T.R | Title | User | Personal Name | Date | Lines |
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1955.1 | See 1777.* | HERON::BLOMBERG | Trapped inside the universe | Thu Mar 23 1995 07:51 | 1 |
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1955.2 | | HANNAH::OSMAN | see HANNAH::IGLOO$:[OSMAN]ERIC.VT240 | Mon Mar 27 1995 18:43 | 14 |
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I'm not sure I agree with your sentence "both are then satisfied".
I split up the beer. You then choose the larger one. I'm unhappy.
I know, you want to know *why* I created a larger one. Well, I didn't mean to.
It's just hard to split beer evenly.
Perhaps we should say that after I split up the beer, you have to even them
up and then I get to choose first.
In other words, I'm reopening the original 2-person question.
/Eric
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1955.3 | non-zero costs | CSSE::NEILSEN | Wally Neilsen-Steinhardt | Tue Mar 28 1995 13:26 | 17 |
| The usual formulation of these problems assumes that the cost of doing the split
and the cost of analyzing the split are both zero.
In the real world, the first assumption is often true, because you can do and
redo the split on paper. However, the second assumption may be false in some
cases. In a large financial deal, millions of dollars may be spent deciding the
value of a package of assets.
When the second assumption is false, the same procedure works, but the result is
only approximate. Person A does a preliminary analysis, then splits, then
analyzes the split, then resplits, continuing until the difference is smaller
than the cost of continuing. Person B analyzes the split until the cost of
further analysis is less than the perceived or more than the expected
difference. Then B chooses. A can be satisfied that B got about the same as A,
plus or minus the value of the last analysis that A did.
When the first assumption is false, I don't see a clean solution.
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1955.4 | Technology of splitting | FLOYD::YODER | MFY | Tue Mar 28 1995 15:20 | 14 |
| If the cost of splitting the loot is high, you can substitute splitting
probability, which can usually be presumed to be cheap. For example, suppose
that there is a single item whose cost of splitting is infinite, or equal to the
item's value: a vase, perhaps. Suppose further that transfer payments of
present or future value between the participants are somehow forbidden. It's
clear that (by symmetry) the solution that yields the best value must be to give
each participant a 1/n chance of getting the entire item. (This assumes they
are not so jealous that they would prefer a certainty of nobody getting it to
the 1/n chance of their getting it!)
Similarly, suppose two people are to split $2.01 in cash. We may reasonably
assume that utility is nearly enough linear in money on the scale of pennies.
So giving each a 1/2 chance of the extra penny splits the value of that penny,
whereas physically splitting it would cost nearly the entire value of the penny.
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1955.5 | | HANNAH::OSMAN | see HANNAH::IGLOO$:[OSMAN]ERIC.VT240 | Wed Mar 29 1995 14:37 | 3 |
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...reminds me of the story of wise King Solomon and the 2 mothers. Anyone
else remember that one ?
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1955.6 | Re: Solomon & 2 mothers | FLOYD::YODER | MFY | Fri Mar 31 1995 11:50 | 8 |
| Of course! Solomon had to divide 2 mothers and a kid with one of his courtiers.
Dividing the mothers was easy, but the kid was a problem, so he ended by calling
for a sword. But one of the mothers pointed out that the kid had eaten three
pomegranate seeds, which the swordsman would have to divide evenly too. This
seemed overly difficult, so the king wisely decided the kid would have to spend
six months of the year with one mother, and six with the other.
Or something like that.
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1955.7 | The don't do justice like that any more | SUBURB::STRANGEWAYS | Andy Strangeways@REO DTN 830-3216 | Fri Mar 31 1995 12:03 | 13 |
| Version I heard was that 2 mothers both claimed the kid and took the
case to court for custody rights. Solomon noted that the situation and
the evidence were copmletely symmetrical between the two mums so
mathematical esthetics demanded a symmetrical solution. He called for
the sword, and one of the mums said "Aaaaargh! No! Let her have the kid
rather than that." So Solomon figured she was the real mum and gave her
the child.
There's a three player non-zero sum game in there somewhere.
Andy.
BTW, I was born in 1955.7 what odds on that coincidence?
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1955.10 | | RTL::GILBERT | | Fri Mar 31 1995 15:02 | 1 |
| I was born in 1955.10. Another coincidence!
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1955.11 | Re: .7 | FLOYD::YODER | MFY | Fri Mar 31 1995 17:23 | 14 |
| Clearly, if Solomon had known modern mathematics, he would have let one mom
divide, and the other choose...
More seriously, I recall reading in my childhood a set of stories about some
famous judge (the term "judge" isn't quite accurate but will do; my recollection
is that the stories came from Chinese culture) who was put in exactly the same
situation as Solomon, and came up with the same solution. Unfortunately, the
false mom was cleverer in this story (or already knew the trick) and both moms
immediately clamored to give the kid to the other rather than kill the child.
So he had to concoct another method, which I don't recall, to tell which was the
real mother.
Can anyone else remember reading a set of stories like this, and supply more
details?
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1955.12 | | HANNAH::OSMAN | see HANNAH::IGLOO$:[OSMAN]ERIC.VT240 | Fri Mar 31 1995 18:39 | 9 |
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Anyone remember a puzzle about how do you divide 17 horses amongst 3 sons ? The answer
was something like borrow a horse from the neighbor so you have a divisible number,
divide them up, then such-and-such is conveniently now true so you return the horse
to the neighbor and you're done ?
