as the lowest number?
mike
Richard Henry
> On 12/28/2010 3:23 PM, A N Niel wrote:
>
>
>
> > In article
> > <ptanenb...@consultant.com> wrote:
I can think of one 36-sided solid with equal probability for each side
- a pencil-shaped cylinder with 18 sides shaved down from the center
to a common point at each end.
Are there others?
k...@att.bizzzzzzzzzzzz
Do you have a 36-sided regular polyhedron?
Richard Henry
Seat of the pants thinking - to squash the probability distribution
out toward the low end (2 or 3) and the high end (11 and 12) suggests
that we increase the probability of 1 and 6 on each die which from a
purely mechanical sense is difficult because they are on opposite
sides of a standars die.
Interesting problem. More responses coming, I am sure.
k...@att.bizzzzzzzzzzzz
wrote:
I didn't think of that; essentially an outy Roulette Wheel, sans bearing.
>Are there others?
ChasBrown
The first that comes to mind is a pair of 18-sided pyramids glued
together at their bases (in the same sense that an octahedron is two 4-
sided pyramids glued together at their bases).
Others?
Cheers - Chas
Bill Bowden
<k...@att.bizzzzzzzzzzzz> wrote:
> On Tue, 28 Dec 2010 19:52:36 -0000, "TTman" <pcw1....@ntlworld.com> wrote:
>
> >"PT" <ptanenb...@consultant.com> wrote in message
I used to like to play the field for one roll of 1:1 payout. Field
bets are 2, 3, 4, 9, 10, 11, or 12 which gives 1+2+3+4+3+2+1 = 16 out
of 36 or 44%. But I think you get a little bit more on 2 and 12, so
maybe it's 46%. Never did figure out if pass or don't pass was
better.
-Bill
k...@att.bizzzzzzzzzzzz
wrote:
For even odds, all sides have to be the same shape and equidistant from the
center. Other than the cylinder, I doubt you'll find a 36-sided solid that'll
fit the bill.
>Others?
>
>Cheers - Chas
Richard Henry
wrote:
The house doesn't care what you bet.
Robert Baer
Vincent Zweije
k...@att.bizzzzzzzzzzzz <k...@att.bizzzzzzzzzzzz> wrote:
Infinitely many, in fact. You can twist the two pyramids slightly, getting
a 36-segment zig-zagging connect between them instead of a 18-hedron,
and with the faces becoming 4-sided instead of 3-sided. Each twist gives
you a different form (mod pi/9 of course).
Vincent Zweije
<vin...@zweije.nl> wrote:
> In article <o0hlh6l9a44j3gvep...@4ax.com>,
> k...@att.bizzzzzzzzzzzz <k...@att.bizzzzzzzzzzzz> wrote:
>
> || On Tue, 28 Dec 2010 19:05:59 -0800 (PST), ChasBrown
> || <cbr...@cbrownsystems.com> wrote:
>
> || >The first that comes to mind is a pair of 18-sided pyramids glued
> || >together at their bases (in the same sense that an octahedron is two 4-
> || >sided pyramids glued together at their bases).
> ||
> || For even odds, all sides have to be the same shape and equidistant from
> || the center. Other than the cylinder, I doubt you'll find a 36-sided
> || solid that'll fit the bill.
> ||
> || >Others?
>
> Infinitely many, in fact. You can twist the two pyramids slightly, getting
> a 36-segment zig-zagging connect between them instead of a 18-hedron,
Make that a regular 18-sided polygon.
alainv...@gmail.com
For dice problem we can work
with polynomials equivalent.
Usual cubic 1/6(t+t^2+...t^6)
Two casts correspond to {1/6(t+t^2+...t^6)}^2 ,
we may isolate r given t^r.
A loaded n faces dice sum(a(i)t^i) , sum(a(i),i 1 to n) = 1
p casts <=> sum(a(i)t^i)^p ....
Alain
The Qurqirish Dragon
Using what I have seen in other posts, I would say this is
impossible.
Reasoning:
there are 36 possible combinations you can roll on the dice. Thus, the
probability of any specific total is n/36, where n is the number of
ways to get that total. Since n is an integer, you cannot have n/36 =
1/11. Thus, no single total can have a probability of 1/11, let alone
all of them. QED.
Now, if you allow 12 different totals, then it is easy: one die is
numbered 1,2,3,4,5,6 and the other is 0,0,0,6,6,6. Every number from 1
to 12 has a 1/12 probability.
