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Another variant of the Pirate Puzzle, inspired by this variant and a bit more complex.

Scenario:

Five pirates of different ages obtain 100 gold pieces in total, and want to decide how to split them.

Procedure:

First each pirate, in order from youngest to oldest, makes a proposition on how to split up the gold (a proposition allocates every pirate a nonnegative integer amount of gold pieces, 100 pieces allocated in total). Then the pirates vote upon the propositions, in the opposite order as the propositions were stated (so the oldest pirate's proposition is voted upon first).

If a proposition is accepted (which requires the agreement of at least 50% of the remaining pirates), the voting procedure ends immediately and the gold is divided according to the proposition.

If a proposition is not accepted, the pirate who made the proposition is tossed overboard, and voting proceeds to the next proposition.

Pirates:

It is common knowledge that all pirates are perfectly intelligent, and that every pirate $X$ has the following decision criteria, in order:

  • Pirate $X$ should not be tossed overboard.
  • Pirate $X$ should receive as much gold as possible.
  • As many pirates as possible should be tossed overboard.
  • The oldest pirate should receive as little gold as possible.
  • The second oldest pirate should receive as little gold as possible.
  • ...
  • The youngest pirate should receive as little gold as possible.

To decide between two outcomes, Pirate $X$ will go down this list and select the first criterium that distinguishes these outcomes, and select the more favourable outcome according to that criterium.

Question:

What will be the outcome of the decision procedure, and why?

Bonus question:

The rules are changed so accepting a proposition requires the agreement of all but at most one of the remaining pirates. What will be the outcome now, and why?

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  • $\begingroup$ You said it was another variant of puzzling.stackexchange.com/questions/54800/…, but I don't see how it's different $\endgroup$ – PotatoLatte Jul 19 at 15:43
  • $\begingroup$ @PotatoLatte all the propositions have to be made in reverse order before they are voted on. I'm working on this right now so I'm not sure if it has a different result but I expect it to. $\endgroup$ – Michael Moschella Jul 19 at 15:57
  • $\begingroup$ @PotatoLatte The decision procedure is completely different, allowing older pirates to outbid younger pirates because of their commitment. You'll find that the three-pirates case is already more involved than in the previous puzzles and has a very different solution. $\endgroup$ – Magma Jul 19 at 18:32
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Oldest Pirate gets 64, second oldest gets 33, youngest gets 3, others get nothing.

Numbering the pirates from 1 to 5, youngest to oldest. A proposal is described as a-b-c-d-e, with youngest pirates listed first (a is the offer to the youngest, e to the oldest).

No matter what Pirates 1, 2, and 3 propose, Pirate 4 can propose one of: 2-33-33-32-0, 33-2-33-32-0, 1-33-34-32-0, 1-34-33-32-0, 34-1-33-32-0. No matter how Pirate 3 splits it, at least one of the pirates is offered not more than 33 gold, and Pirate 4 can offer her better. Since Pirate 4 only needs one vote beside her own, she can ensure a safe offer to herself of at least 32 gold.

After that,

Pirate 5 would offer 3-0-0-33-64, 2-0-0-33-65, 0-3-0-33-64, or 0-2-0-33-65. This would be the cheapest way to get two other votes, which is what Pirate 5 needs.

Since Pirate 3 wants the best split,

she would propose not more than 32 to herself, ensuring Pirate 4 would propose 2-33-33-32-0, and Pirate 5 would propose 3-0-0-33-64. Pirate 3 gets nothing no matter what she proposes, and nobody walks the plank, so the best she can do is ensure the most amount of gold goes to the youngest pirate.

Update for Bonus Question

Split is 0-1-34-34-31.

Each Pirate needs all the votes but one, so they can safely

zero out one other pirate, and increment the others. But they also need to ensure they don't get zeroed out later.

Pirate 3 can always

offer one of the younger two pirates more than Pirate 2 offered, with 51 gold being the most she would have to offer. Thus Pirate 3 can be sure to make an offer at least one of the other two would vote for.

Unfortunately for Pirate 2,

Pirate 3 can always ensure that Pirates 1 and 2 get zeroed out in the proposals by Pirates 4 and 5. The best Pirate 2 can do is ensure she gets zeroed out by Pirate 4, and that Pirate 5 will then offer her one gold.

Pirate 3 will

always offer herself the smallest of the three amounts she proposes. This ensures that Pirates 4 and 5 will zero out Pirates 1 and 2. The most she can offer herself and ensure she has the lowest amount is 32. She can only offer herself that much if she offers the other two (33, 35) or (34, 34).

If Pirate 3 proposes

34-34-32-0-0, Pirate 4 will zero Pirate 1, increment the others, take the rest, and propose 0-35-33-32-0, and Pirate 5 will propose 1-0-34-33-32. Pirate 4 could propose 35-0-33-32-0, followed by 0-1-34-33-32, but Pirate 4 would prefer for the final gold to go to Pirate 1. To avoid Pirate 3 offering 34-34-32-0-0, Pirate 2 must offer herself at least 34, and Pirate 1 even more.

So to get anything,

Pirate 2 must propose 66-34-0-0-0. This leads to Pirate 3 proposing 33-35-32-0-0, followed by 34-0-33-33-0, and 0-1-34-34-31. Offering more to herself would force Pirate 3 to offer less to herself, resulting in more gold for Pirates 4 and 5. For example, 65-35-0-0-0 would lead to 33-36-31-0-0, then to 34-0-32-34-0, and 0-1-33-35-31.

So, the best option is:

Pirate 2 proposes 66-34-0-0-0, followed by proposals 33-35-32-0-0, 34-0-33-33-0, and 0-1-34-34-31.

Unless,

Pirate 2 offers a 50/50 split and Pirates 1 and 2 agree to vote no on all other proposals. But Pirates aren't known for their reliability, so when Pirate 3 offered one of them 51, she would take it.

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  • $\begingroup$ Well done. What do you make of the bonus question? $\endgroup$ – Magma Jul 19 at 19:55
  • $\begingroup$ Good job on the bonus as well. $\endgroup$ – Magma Jul 19 at 20:57
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I believe:

The oldest pirate will get 50 coins, the second will get 26, and the third 24.

We actually don't want to work backwards for this one.

Starting from the eldest pirate, he will obviously give the most coins possible to himself while giving one more coin to the two pirates with the least amounts of coins. Therefore, the second oldest pirate will want to have the second least amount, maximizing that number if possible. The third pirate wants to get as much as he can from the second oldest, so he will put himself in a position that the second oldest can give him less than the two youngest to his benefit. He will give 22 to himself and split the remaining 78 between the other two (Doesn't matter how, as long as he gives each of them at least 23). It doesn't matter what the fourth pirate proposes and the youngest will always give 100 coins to himself. The reason why the third pirate must give himself only 22 instead of 23 to get 24 from the second pirate is to ensure the second pirate doesn't split 24-26-25-25 instead, which is actually in his best interest, as it will ultimately minimize the number of coins the oldest pirate and third pirate will get while giving more to the youngest instead. Since the third pirate gave himself only 22, however, the second pirate can give him 23 and get an extra coin in the process.

So:

Since the third pirate gave himself 21, the second pirate will give himself 25 and the third pirate 23, giving the remaining two 26 each. Now the oldest pirate will have to give him 26 and the third pirate 24, leaving 50 coins to himself.

My explanation is a bit messy, I know, and I very well may be missing a possibility, but I think this is at least pretty close.

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  • $\begingroup$ The second oldest pirate can do better than 26 coins. $\endgroup$ – Magma Jul 19 at 18:34

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