What's the part I'm forgetting ?
/Eric
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1955.13 | .12 reformatted into 80-column mode | EVMS::HALLYB | Fish have no concept of fire | Mon Apr 03 1995 09:07 | 13 |
| Anyone remember a puzzle about how do you divide 17 horses amongst 3
sons ? The answer was something like borrow a horse from the neighbor
so you have a divisible number, divide them up, then such-and-such is
conveniently now true so you return the horse to the neighbor and
you're done ?
> What's the part I'm forgetting ?
The part about keeping your entries to 80 columns or less :-)
(Actually I remember the puzzle too, but no more than what is cited above)
John
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1955.14 | 1/2, 1/3, 1/9 ? | HERON::BLOMBERG | Trapped inside the universe | Mon Apr 03 1995 09:28 | 3 |
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Wasn't it that the sons should get 1/2, 1/3 and 1/9 of the horses?
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1955.15 | | HANNAH::OSMAN | see HANNAH::IGLOO$:[OSMAN]ERIC.VT240 | Mon Apr 03 1995 10:04 | 18 |
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01234567890123456789012345678901234567890123456789012345678901234567890123456789
Yes, that was it, thanks. I'll try to keep this reply to 80 columns. I'm
on decwindows notes so I must have dragged the window size to something
large before.
Yes, the problem was he had to divide his 17 horses amoungts his 3 sons,
in the amounts 1/2 1/3 1/9. So he borrowed a neighbor's horse to make 18.
He gave out 9 horses, 6 horses, and 2 horses, then returned the neighbor's
horse !
/Eric
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1955.16 | one of a class | CSSE::NEILSEN | Wally Neilsen-Steinhardt | Mon Apr 03 1995 14:05 | 9 |
| I heard this as an inheritance problem, in which the will specified the
fractions. The three sons were quarreling when a dervish wandered by with an
extra horse and the solution.
I've heard this told so that the dervish makes a profit on the deal. I can't
remember the math.
I've also heard a variation with coconuts and a monkey. Was that in this
conference?
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1955.17 | | HANNAH::OSMAN | see HANNAH::IGLOO$:[OSMAN]ERIC.VT240 | Mon Apr 03 1995 16:02 | 18 |
| I can tell you the coconut and monkey one, and someone else can tell
you what note it's in (and whether it's in this conf or the brain
bogglers one)
Anyway, 5 sailors on an island collect coconuts with intention to then
sleep and divide up the lot in the morning.
Each of the 5 sailors wakes up in the middle of the night, and not
trusting his buddies, he takes his 1/5 of the coconuts and hides them,
except that there's always 1 coconut left over, which each sailor
gives to a monkey.
In the morning, they divide the remaining coconuts, and of course
there's again 1 left over for the monkey.
How many did they start with ?
/Eric
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1955.18 | 1094 | EVMS::HALLYB | Anything you can do, you can do better | Mon Apr 03 1995 16:16 | 1 |
| It's note 1094 where they discuss "cocoanuts" and monkeys.
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1955.19 | | CSC32::D_DERAMO | Dan D'Eramo, Customer Support Center | Mon Apr 03 1995 16:33 | 18 |
| In 1094 the monkey didn't get another coconut in the
morning.
>It's note 1094 where they discuss "cocoanuts" and monkeys.
The dictionary does give "cocoanuts" as an alternative
spelling of "coconut" so that is okay. But some titles
in this notes conference are so much fun. I'll never
forget how I searched and searched before finding the
topic on how freqently a calendar month has two full
moons in it:
Notes> dir/title="you saw me standing alone"
:-)
Dan
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1955.20 | March '95 had two NEW moons | EVMS::HALLYB | Anything you can do, you can do better | Tue Apr 04 1995 11:17 | 9 |
| > topic on how freqently a calendar month has two full
> moons in it:
>
> Notes> dir/title="you saw me standing alone"
Obviously a deficiency in NOTES. Clearly we need two type of titles:
a "display" title and a "reference" title.
Anyhow, methinks any nuts'n'monkeys puzzles oughta go in 1094.
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1955.21 | | HANNAH::OSMAN | see HANNAH::IGLOO$:[OSMAN]ERIC.VT240 | Tue Apr 04 1995 14:58 | 4 |
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Would a month with 2 full moons tend to follow one with 2 new moons ?
/Eric
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1955.22 | | EVMS::HALLYB | Anything you can do, you can do better | Tue Apr 04 1995 16:57 | 8 |
| > Would a month with 2 full moons tend to follow one with 2 new moons ?
No. With 2 new moons in a month the full moon is around the 15th of
that month. Given a period averaging 29.53053 days between full moons,
the full moon the following month would be around the 13th-15th.
Exactly the opposite of what you want, i.e., the full moon on the 1st.
John
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1955.23 | split a topic 23 ways | HERON::BUCHANAN | Et tout sera bien et | Sat Apr 15 1995 08:10 | 6 |
| I love this note. It's like concept association football.
The infant bisection problem cropped up in Cervantes' Don Quixote
as well, didn't it?
Andrew.
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