Using my argument, with 11values, it is theoretically possible to have
8 sums with probability 1/12 and 3 with probability 1/9, which is the
closest you can get to the original problem. The question is: can you
actually make dice to fit this probability distribution?
Willem
) On Dec 28, 6:51?pm, Richard Henry <pomer...@hotmail.com> wrote:
)> On Dec 28, 10:36?am, PT <ptanenb...@consultant.com> wrote:
)>
)> > Toss two fair dice, they will sum to 2, 3, ... 12.
)> > 7 is most likely, followed by 6 and 8, etc, where
)> > all arithmetic is performed in fractions of 36.
)> > This group requires no elaboration.
)>
)> > Your mission, Jim, should you decide to accept,
)> > follows: weight a pair of dice, such that the 11
)> > sums are equiprobable. ?i.e. denote them A and
)> > B, and weight (probabilistically, not gravity) the
)> > 6 faces of each, appropriately.
)
) Using what I have seen in other posts, I would say this is
) impossible.
) Reasoning:
) there are 36 possible combinations you can roll on the dice. Thus, the
) probability of any specific total is n/36, where n is the number of
) ways to get that total.
Huh ?
Each side of each of the dice can have a different probability
of coming up, so the probability of each combination can be different.
SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT
alainv...@gmail.com
Good evening,
Since poly 1/11(t^2+t^3+...t^12) can't be split into
two equal real six degree polys we can not get a uniform
distribution with two identical loaded dice or with
two casts of a same dice.
Anyway we may represent two different dices by two polys:
dice one p faces sum(a(i)t^i , i 1 to p ) , sum(a(i))=1 ,
dice two q faces sum(b(j)(ut)^j , j 1 to q ) , sum(b(j))=1 ,
A common cast <=>sum(a(i)t^i)*sum(b(j)(ut)^j)
A specified value r corresponds to sum of all terms
a(i)*b(j)*u^j*t^(i+j) with (i+j)= r ,
Alain
Don Del Grande
> krw wrote:
>> TTman wrote:
>>
>> >So why is 7 more likely than any other combination ?
>>
>> # Possible rolls #Possible rolls
>> 2: 1-1 = 1
>> 3: 1-2, 2-1 = 2
>> 4: 1-3, 2-2, 3-1 = 3
>> 5: 1-4, 2-3, 3-2, 4-1 = 4
>> 6: 1-5, 2-4, 3-3, 4-2, 5-1 = 5
>> 7: 1-6, 2-5, 3-4, 4-3, 5-2, 6-1 = 6
>> 8: 2-6, 3-5, 4-4, 5-3, 6-2 = 5
>> 9: 3-6, 4-5, 5-4, 6-3 = 4
>> 10: 4-6, 6-6, 6-4 = 3
>> 11: 5-6, 6-5 = 2
>> 12: 6-6 = 1
>> +---
>> 36
>
>I used to like to play the field for one roll of 1:1 payout. Field
>bets are 2, 3, 4, 9, 10, 11, or 12 which gives 1+2+3+4+3+2+1 = 16 out
>of 36 or 44%. But I think you get a little bit more on 2 and 12, so
>maybe it's 46%. Never did figure out if pass or don't pass was
>better.
Don't pass is slightly better, even if a 2 or 12 on the come-out roll
is a push. (If the push number is 3, I think pass becomes better, but
I'm not sure.)
Going back to the original problem, if the dice are labeled A and B,
and An and Bn are the probabilities of side n coming up on die A or B
respectively, and X = 1/11, you get:
X = A1 B1
X = A1 B2 + A2 B1
X = A1 B3 + A2 B2 + A3 B1
X = A1 B4 + A2 B3 + A3 B2 + A4 B1
X = A1 B5 + A2 B4 + A3 B3 + A4 B2 + A5 B1
X = A1 B6 + A2 B5 + A3 B4 + A4 B3 + A5 B2 + A6 B1
Working from top to bottom, you can get each B in terms of X and As,
but B6 has a term of A1^6 in it somewhere.
-- Don
Herman Rubin
> On Dec 28, 6:51 pm, Richard Henry <pomer...@hotmail.com> wrote:
>> On Dec 28, 10:36 am, PT <ptanenb...@consultant.com> wrote:
>> > Toss two fair dice, they will sum to 2, 3, ... 12.
>> > 7 is most likely, followed by 6 and 8, etc, where
>> > all arithmetic is performed in fractions of 36.
>> > This group requires no elaboration.
>> > Your mission, Jim, should you decide to accept,
>> > follows: weight a pair of dice, such that the 11
>> > sums are equiprobable. i.e. denote them A and
>> > B, and weight (probabilistically, not gravity) the
>> > 6 faces of each, appropriately.
