[Other] Word problems for fun!

[Red Herring]

I really have to get back to work, but want a shot at this one later on.

A short question though beforehand: In the theoretical case that the highest rankimg pirate does get killed, how do the remaining 4 (or any even number of) pirates determine a majority?

 

[Jonny]

Even - n/2 + 1
Odd - n/2 + .5

 

[Red Herring]

...and where do I get half a pirate?

 

[Jonny]

lol, you don't. I just mean that when there are an odd number of pirates, say 5, the majority is n/2 + .5 → 5/2 + .5 → 2.5 + .5 → 3; and when there are an even number of pirates, say 4, the majority is n/2 + 1 → 4/2 + 1 → 2 + 1 → 3.

 

[Red Herring]

So the proposal is considered rejected in case of a 2:2 stalemate?

 

[esidebill]

...and where do I get half a pirate?

Rofl. Most hilarious response.

 

[Jonny]

So the proposal is considered rejected in case of a 2:2 stalemate?

Precisely. It is accepted if and only if a majority of pirates vote for it; rejected otherwise.

 

[matmos]

@Johnny

My bad - I totally misread your OP. Apologies, good sir.

If I understand your take, the results are:

Pirate 5: 97
Pirate 4: 0
Pirate 3: 1
Pirate 2: 0
Pirate 1: 2

However, there's a problem. We assume that all 5 agents are "perfectly rational". Therefore all five agents would be able to independently work out the mathematical equilibrium you posit. Agents 1-4 would be anxious to avoid this scenario as the possibility of more gold for each is available. This would be perfectly rational.

All know this is the likely outcome - because all are brilliant game theorists. Therefore the perfectly rational approach would be for pirates 1 - 4 to feign irrationality and demand nothing less than a fair split (20 coins is significantly better than zero, one or two) at the cost of killing pirate 5 and obtaining a larger, equal, split between the remaining four pirates.

Pirates 1-4, being perfectly rational, realise that Pirate 5 will realise this and, rather than risk rejection, accept an equal cut or a bargained offer, depending on how highly rank is enforced.

So a "rational" agent at best can hope for zero, one or two coins, but an "irrational" agent can obtain 20 or even 25. - Some rationality!

The rational approach would be to forget the rational approach.

 

[onemoretime]

Are you taking the LSAT or something?

 

[esidebill]

Are you taking the LSAT or something?

I used to think I would but no.

 

[esidebill]

Harry: Airlines have made it possible for anyone to
travel around the world in much less time than
was formerly possible.
Judith: That is not true. Many flights are too expensive
for all but the rich.
Judith’s response shows that she interprets Harry’s
statement to imply that

(A) the majority of people are rich

(B) everyone has an equal right to experience world
travel

(C) world travel is only possible via routes serviced
by airlines

(D) most forms of world travel are not affordable
for most people

(E) anyone can travel

 

[onemoretime]

I used to think I would but no.

You are a lot smarter than I am. Rest assured in that.

 

[Little Linguist]

E. Now gimme another one, please. ;D

 

[Red Herring]

What's with the reading comprehension tests? Sounds like the PISA study*

*which, by the way, showed that the kids in my country have pretty bad reading skills.

 

[Jonny]

Prove that there are infinitely many twin primes.

 

[Little Linguist]

Prove that there are infinitely many twin primes.

Oh my god, not more numbers.

 

[Red Herring]

Oh my god, not more numbers.

Agreed.

But relax, he is kidding.

Counter question to Jonnyboy: is there always at least one prime number between n2 and (n + 1)2 ?
(n being a natural number)

 

[Jonny]

There is not. Consider the specific case of n=4. In this case, our range (2n, 2n+2) is defined by (8, 10). Since the only whole number to fall within that range is 9, and since 9 is not a prime number, there is not always at least a prime number between 2n and 2n+2.

Did you mean n^2 and (n+1)^2?

 

[esidebill]

There is not. Consider the specific case of n=4. In this case, our range (2n, 2n+2) is defined by (8, 10). Since the only whole number to fall within that range is 9, and since 9 is not a prime number, there is not always at least a prime number between 2n and 2n+2.

Did you mean n^2 and (n+1)^2?

I remember this is comp. sci. theory >.>

 

[Red Herring]

It's n squared, not two n! Sorry if that wasn't readable.
So in your example of n=4 we would have a range from 16 to 25, that gives us even three prime numbers: 17, 19 and 23!

It's Legendre's conjecture. I was just kidding because you (jokingly, I assume) threw the twin prime thing at us which according to a quick research is one of the remaining mysteries of number theory.