# Off Topic > The Water Cooler >  >  OT #2 -- week ending Sun 20-Dec-09

## darkyam

OK, I guess since no one else has posted one of these for this week so far, I get to.  :Smilie: 

Topics for this week are:
1. Post your favorite riddle.  It should be a riddle that actually has a logical solution and not just a paradox, like, "Can God make a rock so big He can't lift it?"

2. What's the funniest thing that happened to you this year?

I'll start off with #1.  This isn't the hardest riddle I've heard, but it is still my favorite.  

There are five pirates, ranked in order from captain (#1) to new recruit (#5).  They find a chest with 1,000 gold coins.  According to their code, they each take turns proposing a way to split the treasure.  #5 goes first and then they vote on his proposal.  If they agree with a simple majority (including #5's vote), then they split the treasure according to the plan.  Otherwise, they slit his throat and #4 presents, and so on.  

All of the pirates have several things in common: they are perfectly logical, they are incredibly greedy, and they will not kill if it gains them nothing.  In other words, they'll gladly vote to kill a pirate if it means a single gold coin more for them, but won't if the offer they're made is the best they can get.

What should #5's proposal be to both keep him alive and get him as much of the treasure as possible, if any?

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## martindwilson

hmm the old pirates riddle spawning myriads of internet threads. the main flaw is if that were really smart/logical they would never propose this system in the first place!

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## pike

Maybe one thrid share for prirates 3,4 and himself?

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## darkyam

That is one possibility, and it will keep him alive, but it is not the optimal solution.

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## pike

changed my mind 1/3 won't do it 3,4,5 split may have to be 50 30 20 and he stays alive

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## pike

But fours dead so 50 , 0 50

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## darkyam

Getting warmer, Pike, but not there just yet.

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## pike

0,0,0,0,100       as  3 ia also as good as dead

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## darkyam

Right answer, dubious logic.  Start from the captain.  If it gets to #2, #2 has to give it all to the captain, which means that #2 can't possibly get a single coin and so would vote yes on any plan.  #3 knows this (being perfectly logical), so if it ever got to #3, #3 could claim all the treasure.  He'd have his vote and #2's vote.  The captain, knowing that #2 will vote for #3's plan if it got that far, realizes that he can't have any of the treasure, either, and so will vote for any plan.

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## teylyn

Is that like the "prisoner's dilemma"?? Always good as an ice-breaker excercise in a big group of people who don't know each other and have to work together for a stretch of time.

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## teylyn

It's a long time since I touched a PC magazine. This weekend I did. It will be a long time until I do again. 

Took the boys to the library. They practically *live* there and it's not a case of "in - grab a few books - check out - leave". They need *time*, big time.  So, stuck in the library for a couple of hours. 

How to kill the time. Lib'ry catalog on the 'puter. One Walkenbach book. Office 2000 or some such, published 1998 or so. Right. Provincial NZ. Next. Stephen Few. Not found. Edward Tufte. Nada. Andy Pope. Not published.

OK.

On to the magazine rack. Browse around the current PC mags. Not much choice. Some NZ'ised titles of Oz stuff. Leafed through one. 

Used to be that in PC magazines you could actually find information about PCs and applications that run on them. No more. 
- iPhone
- digital Cameras
- home theatre systems
- external usb hard drives for backing up photos and songs (ok, that one half way qualifies)
- more iPhone
- D_VD players and stuff
- more iPhone
- other mobile phones
- virus hypes
- more iPhone

Not a single PC application or tricks or usability hints in the whole mag. 

Useless.

Went to the third floor internet pool and spent 2 hours in Excelforum.

Ahhhhh.

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## darkyam

What's the prisoners' dilemma?  I know one about prisoners that involves turning on and off a light.  Is that the one you're thinking of?  If not (or even if so, since that's a good riddle, too), could you please post?

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## DonkeyOte

http://en.wikipedia.org/wiki/Prisoner's_dilemma

As I recall my economics lecturer used this when starting discussions on cartel / oligopoly etc... (ie OPEC!)

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## sweep

A man is in a boat in a pool.  In the boat there is a rock.  He rows out to the centre of the pool, and throws the rock overboard.  It (obviously) sinks.  

What happens to the level of the pool?

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## darkyam

What lake?  He's in a pool.

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## sweep

No, it doesn't!

When in the boat the rock is displacing the level of the lake due to its mass, when it's thrown in, it's displacing by volume, so the level of the lake drops, albeit very slightly.

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## martindwilson

id go along with that

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## pike

A dog runs into a heavyily timbered forest. How far can it run into it?

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## darkyam

Why do I sense this is a bad pun and not a riddle?

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## teylyn

.... because you're clever and the challenger is pike, maybe?? <duck>

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## darkyam

Of course, if it is a real riddle, one answer could be that he can only go as far into it as he is long, because once he's in it, if he continues to run, he's getting closer to running out of the forest on the other side.

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## rwgrietveld

Darkyam,

Continuing on your quest I would say "till he is in the middle of the forest", but this is completely dependant on the shape of the forest.