This is easily seen to be impossible. It is impossible
to load them so that, if the probabilities of 2 and 12
are equal, the probability of 7 is not at least twice
as large.
Let the probabilities of 1 be hb and h/b, and that of
6 be hf and h/f; this must be the case if the probablilities
of 2 and 12 are equal. Then the contribution of these
to a sum of 7, divided by h^2, is b/f + f/b, and as
this is of the form x + 1/x, it is at least 2.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
bill
> Toss two fair dice, they will sum to 2, 3, ... 12.
> 7 is most likely, followed by 6 and 8, etc, where
> all arithmetic is performed in fractions of 36.
> This group requires no elaboration.
>
> Your mission, Jim, should you decide to accept,
> follows: weight a pair of dice, such that the 11
> sums are equiprobable. i.e. denote them A and
> B, and weight (probabilistically, not gravity) the
> 6 faces of each, appropriately.
>
>
> --
> Paul T.
If all sums are equiprobable and the dice are
identical, then
1) P(1,1)=P(1)*P(1)=1/11
2) P(1,2 or 2,1) = 2*P(1)* P(2) = 1/11
3) P(1,3 or 2,2 or 3,1)= 2*P(1)*P(3)+P(2)^2 = 1/11
4) P(1,4 or 2,3 or 3,2 or 1,4)= 2*P(1)*P(4)+
2*P(2)*P(3) = 1/11
5) P(Sum=6) = 2*P(1)*P(5)+2*P(2)*P(4)+P(3)^2 = 1/11
6) P(Sum=7)=2*P(1)*P(6)+2*P(2)*P(5)+2*P(3)*p(4)= 1/11
Bill J
James Dow Allen
Using two balanced six-sided dice, label the faces with
positive integers to get the same probabilities for each
sum 2 to 12 as with ordinary dice.
(1,2,3,4,5,6), (1,2,3,4,5,6) works by definition.
Is there another way?
James Dow Allen
George Herold
That look much easier. 0,1,2,3,4,5 and 2,3,4,5,6,7.
(Is there a less trivial solution?)
George H.
Jasen Betts
> Toss two fair dice, they will sum to 2, 3, ... 12.
> 7 is most likely, followed by 6 and 8, etc, where
> all arithmetic is performed in fractions of 36.
> This group requires no elaboration.
>
> Your mission, Jim, should you decide to accept,
> follows: weight a pair of dice, such that the 11
> sums are equiprobable. i.e. denote them A and
> B, and weight (probabilistically, not gravity) the
> 6 faces of each, appropriately.
you hot the phrasing right, the task set is impossible unless the dice
can communicate in some way.
there's no way to weight two independant dice to give totals of 2,
7, and 12 with equal probability.
> Good luck, Jim -
>
> --
> Paul T.
--
⚂⚃ 100% natural
--- news://freenews.netfront.net/ - complaints: ne...@netfront.net ---
ChasBrown
>
> >> > Please visit my GURU's LAIR web site athttp://www.tinaja.com
>
> >> I can think of one 36-sided solid with equal probability for each side
> >> - a pencil-shaped cylinder with 18 sides shaved down from the center
> >> to a common point at each end.
>
> >> Are there others?
>
> >The first that comes to mind is a pair of 18-sided pyramids glued
> >together at their bases (in the same sense that an octahedron is two 4-
> >sided pyramids glued together at their bases).
>
> For even odds, all sides have to be the same shape and equidistant from the
> center. Other than the cylinder, I doubt you'll find a 36-sided solid that'll
> fit the bill.
>
The example I gave fits those requirements.
However, your requirements are a bit excessive; they just make the
problem tractable. Imagine a 35 sided pyramid, with a base the shape
of a regular 35-gon. If the base is very small compared to the size of
a pyramid triangle, it will come up far less than any of the 35 sides.
If it is really big (e.g., the pyramid is extremely short, approaching
a flat shape) it will come up very often. Somewhere in between those
two extremes, it comes up just as often as any the 35 sides of the
pyramid; it's just annoying to figure out exactly where that might
happen.
Cheers - Chas
> >Others?
>
> >Cheers - Chas
>
>
Jasen Betts
zero is not usually considered a positive integer.
> (Is there a less trivial solution?)
there is, I googled it so I won't post the spoiler so soon after it
was posed.
--
⚂⚃ 100% natural
Don Del Grande
> James Dow Allen wrote:
>> Here's a variant puzzle that might be fun if you've never seen it.
