How Can Pirate #5 Maximize His Gold Share in a Clever Division Strategy?

• sandwhales
In summary, the problem of dividing the gold among an infinite number of infinitely greedy, ruthless, and intelligent pirates becomes increasingly complex and ultimately impossible as the number of pirates increases. With a finite number of pirates, the solution involves strategic bribes and alliances, but with an infinite number, it is impossible to determine who is next in line to make a proposal. The gold would likely run out before a solution is reached, and even if it didn't, the division of gold would become increasingly unfair as the number of pirates increases. This problem becomes even more complicated if gold pieces are allowed to be divided or if random lotteries are allowed.
sandwhales
Five pirates are trying to split up 1000 gold pieces. The rules are as follows:

Pirate #5 must divide the gold up in such a way that a majority of the pirates (including himself) agree to. If he does not get a majority vote, he will be killed, and pirate #4 will get to propose a solution to the remaining 3 pirates, and the same rules will follow. This continues until someone comes up with a plan that earns a majority vote.
What is the most amount of gold pieces that pirate #5 can keep to himself, and what would his proposal be?

The pirates are infinitely greedy, infinitely ruthless (the more dead pirates the better), and infinitely intelligent.

Assume that the pirates value their own life over gold, but would love to see their comrades die even if for no apparent reason. A tiebreaker means the pirate that made the proposal is thrown overboard.

Now I solved the riddle, but I wanted to further develop it. For those of you that are interested, what would happen if there were an infinite amount of pirates? (Granted it can't be infinite as the gold runs out at some point)

This is where I get stuck, i can't seem to figure the riddle out for an infinite amount of pirates. What happens to the gold then?

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sandwhales said:
This is where I get stuck, i can't seem to figure the riddle out for an infinite amount of pirates. What happens to the gold then?

What is a majority vote in the countable case? It depends on how you measure the vote distribution. You may be familiar with the difficulty of this. One suggestion could be natural density, but then not all distributions of votes would be measured.

If one considers a continuum of pirates, one might consider the first vote the measure (for some measure on the continuum) of the subset of pirates voting for the first one, making it a majority vote if the measure is larger than 1/2. But who is the second pirate?

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Jarle said:
But who is the second pirate?

What do you mean by the second pirate?

sandwhales said:
What do you mean by the second pirate?

Obviously, the pirate who proposes a distribution if the first is thrown overboard. If you make a decision, who is the third? And so on. The point is, how are you to decide "who's next" in a given situation? You will have to invoke the well-ordering theorem.

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If priate #1 is the last surviving and gets to a vote he will automatically accept his own choice and win all the gold.

But if pirate #2 makes a proposal than he will surely be killed because pirate #1 wants him to die no matter what. Hence #2 must agree with anything #3 proposes.

#3 would give pirates #1 and #2 nothing and he will win all the gold. #2 dare not vote against him lest he will himself die.

#4 can thus secure #1 and #2's votes with a bribe of 1 gold. #3 is powerless to stop him and gets zilch. The distribution of gold for each pirate would be (1,1,0,998)

#5 need only offer 1 more gold to a pair of the pirates 1 through 3 to win enough votes. Thus the distribution is (2,2,0,0,996),(0,2,1,0,997), or (2,0,1,0,997). Pirate #5 will choose one of the latter two distributions since it maximizes his take.

Let's try more pirates. A sixth pirate would need to secure three other votes. He needs to beat 5's offer with expectation value (1,1,1,0,997). (2,0,2,1,0,995), (0,2,2,1,0,995), or (2,2,0,1,0,995) would work. The expectation value is (4/3,4/3,4/3,1,0,995).

A seventh pirate will offer (2,0,0,2,1,0,995), (0,2,0,2,1,0,995), or (0,0,2,2,1,0,995). The algorithm at this point should be clear.

At some point pirate #X (around 1000, I think?) will offer a pittance to X/2 other pirates and will take only 1 gold himself.
Many pirates past that point are doomed, since they can offer no better. The gold has run out.

Once there around 2000 pirates the latter "doomed" pirates will vote only to save their own skin. The lucky 2000th pirate can steal all the gold in this case because he has enough broke pirates that are forced into saving their own lives. I'm not sure what happens after that... I think pirates 2001-3999 will be sentenced to death by pirates 0 - 2000. But when there are 4000 pirates all will survive and the last will get all the gold. Then pirates 4001-7999 will be killed. And so on. My numbers are estimates and I could be wrong though. This is a pretty interesting problem.

Note: the solution changes radically if gold pieces are allowed to be divided, or if a pirate can propose random lotteries (which are essentially the same as dividing gold pieces). Then it becomes much easier to bribe a large number of pirates into not killing you, as each pirate could have a small chance of winning.

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