I think you should keep away from specifying a unit such as metres [m].

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## rwgrietveld

The one I always like is the following:

Asume you would put a rope around the world (tightly).
Now you would like to lift the rope 1 meter [m] above the ground. What is the new length of the rope?

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## darkyam

If it's the same rope, it's the same length.  Ropes don't magically grow when you lift them.  :Wink:

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## rwgrietveld

Very smart: How much would you need to add to THAT rope to be able to lift it 1 metre of the ground?

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## darkyam

Assuming you mean 1 meter all around the world, that's easy: 2m*pi.  You're just increasing the diameter of a circle by two meters, so the circumference increases by that *pi.

If you mean you as an individual lifting it, then less than 2m, because it goes 1m up and 1m back down, but you'd be dealing with essentially a hypotenuse at that point, so you'd have to add very little.

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## ChemistB

A farmer has 4 sheep.  One day he notices that the 4 sheep are all standing equal distance from each other.  That is, that the distance between any 2 sheep is the same as any other two sheep.  How is this so?

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## ChemistB

At a family reunion, there are represented a mother, father, uncle, aunt, son, daughter, cousin, nephew, niece, brother and sister.  What's the smallest amount of individuals that can be at that reunion?  Explain. (no incestial relationships or gender change operations)

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## rwgrietveld

About the sheep: if it were three sheep they must be in a triangle.

Adding a fouth, one must go 3D, so somehow the fourth sheep must have climbed on the wagon and they form some sort of piramid.

The shape is like putting 3 balls against each other and putting one on top.

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## rwgrietveld

... and the family. It must be the brother and the sister who are all the others (mother, father, uncle, aunt, son, daughter, cousin, nephew, niece) to someone else.

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## darkyam

There are 100 pirates on an island.  Half have blue eyes, half have brown eyes.  All of these pirates have two things in common: they live by the same code and they are perfectly logical (save for living by the code  :Wink:  ).  The code states that all brown-eyed pirates must throw themselves into the island's volcano at midnight if they know they're brown-eyed.  However, no pirate is allowed to tell another pirate what color their eyes are (through speech, sounds, signals, writing, anything at all) and no pirate is allowed to look at anything reflective to determine their own eye color.  Things are fine for years because no one can tell for certain whether they themselves have brown eyes, although everyone can see everyone else's eye color.

One day a parrot flies to the island, takes a look around, and says, "At least one of you has brown eyes."  Then the parrot flies away.  What happens with the pirates?

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## teylyn

On that families topic: Way back in my family tree

John is a widower with children, one of them is Sue, grown up
Jack is a widower with children, one of them is Sarah, grown up

John marries Jack's daughter Sarah
Jack marries John's daughter Sue

Both marriages are blessed with additional children. 

I've spent ages sorting out who is his own uncle or her own mother in law.

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## martindwilson

since pirates are perfectly logical and stick to the code no one would have said "at least one of you has brown eyes" in the hearing of the parrot for it to mimic . therefore the parrot must have overheard it somewhere else and so the pirates can quite happily ignore it!

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## darkyam

Nice try, Martin, but the parrot is merely a convention in here.  It could be a traveler that comes ashore for a second and leaves, a wizard that appears and vanishes, you beaming yourself down as in Star Trek, etc.

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## pike

if one has brown eyes - every one know who that is 
Every knows the priate with brown eye because 
He/She can only find blue eyed priates or one brown eyed priate

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## darkyam

A start...but not an end...

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## teylyn

There's only one pirate who sees that all the others have blue eyes. So that night that pirate will throw himself into the crater.

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## darkyam

You seem to be misreading it. I said half the pirates are brown-eyed and half are blue-eyed.

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## teylyn

yup ... was a bit confused ... going back into my cage now

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## pike

Lets look at this
Whey can count all the priate eye colours 
so they know that it is close too fifty/fifty or do they know it fifty fifty? 
if they know its fifty fifty  then they know their own eye colour 
If they dont know its fifty fifty then hmmm.

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## darkyam

They don't know it's fifty/fifty precisely because they don't know their own eye color.

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## shg

> One day a parrot flies to the island, takes a look around, and says, "At least one of you has brown eyes."



I don't see how the parrot added any information unless he was heard only by a group containing a single brown-eye (who would then know he had a hot date later in the evening). Everyone knows there are at least 49 brown-eyes.

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## pike

i dont know, they all be come paranoid and jump into volcano

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## teylyn

maybe one of them gets wise and starts to count the blue eyes. If he finds 50 pairs of blue eyes it means that he himself has brown eyes --- off to the crater.

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## pike

That’s been thought of teylyn  ~ they don’t know that its fifty fifty 
I going for every pirate becomes mad thinking logically and they all jump in

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## ConneXionLost

I agree with shg on this.  Every pirate sees at least one other pirate with brown eyes.  There is no new information added by the parrot; therefore nothing happens.

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## rwgrietveld

> There are 100 pirates on an island. Half have blue eyes, half have brown eyes



If this is a given then we must say it is obvious if they can count.