>>
>> Using two balanced six-sided dice, label the faces with
>> positive integers to get the same probabilities for each
>> sum 2 to 12 as with ordinary dice.
>>
>> (1,2,3,4,5,6), (1,2,3,4,5,6) works by definition.
>> Is there another way?
>>
>> James Dow Allen
>
>
>That look much easier. 0,1,2,3,4,5 and 2,3,4,5,6,7.
>(Is there a less trivial solution?)
There may be a more correct solution: the problem asked for positive
integers.
(I hate it when I do that, too.)
-- Don
Bill Bowden
Yes, the house always wins except in the stock market. Never did
understand why gamblers insist on losing money playing card games and
craps when they can just stay home and play the stock market on their
PC for better odds. Buy a few call options or whatever and watch the
action.
-Bill
k...@att.bizzzzzzzzzzzz
The house always wins in the stock market, too.
>Never did
>understand why gamblers insist on losing money playing card games and
>craps when they can just stay home and play the stock market on their
>PC for better odds. Buy a few call options or whatever and watch the
>action.
Easy. A few gamblers win. Most think they're part of that few. OTOH, horse
racing can be a skill game.
Richard Henry
<k...@att.bizzzzzzzzzzzz> wrote:
> On Wed, 29 Dec 2010 21:40:06 -0800 (PST), Bill Bowden
>
>
>
Horse race gambling is an interesting skill, involving not only
judgement of the horses' ability, but also identifying defects in the
popular opinion of the horses' ability.
Bill Bowden
<k...@att.bizzzzzzzzzzzz> wrote:
> On Wed, 29 Dec 2010 21:40:06 -0800 (PST), Bill Bowden
>
>
>
Yes, if you want to invest a lot of time, you might beat the horses.
The same could be said for a stock. What's the difference of a horse
or a stock? Who the hell cares if it makes money?
-Bill
James Dow Allen
> >> Using two balanced six-sided dice, label the faces with
> >> positive integers to get the same probabilities for each
> >> sum 2 to 12 as with ordinary dice.
Yes. I almost wrote "characterize all solutions", but decided
writing "positive integers" severed the trivia most tersely.
> I googled it so I won't post the spoiler so soon after it
> was posed.
Please do tell what your search terms were. My Googling
skill is very poor; this is one I wouldn't even attempt.
On Dec 30, 8:31 am, Jasen Betts <ja...@xnet.co.nz> wrote:
> > Your mission, Jim, should you decide to accept, ...
>
> you hot the phrasing right, the task set is impossible unless the dice
> can communicate in some way.
Communicating dice? Is this one of those quantum physics
puzzles? Or are these those special dice I've seen in
operation in Asian gambling dens! !?!?!
James Dow Allen
alainv...@gmail.com
Good morning,
May I propose you unloaded dices ,
dice1 2,3,5,6
dice2 2,9,14,21
common cast gives 16 equiprobable values:
4,5,7,8,11,12,14,15,16,17,19,20,23,24,26,27
Just have the right polys...
Alain
Herman Rubin
> Good morning,
> Alain
The simplest way to get 36 equally likely values is
to use the digital method.
For 0-35, have one die 0,1,2,3,4,5 and the other
0,6,12,18,24,30, to express the sum in base 6.
Martin Brown
> In article
> <ptane...@consultant.com> wrote:
>
>> Toss two fair dice, they will sum to 2, 3, ... 12.
>> 7 is most likely, followed by 6 and 8, etc, where
>> all arithmetic is performed in fractions of 36.
>> This group requires no elaboration.
>>
>> sums are equiprobable. i.e. denote them A and
>> B, and weight (probabilistically, not gravity) the
>> 6 faces of each, appropriately.
>>
>>
>> --
>> Paul T.
>
> Try this...
>
> http://mathoverflow.net/questions/41310/
>
> Now, PT, your assignment is to answer honestly: Was this your homework
> problem?
An exact solution is impossible as the above URL proves.
The next question is what is the closest to the target that can be
achieved in practice with P>=0. The closest solution I can see gives
P(2)=P(12)= 1/16
P(7)=1/8
P(3,4,5,6,8,9,10,11) = 3/32
Dice 1 : 1/2, 0, 0, 0, 0, 1/2
Dice 2 : 1/8, 3/16, 3/16, 3/16, 3/16, 1/8
If I am allowed to use non-physical negative probabilities then the
optimum solution is a bit better :
P(Even) = 1/12
P(Odd) = 1/10
Dice 1 : 0.66679, -0.23621, 0.069423, ...
