Can Greedy Pirates Outsmart Each Other? Find Out!

[Raghav Gupta]

The Puzzle:
5 Pirates
5 pirates of different ages have a treasure of 100 gold coins.

On their ship, they decide to split the coins using this scheme:

The oldest pirate proposes how to share the coins, and ALL pirates (including the oldest) vote for or against it.

If 50% or more of the pirates vote for it, then the coins will be shared that way. Otherwise, the pirate proposing the scheme will be thrown overboard, and the process is repeated with the pirates that remain.

As pirates tend to be a bloodthirsty bunch, if a pirate would get the same number of coins if he voted for or against a proposal, he will vote against so that the pirate who proposed the plan will be thrown overboard.

Assuming that all 5 pirates are intelligent, rational, greedy, and do not wish to die, (and are rather good at math for pirates) what will happen?

answer it.......

 

[Enigman]

Done, I think.

Spoiler

Let the pirates be 5); 4); 3); 2); 1):: 5) being the oldest
Proposal by 5)
5) - 96
4) - 0
3) - 0
2) - 3
1) - 1

Case A deciding votes 2) and 1) say yes.
Case B
2) or 1) says no, 5 thrown out.
Proposal by 4
4) - 98
3) - 0
2) - 2
1) - 0

Case B.a deciding vote 2) agrees
Case B.b deciding vote 2) disagrees
4) thrown out
Proposal by 3
3) - 99
2) - 0
1) - 1
Deciding vote : 1 has to agree or he gets nothing.

Because proposal 3 gives 2) nothing (s)he has to agree to proposal 4.
proposal 4 > 2) gets 2 coins 1) gets none
:. 1) needs to agree to proposal 5
also proposal 5 gives 2) three coins which is the highest he can get.

therefore
5) - 96
4) - 0
3) - 0
2) - 3
1) - 1

 

[Raghav Gupta]

Why not 98:0:1:0:1? Since the pirate is greedy.

 

[micromass]

If I only get 1 gold coin, and somebody else gets 98, then I would vote no on principle.

 

[gmax137]

If I only get 1 gold coin, and somebody else gets 98, then I would vote no on principle.

Really, me too. If the proposer gets "too greedy" the others will vote him out. Why wouldn't pirate (5) (the eldest) propose: 34 coins for (5), 33 for (4) and 33 for (3)? The vote passes with (2) and (1) voting against; and number (5) gets 34 instead of zero. Can (4) and (3) prove that they will do better than 33 coins by voting (5) off?

 

[micromass]

Really, me too. If the proposer gets "too greedy" the others will vote him out. Why wouldn't pirate (5) (the eldest) propose: 34 coins for (5), 33 for (4) and 33 for (3)? The vote passes with (2) and (1) voting against; and number (5) gets 34 instead of zero. Can (4) and (3) prove that they will do better than 33 coins by voting (5) off?

I guess an interesting problem would be how much pirate (5) can get away with. I think almost everybody would vote him out if he proposes 98, while an agreement of 34 would be acceptable for most. So what is the maximum he can get out of this game. This isn't logically solvable, but there has been research on it: https://en.wikipedia.org/wiki/Ultimatum_game

 

[Enigman]

Why not 98:0:1:0:1? Since the pirate is greedy.

Yes, my mistake. I did that to counteract proposal 3 but I forgot that proposal 3 wouldn't come around as proposal by 4) would be forced by 2).

But it seems micromass is planing a mutiny?!?

 

[Raghav Gupta]

Here is my logic,
In combination 98:0:1:0:1 the 2nd and 4th pirate would definitely disagree but if 3rd disagrees, then we have a bad scenario for 3rd .
Cause then we going to have
1st gone overboard : 99:0:1:0

 

[Enigman]

Really, me too. If the proposer gets "too greedy" the others will vote him out. Why wouldn't pirate (5) (the eldest) propose: 34 coins for (5), 33 for (4) and 33 for (3)? The vote passes with (2) and (1) voting against; and number (5) gets 34 instead of zero. Can (4) and (3) prove that they will do better than 33 coins by voting (5) off?

Taking it from the back and beginning at the end
#2 and #1 100 : 0
.'. #1 agrees to previous proposal if it gives him > 0 coin
so #3 #2 #1 99:0:1
#2 gets nothing, so he has to agree to the previous proposal if it gives him something.
#4 #3 #2 #1 99:0:1:0
#3 and #1 need to prevent this. So, they will agree to the previous proposal if it gives them something
So #5 #4 #3 #2 #1 we get 98:0:1:0:1But by your argument,
#5 #4 #3 #2 #1 34 : 33 : 33 : 0 : 0
#4 may say no and say
#4 #3 #2 #1 90 : 0 : 10: 0 (90:10 is just an example #2 should accept it even if his share goes up to 1 as after this he gets nothing as #1 is forced to accept even at 1 coin in the next proposal.)

 

[Enigman]

If I only get 1 gold coin, and somebody else gets 98, then I would vote no on principle.

I wouldn't risk it.