I think the pirates don't know the fraction, but as the fraction is more or less 50/50 the parot does not bring any new information either ???

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## darkyam

> I going for every pirate becomes mad thinking logically and they all jump in



LOL, Pike!  I promise that there is a logical answer.  And if no one solves it, I will post the solution Saturday evening.

The parrot does not bring _new_ information, but it is necessary to get the ball rolling (or the conga line to the mouth of the volcano started).

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## ChemistB

> About the sheep: if it were three sheep they must be in a triangle.
> Adding a Fourth, one must go 3D, so somehow the fourth sheep must have climbed on the wagon and they form some sort of pyramid.



Very good.  Three were forming an equilateral triangle and there was a mound or small hill in the center upon which the 4th was perched equal distance from the other three.




> ... and the family. It must be the brother and the sister who are all the others (mother, father, uncle, aunt, son, daughter, cousin, nephew, niece) to someone else.



I think I need to be clearer.  Someone at the reunion could use one of the above labels to refer to another at the reunion. (e.g. person 1 could refer to another person there as "sister"  person 2 could refer to another person there as "cousin").

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## pike

Hi darkyam
Its saturday

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## darkyam

Hi, Pike, not where I am.  It's only Friday morning.

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## ChemistB

Concerning the pirates; if there are 100 pirates and 50% are blue eyed and 50% are brown eyed, then every pirate on the island knows that there is at least 1 pirate that has brown eyes.  I can't see that the parrot is bringing any new information to the group.   :Confused:   I guess I'll wait till Saturday.

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## martindwilson

you're both wrong its now friday afternoon! GMT /Zulu thats it ! no argument
"the waves  Britannia rules" rearrange to make the appropriate statement lol

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## teylyn

I vote that it's Saturday now, too. 2:20 am at that! So, part with it, will you? Can't keep me in sleepless suspense for another night!! <no, I won't tell the rest of you what the sleepless suspense for the other night was!>

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## teylyn

Pike, d'you realise when the aardvark says "saturday evening" it'll be actually Sunday morning for us? Ohmyloooord, the suspense ---- Schnaps!

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## darkyam

Martin, hate to break it to you, but the old British slogan, "The sun never sets on the British Empire," hasn't been true for years now.  

How about a compromise?  If I remember, I'll post tonight before I go to bed.

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## pike

Might as well compromise 50% of the poms live here anyway
I'm still going for they must all DIE. It's the only lodgical end

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## martindwilson

i'm sitting 2 miles away from the Prime meridian  , its in the name aint it!
anyway the empire was a mistake we only went looking for prime places to play cricket the rest was an accident.

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## darkyam

By that logic, I'm sitting a half mile from a TGIFriday's, so it must always be Friday.

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## martindwilson

well your right at this precise time then !

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## teylyn

It's Sunday. Here and now. Even in Australia, well, almost. Can't wait forever for the rest of the world to catch up. 

*Darkyam* ???

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## darkyam

Mea culpa for not posting last night.  In my defense, a blizzard canceled my flight, so I spent the day driving nearly 1500 km, nearly 500 of it in a snowstorm and fighting bad roads.

For the answer, if there are two pirates, and one is brown-eyed and one is blue-eyed, the brown-eyed pirate would know he is brown-eyed after the parrot said something.  Before the parrot said anything, he would not know he had brown eyes, hence the necessity of the parrot: to get the ball rolling.  If both were brown-eyed, they would wait one night, expecting the other to throw themselves in.  When the other didn't, they would know they both had brown eyes and so would both throw themselves in the second night.  You would just extrapolate that logic out to fifty nights for the fifty pirates, so they all fifty brown-eyed ones throw themselves in on the fiftieth night.

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## pike

I dont think thats quite right as there are 100 pirates not two
every pirate knows half a dozzen brown or blue eyed pirates
time to google

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## pike

Part answer 

There's an inductive solution. I think there is a fundamental common sense error to this solution. But, since these these being are so buried in their own logic. They probably won;t see it. Any how it works like this.

There are 100 people on the island with either blue or brown eyes.

If only one had brown eyes then he would have concluded from the elder is own color. So, he would have left. Since, no one leaves, everyone on the island assumes there is at least two people with brown eyes.

If we assume that N-people with brown eyes would leave on day N, then on day N+1 would could conclusion that the number of brown eysers is at least N.

Anyone with browns eyes counts 99 pairs of Brown eyes. So, when no leaves on day 99, all the brown eyesers conclude their own eye-color and depart on day 100. The blue eyesers pack and leave the next day.

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## darkyam

> If we assume that N-people with brown eyes would leave on day N, then on day N+1 would could conclusion that the number of brown eysers is at least N.



Huh?

Anyway, it *sounds* as if you're just re-explaining the logic.  I never said there were only two pirates with brown-eyes.  I was saying, if there were, neither would know their own color on the first day and both would know it on the second.  Extrapolating that logic over 50 pirates means that on the 49th day, none of them knows their own eye color, but on the 50th, every brown-eyed pirate knows their own eye color.

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## martindwilson

to much like a crystal gale song if you ask me lol

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