Dice 2 : 0.12498, 0.19425 0.180777, ...
(and symmetric)
Are there any better solutions obtainable numerically?
Regards,
Martin Brown
The Qurqirish Dragon
Sorry, I misread the question. I read "weight the dice" as "renumber
the dice" For example, a die numbered 1-2-3-5-5-6, even if fair, is
still a weighted die in that 5 is more likely.
The Qurqirish Dragon
Why be so fancy? using 4-sided dice as you did, use the numbers:
die 1: 1,2,3,4
die 2: 1,5,9,13
Gets all numbers from 2 to 17 with equal probabilities.
tc...@lsa.umich.edu
James Dow Allen <jdall...@yahoo.com> wrote:
>On Dec 30, 8:53�am, Jasen Betts <ja...@xnet.co.nz> wrote:
>> I googled it so I won't post the spoiler so soon after it
>> was posed.
>
>Please do tell what your search terms were. My Googling
>skill is very poor; this is one I wouldn't even attempt.
I'm not Jasen, but I Googled "Sicherman dice" since that's what I
(correctly, as it turned out) thought these dice were called.
--
Tim Chow tchow-at-alum-dot-mit-dot-edu
The range of our projectiles---even ... the artillery---however great, will
never exceed four of those miles of which as many thousand separate us from
the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences
k...@att.bizzzzzzzzzzzz
It's more of a career.
>The same could be said for a stock. What's the difference of a horse
>or a stock? Who the hell cares if it makes money?
Really, not much, I suppose. Both are pari-mutuel betting, sort of (with
stocks not having a post-time). With stocks, the house's cut is a *lot* less
and taxes are simpler for the speculator. Perhaps that's some insight into
day traders and the stock market's boom and bust cycles?
Richard Henry
> 7 is most likely, followed by 6 and 8, etc, where
> all arithmetic is performed in fractions of 36.
> This group requires no elaboration.
>
> Your mission, Jim, should you decide to accept,
> follows: weight a pair of dice, such that the 11
> sums are equiprobable. i.e. denote them A and
> B, and weight (probabilistically, not gravity) the
> 6 faces of each, appropriately.
>
>
> --
> Paul T.
While fiddling with possibilities in a spreadsheet, I came across this
interesting result. If you use a regular balanced die combined with
one which is loaded so that it can only come up 1 or 6 (essentially
flipping a coin marked 1 and 6 on opposite sides) the results are even
at 1/12 for all numbers except 7, which gets a double dose at 1/6.
Andrew B.
> On Dec 30, 8:53 am, Jasen Betts <ja...@xnet.co.nz> wrote:
>
> > >> Using two balanced six-sided dice, label the faces with
> > >> positive integers to get the same probabilities for each
> > >> sum 2 to 12 as with ordinary dice.
> > zero is not usually considered a positive integer.
>
> Yes. I almost wrote "characterize all solutions", but decided
> writing "positive integers" severed the trivia most tersely.
(x + x^2 + x^3 + x^4 + x^5 + x^6)^2 = [x + 2x^2 + 2x^3 + x^4][x + x^3
+ x^4 + x^5 + x^6 + x^8].
Frithiof Andreas Jensen
>> The house doesn't care what you bet.
>
> Yes, the house always wins except in the stock market.
Goldman Sachs always win in the market too. Actually, "Everybody" wins
these days: The people who blew up the mortgage bond market and the
people who shorted it all retired RICH, thanks to good olde Uncle Sap!
Zero-sum games are soo deflationary.
They are at it again BTW - the shops are full of Cash-Back deals, its so
1990'ish.
> Never did
> understand why gamblers insist on losing money playing card games and
> craps when they can just stay home and play the stock market on their
> PC for better odds. Buy a few call options or whatever and watch the
> action.
LEAP Options are the best bets around - Black Scholes modelling does not
work too well when extended to 2 year periods (Except they probably
still cancel your trades if GS were to lose).
>
> -Bill
Robert Israel
Closest in what sense? Your solution above is, I think, optimal in the
least-squares sense, i.e. minimizing sum_{j=2}^{12} (P(j) - 1/11)^2.
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Frithiof Andreas Jensen
>> The same could be said for a stock. What's the difference of a horse
>> or a stock? Who the hell cares if it makes money?
>
> Really, not much, I suppose. Both are pari-mutuel betting, sort of (with
> stocks not having a post-time). With stocks, the house's cut is a *lot* less
> and taxes are simpler for the speculator. Perhaps that's some insight into
> day traders and the stock market's boom and bust cycles?
