# Thread: Word problems for fun!

1. 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?

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

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

4. 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.

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

6. Originally Posted by Red Herring
...and where do I get half a pirate?
Rofl. Most hilarious response.

7. Originally Posted by Red Herring
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.

8. @Johnny

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.

9. Are you taking the LSAT or something?

10. Originally Posted by onemoretime
Are you taking the LSAT or something?
I used to think I would :p but no.

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