There are 5 pirates numbered 1 through 5

[Oblivion]

There are 5 pirates numbered 1 through 5. They have to divide 1,000 gold coins amongst themselves. Any proposal on how to divide the coins has to be passed by a majority, otherwise the pirate proposing it is thrown overboard. All the pirates are assumed to be greedy and intelligent . They have to make their proposals in the order 5,4,3,2,1. What should pirate #5 propose?

 

[dextercioby]

Let pirate labeled #4 choose before me

Daniel.

 

[Galileo]

This is a very tricky problem. I've made the assumption that, if a proposal will lead a pirate to an equal share whether he agrees or not, he will disagree as to throw the other pirate overboard. Since they are pirates after all, arrrr.

Here's my solution:


Pirate 5 has to maximize his profit, but wants to stay alive most of all. The idea is that the other pirates will only agree with him if they get more than what they would get if they threw number 5 overboard, in which case there are 4 pirates left and they start again. This suggests we reason backwards.

Suppose 1 and 2 are left. Then 2 can make any suggestion whatsoever, 1 will disagree, throw 2 overboard and take all the loot (even if the proposal is that 1 gets all, since he IS a pirate ).

Suppose 3 pirates are left. 2 will agree with three anyway, since he knows he's off worse (death) if 3 doesn't get a majority. 1 will disagree ofcourse, so 3 can suggest to keep all the money himself.

Suppose 4 pirates are left. 3 will disagree with him, since he can get all the loot if 4 is gone, so 3 has to make sure 1 and 2 agree. Therefore he should give 1 coin to pirate 2 and 1 coin to pirate 1.

Now we can deduce pirate 5's decisicion. He should get 2 pirates to agree with him. Pirate 3 doesn't get anything if 5 is overboard, so he should get 1 coin. Then pirate 1 or 2 should get 2 coins. 5 can keep the rest himself.

Conclusion: The proposal will be:
Pirate 5: 997 coins (agree) Pirate 5: 997 coins (agree)
Pirate 4: 0 coins (disagree) Pirate 4: 0 coins (disagree)
Pirate 3: 2 coins (agree) or Pirate 3: 2 coins (agree)
Pirate 2: 0 coins (disagree) Pirate 2: 1 coins (agree)
Pirate 1: 1 coin (agree) Pirate 1: 0 coins (disagree)

 

[dextercioby]

Sorry,Galileo,but doesn't make any sense to me how a pirate would agree to get less than he'd get by sharing the coins equally,200 each.Unless,there's somekind of a trick to throw the 2 (which don't agree) overboard...Still,1000/3 is not exact.

Daniel.

 

[xJuggleboy]

Here we go...

Knowing that pirate 1 will never agree to anything, as he will never have to worrie about being chucked and get it all. You should suggest that 2 other pirates say 3&4 get 500 each. They all vote.. 3&4 should like getting 500Gold and not getting thrown overboard, they agree and you agree. 3 to 2 is a majority... You get Nada... but live yay! =-D

There are lots of explanations which I have left out but that shuld be pritty close to right...

 

[vikasj007]

All the pirates are assumed to be greedy and intelligent .

intelligent or not, the pirates are greedy, and with all that gold in front of them what is the need to spend so much time on such solutions, specially when #5 is getting everything. they will just take out there swords/guns and decide it the old way...



(1 down)
(2 down)
(3 down)
(4 down)


the last pirate standing gets all the gold.

 

[Galileo]

Sorry,Galileo,but doesn't make any sense to me how a pirate would agree to get less than he'd get by sharing the coins equally,200 each.Unless,there's somekind of a trick to throw the 2 (which don't agree) overboard...Still,1000/3 is not exact.

Daniel.

No man, it makes perfect sense. From the way the problem is posted I figured the idea is the following:
Pirate 5 makes a proposal on how to divide the coins. If a majority agrees with him it will be done, if not number 5 becomes shark bait.
Then pirate 4 can make a proposal on how to divide the cash etc.

Although, for example pirate 3 agrees with 5 while he gets way less than a fair share, if he would disagree, 5 would go overboard and pirate 4 will make a proposal that will get him even less! That's the reason why he (the same reasoning applies to all of them) agrees with the proposal.

 

[Oblivion]

Here is the answer

Pirate #5 should propose the following distribution for (5,4,3,2,1):
997,0,1,0,2

Let us look at this problem in reverse order:

1. If there is only one pirate left, then he gets all the gold.

2. If there are two pirates left, then pirate 4 has to offer all the gold
to pirate 5 - because otherwise pirate 5 would simply refuse the offer,
and get all the gold anyway.

Therefore, in this case: Pirate 5 Pirate 4
1000 0

and the offer would be passed.

3. If there are three pirates left, pirate 3 would make the following
proposal:

Pirate 5 Pirate 4 Pirate 3
0 1 999

Pirate 5 is going to veto this but pirate 4 would have to agree because
otherwise he is going to get 0 gold pieces. So the offer would be passed
by pirates 4 and 3.

4. If there are four pirate left, pirate 2 would make the following
proposal:

Pirate 5 Pirate 4 Pirate 3 Pirate 2
1 2 0 997

Pirates 4 and 5 would agree because otherwise they would get less gold.
Pirate 3 would disagree, but the proposal would be passed by pirates 5, 4
and 2.

5. If there are 5 pirates left, pirate 1 would make the following
proposal:

Pirate 5 Pirate 4 Pirate 3 Pirate 2 Pirate 1
2 0 1 0 997

Pirates 5 and 3 would agree, and pirates 2 and 4 would disagree, so the
proposal would be passed by pirates 5, 3 and 1.