In Denmark the horses are much more tax efficient than bonds or stock -
One suspects from the huge bets on "place" (a horse is 1,2 or 3 -
usually barely returns the money) or "Win" on favorites at odds 1:1.10
that a lot of "black" money is laundered via the race track and the
casino; you get a recipt. Then pay 18% gambling tax on your drugs money
- which is better than 50% income tax and jail.
k...@att.bizzzzzzzzzzzz
Yes, that's been done here, but the "bosses" tend to get upset when #2 breaks
a leg out of the gate. Taxes on winnings are the issue. The IRS tends to
believe brokerage statements and losses are allowed to exceed winnings. Not
so much for track tickets.
Richard Henry
My brother-in-law, a former accountant with a large hotel, pointed out
that empty hotel rooms can be "sold" to cash customers who exist only
on paper. A few dozen $1000/night rooms every week can chew through
a wad of drug money pretty quickly.
ehsjr
Pace. In the stock market, if you doubled your money in say a year,
that would be an extraordinary accomplishment. If you did it in
the casino on one hand of cards, it would take what - a couple of
minutes? The casinos create, and appeal to, excitement, and some
people love that. My guess is that stock market speculators like
the same excitement of betting, but don't require that fast pace
to get the thrill. Of course, some people probably do both.
In either case, the activity is not based on sound financial
reasoning. Most gamblers, and most day traders, lose.
Ed
Jasen Betts
> On Dec 30, 8:53 am, Jasen Betts <ja...@xnet.co.nz> wrote:
>> >> Using two balanced six-sided dice, label the faces with
>> >> positive integers to get the same probabilities for each
>> >> sum 2 to 12 as with ordinary dice.
>> zero is not usually considered a positive integer.
>
> Yes. I almost wrote "characterize all solutions", but decided
> writing "positive integers" severed the trivia most tersely.
>
>> I googled it so I won't post the spoiler so soon after it
>> was posed.
>
> Please do tell what your search terms were. My Googling
> skill is very poor; this is one I wouldn't even attempt.
reasoning that this is a well-known puzzle (I have encountered it
before in dead-tree form) I chose the following:
alternative dice numbering
(result number 4 when I tried it)
James Dow Allen
> reasoning that this is a well-known puzzle (I have encountered it
> before in dead-tree form) I chose the following:
>
> alternative dice numbering
>
> (result number 4 when I tried it)
Nifty. I need to work on my Googling skills.
BTW, I don't think I'd seen these "Sicherman dice";
I remembered a solution using one 4-sided die and one
9-sided die, realized they could be decomposed into
respectively two 2-siders and two 3-siders; that those
could then be reassembled into the "Sicherman" form.
James
RichD
> > One suspects from the huge bets on "place" (a horse is 1,2 or 3 -
> > usually barely returns the money) or "Win" on favorites at odds 1:1.10
> > that a lot of "black" money is laundered via the race track and the
> > casino; you get a recipt. Then pay 18% gambling tax on your
> > drugs money - which is better than 50% income tax and jail.
>
> My brother-in-law, a former accountant with a large hotel, pointed out
> that empty hotel rooms can be "sold" to cash customers who exist only
> on paper. A few dozen $1000/night rooms every week can chew
> through a wad of drug money pretty quickly.
?
Elaborate please.
--
Rich
Richard Henry
Richard Henry
Consider, as an example, a hotel with rooms that rent for $500 a
night. Some customers pay cash, and are required only to sign a
register as a form of identification. For any such room that is not
rented for a night, the money launderer can create a fictitious
registration and transfer $500 from the dirty pile to the clean pile.
PT
> >> > Toss two fair dice, they will sum to 2, 3, ... 12.
> >> > 7 is most likely, followed by 6 and 8, etc,
> >> > where all arithmetic is performed in fractions
> >> > of 36.
> >> > follows: weight a pair of dice, such that the 11
> >> > sums are equiprobable. i.e. denote them A and
> >> > B, and weight (probabilistically, not gravity)
> >> > the 6 faces of each, appropriately.
>
> probabilities of 2 and 12 are equal, the probability
> of 7 is not at least twice as large.
>
> Let the probabilities of 1 be hb and h/b, and that
> of 6 be hf and h/f; this must be the case if the
> probablilities of 2 and 12 are equal. Then the
> contribution of these to a sum of 7, divided by
> h^2, is b/f + f/b, and as
> this is of the form x + 1/x, it is at least 2.
I must confess to befuddlement.