 

[Raghav Gupta]

If I only get 1 gold coin, and somebody else gets 98, then I would vote no on principle.

Well this is a theoretical question.

I wouldn't risk it.

It practically also makes a bit sense.

 

[reenmachine]

Here's a twist, what if we change the rule such that if 50% (or more) pirates refuse the deal, then the pirate is thrown overboard, can you tell me what happens then?

We can also rank the priorities of the pirates from top priority to last priority:

Situation A

1)Survival
2)Greed
3)Blood Thirst

Situation B

1)Survival
2)Blood Thirst
3)Greed

 

[Enigman]

Well, let's see.

Spoiler

Supposing it gets to the point where #2 is giving the proposal.
proposal by #2 [#2 #1 :: 0:100] #1 gets #2 killed because of blood thirst.
#2 needs to agree to previous proposal by #3 if he wants to survive.
#1 will say no to proposal by #3 as this way he gets it all.
proposal by #3 [#3 #2 #1 :: 100:0:0] (#3 gets away with this because #2 HAS to agree to this if he wants to survive.)
#2 and #1 get nothing and therefore will try to preempt this proposal by agreeing to proposal by #4 while #3 in a bid to get it all will disagree with proposal by #4
proposal by #4 [#4 #3 #2 #1 :: 98:0:1:1] as #3 disagrees regardless of money and #2 and #1 will take anything more than 0.
proposal by #5 needs two votes to be passed except that of #5
as the endgame for #3 is at 0 coins and #2 is at 1 coin if they get to proposal by #4, #5 will propose a division like-
97:0:1:2:0

 

[reenmachine]

Your solution looks to be the best one , good job.

 

[collinsmark]

If I only get 1 gold coin, and somebody else gets 98, then I would vote no on principle.

While I like your way of thinking, the pirates -- albeit all completely logical -- are also completely greedy. Their bloodthirstiness is only surpassed by their greed. (And nothing surpasses their logic.)

The correct answer, as already mentioned, is

Spoiler

98:0:1:0:1

 

[Czcibor]

I guess an interesting problem would be how much pirate (5) can get away with. I think almost everybody would vote him out if he proposes 98, while an agreement of 34 would be acceptable for most. So what is the maximum he can get out of this game. This isn't logically solvable, but there has been research on it: https://en.wikipedia.org/wiki/Ultimatum_game

I'm not convinced that it is so related to ultimatum game. After all here is a possibility to throw overboard person making ultimatum and hope for better deal in next round ;)

While I like your way of thinking, the pirates -- albeit all completely logical -- are also completely greedy. Their bloodthirstiness is only surpassed by their greed. (And nothing surpasses their logic.)

The correct answer, as already mentioned, is

Spoiler

98:0:1:0:1

Wouldn't in such case be more rational to throw such person overboard and take chances in next split to get much more than 1? Throwing overboard 5, means that 4 has to make at least one pirate really happy, he may pick any. If he fails to do so, then he would share the fate of 5.
If the quantity 4 would have to use to save his life would be greater than 2, then such move would be rational.

There is also one more thing that bugs me. Shouldn't pirate with low numbers actually try to throw overboard anyone giving them even an equal deal? I mean reducing the number of players could increase the share per head.

 

[collinsmark]

I'm not convinced that it is so related to ultimatum game. After all here is a possibility to throw overboard person making ultimatum and hope for better deal in next round ;)

Wouldn't in such case be more rational to throw such person overboard and take chances in next split to get much more than 1?

When applying logic, and knowing that all pirates are completely logical, there's no reason to rely on chance at all. The results are certain.

Throwing overboard 5, means that 4 has to make at least one pirate really happy, he may pick any. If he fails to do so, then he would share the fate of 5.
If the quantity 4 would have to use to save his life would be greater than 2, then such move would be rational.

There is also one more thing that bugs me. Shouldn't pirate with low numbers actually try to throw overboard anyone giving them even an equal deal? I mean reducing the number of players could increase the share per head.

As a guide to determine the results, I suggest @Enigman post in post #9, where we start with just 2 pirates and move up from there.

But first we must start with the initial assumptions that all pirates are completely logical above all else, are all greedy, and then if in doubt, are all bloodthirsty.

If there are only two pirates, and you are the oldest pirate, you can keep all 100 coins for yourself. It doesn't matter if the younger pirate votes to throw you overboard. If he does, there still is not the > 50% vote needed to throw you overboard so you get to stay alive and keep all the coins. I should point out though that the necessary vote is greater than fifty percent, not greater than or equal to 50%. Had the condition been \geq 50% instead, it would change the answer to the riddle considerably.

Now add in a third pirate. The youngest pirate, being completely logical, knows that he doesn't want to get into the situation where he's the youngest of two pirates, else he will get nothing. He knows that the 1 coin you offer him is better than the alternative of 0 coins (0 coins would be the case if you get thrown overboard, as outlined above) so he'll vote to accept your offer. That's enough to guarantee the 99:0:1 offer.

Then keep adding more pirates, continuing the same sort of logic, until you reach 5 pirates.