Your proof does indeed look correct.
The problem originates at a lecture by
Don Knuth I attended last month. He's
retired - 'emeritus' - but still does his
own research - he calls it "experimental
mathematics". As old as Methuselah, but
still sharp as Gillette.
The lecture primarily discussed the
appearance of pi in Sterling's approximation
of the factorial function. This dice
problem was the jumpoff point, then segue'd
down various other paths, quite marvelously.
He indicated the problem was difficult, and
its refutation involved some advanced math.
But your proof is almost trivial, which
leaves me scratching my head.
I will write to the perfessor, and post
his reply here, if any.
---
Paul T.
James Dow Allen
> On Dec 29 2010, Herman Rubin <hru...@skew.stat.purdue.edu> wrote:
> > this is of the form x + 1/x, it is at least 2.
>
> still sharp as Gillette.
Three comments.
(1) Robert Israel gave the same correct answer before Herman did.
(Although Prof. Israel's response was largely ignored(?))
(2) Herman removed rec.puzzles from the newsgroup list (why?),
so I didn't see his post. PT (carefully?) restored the newsgroup.
(3) Knuth is indeed a great treasure! I'm afraid to click
his website as it often leads to digressions of long hours!
James
RichD
> > > >One suspects from the huge bets on "place"
> > > >that a lot of "black" money is laundered via the
> > > >race track and the casino; you get a recipt. Then
> > > >pay 18% gambling tax on your drugs money -
> > > >which is better than 50% income tax and jail.
>
> > > My brother-in-law, a former accountant with a large hotel,
> > > pointed out that empty hotel rooms can be "sold" to cash
> > > customers who exist only on paper. A few dozen
> > > $1000/night rooms every week can chew
> > > through a wad of drug money pretty quickly.
>
>
> Consider, as an example, a hotel with rooms that rent for
> $500 a night. Some customers pay cash, and are required
> only to sign a register as a form of identification.
> For any such room that is not rented for a night, the
> money launderer can create a fictitious registration
> and transfer $500 from the dirty pile to the clean pile.
I've heard this money laudering stuff many times
on the news, but never sussed it.
So let's see, in your example, you have a big
wad, you're afraid to deposit it or send it to your
stockbroker, as they are all agents of Big Brother.
So you buy a hotel, and rent rooms to yourself,
and the cash becomes business income. (taxed, though)
Clever (except for the tax). Any other ideas?
I'm in the mood for a career change - financial
advisor seems promising, especially with Obamanomics -
--
Rich
k...@att.bizzzzzzzzzzzz
wrote:
That's the point of laundering money. It's now legit.
>Any other ideas?
Any business that deals in cash. I know someone who made a *pile* of un-taxed
money (kinda the reverse of the problem) with a sandwich truck.
>I'm in the mood for a career change - financial
>advisor seems promising, especially with Obamanomics -
Mafia or drug cartel financial adviser is a rather risky occupation. OTOH,
Demonicrat financial adviser is easy. Just think of ways to piss away China's
money.
PT
> > >> > Toss two fair dice, they will sum to 2, 3, ... 12.
> > >> > 7 is most likely, followed by 6 and 8, etc,
> > >> > where all arithmetic is performed in fractions
> > >> > of 36.
> > >> > Your mission, Jim, should you decide to accept,
> > >> > follows: weight a pair of dice, such that the 11
> > >> > sums are equiprobable. i.e. denote them A and
> > >> > B, and weight (probabilistically, not gravity)
> > >> > the 6 faces of each, appropriately.
>
> > It is impossible to load them so that, if the
> > probabilities of 2 and 12 are equal, the probability
> > of 7 is not at least twice as large.
>
> > Let the probabilities of 1 be hb and h/b, and that
> > of 6 be hf and h/f; this must be the case if the
> > probablilities of 2 and 12 are equal. Then the
> > contribution of these to a sum of 7, divided by
> > h^2, is b/f + f/b, and as
> > this is of the form x + 1/x, it is at least 2.
>
> I must confess to befuddlement.
> Your proof does indeed look correct.
>
> The problem originates at a lecture by
> Don Knuth I attended last month.
>
> appearance of pi in Sterling's approximation
> of the factorial function. This dice
> problem was the jumpoff point, then segue'd
> down various other paths, quite marvelously.
> He indicated the problem was difficult, and
> its refutation involved some advanced math.
>
> But your proof is almost trivial, which
> leaves me scratching my head.
>
> I will write to the perfessor, and post
> his reply here, if any.
Couldn't contact him, but Stanford has made
the lecture available:
http://scpd.stanford.edu/knuth/index.jsp
click "why pi?"
My Windows player choked on it.
What decoder do I need for this file type?
--
Paul T.
BartC
news:e13092e6-fc5d-41ad...@l7g2000vbv.googlegroups.com...
Seems to work with Windows Media Player (W7). (Although WMP doesn't work on
much else...)
But, I'm sure VideoLAN (VLC) will probably cope with it, if you can get a
download.
> click "why pi?"
Quite interesting actually, apart from the mild surprise of finding he was
still alive, and using Google (about as badly as me), and finding GUIs
fiddly (also like me).
--
Bartc
Rob Johnson
PT <ptane...@consultant.com> wrote:
>On Jan 6, PT <ptanenb...@consultant.com> wrote:
>> > >> > Toss two fair dice, they will sum to 2, 3, ... 12.
>> > >> > 7 is most likely, followed by 6 and 8, etc,
>> > >> > where all arithmetic is performed in fractions
>> > >> > of 36.
>> > >> > Your mission, Jim, should you decide to accept,
>> > >> > follows: weight a pair of dice, such that the 11
>> > >> > B, and weight (probabilistically, not gravity)
>> > >> > the 6 faces of each, appropriately.
>>
>> > It is impossible to load them so that, if the
>> > probabilities of 2 and 12 are equal, the probability
>> > of 7 is not at least twice as large.
>>
>> > Let the probabilities of 1 be hb and h/b, and that
>> > of 6 be hf and h/f; this must be the case if the
>> > contribution of these to a sum of 7, divided by
>> > h^2, is b/f + f/b, and as
>> > this is of the form x + 1/x, it is at least 2.
>>
>> I must confess to befuddlement.
>> Your proof does indeed look correct.
>>
>> The problem originates at a lecture by
>>
>> The lecture primarily discussed the
>> appearance of pi in Sterling's approximation
>> problem was the jumpoff point, then segue'd
>> down various other paths, quite marvelously.
>> He indicated the problem was difficult, and
>> its refutation involved some advanced math.
>>
>> But your proof is almost trivial, which
>> leaves me scratching my head.
>>
>> I will write to the perfessor, and post
>> his reply here, if any.
>
>Couldn't contact him, but Stanford has made
>the lecture available:
>http://scpd.stanford.edu/knuth/index.jsp
>
>click "why pi?"
>
>My Windows player choked on it.
>What decoder do I need for this file type?
The video format seems to be WMV. From the headers to your message,
I assume you are running Windows. This being the case, I am sort of
surprised that your player choked on it. What browser are you using
and what player plugins do you have? It works fine on my Mac using
the Flip4Mac plug-in.
Rob Johnson <r...@trash.whim.org>
take out the trash before replying
to view any ASCII art, display article in a monospaced font
Esra Sdrawkcab
couldn't find it - but this seems to be it:
http://stanford-online.stanford.edu/seminars/knuth/101206-knuth-500.asx
>
> My Windows player choked on it.
> What decoder do I need for this file type?
>
> --
> Paul T.
>
--
"Nuns! NUNS! Reverse! Reverse!"
falesia
>
>>Any other ideas?
>
> Any business that deals in cash. I know someone who made a *pile* of
> un-taxed money (kinda the reverse of the problem) with a sandwich
truck.
>
Someone I know here in the UK strongly suspects that her boss, a
solicitor (=lawyer) is involved in a carousel fraud. She said that even
a very small and unobtrusive one could easily get 50K (UK pounds) /
year.
We started idly chatting about it but never did anything about it, but
it doesn't seem too difficult.
http://www.moneyweek.com/personal-finance/how-carousel-fraud-is-
putting-the-vat-system-in-a-spin.aspx
--- news://freenews.netfront.net/ - complaints: ne...@netfront.net ---
Ostap Bender
> On 10-12-28 04:25 PM, Don Lancaster wrote:
>
> > Back when I first ran this dice project <
> >http://www.tinaja.com/glib/eldicepe.pdf>, a bunch of readers really got
> > bent out of shape when they did not realize that shooting two six sided
> > dice at once was exactly the same as shooting one thirty six sided dice.
>
> > "But they are locked together!"
>
> > Golly Gee mister science.
>
> Would that 36 sided dice have a '2' as the lowest number?
Evidently. He probably labeled the sides of his 36-sided die from 2 to
12, with the according frequency to make the outcomes look like those
for two 6-sided dice. This man is a walking genius... Too bad he
can't walk and think at the same time.