Riddles and puzzles

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The pirate with the highest rank proposes a way to split that loot. If at least half of the pirates vote to approve that plan

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Is it half of the other pirate? or does the pirate who proposes the plan get to vote?

My pirate solution is as follows. First we number and order the pirates' ranks as follows P5>P4>P3>P2>P1. I'd say the answer is:

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P5 gets 96 gold, P4 gets 0, P3 gets 0, P2 gets 2 gold, P1 gets 2 gold.

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Explanation. We build up from the 1 person pirate crew, up to the 5 person pirate crew.

Solution to a crew of 1:

A pirate finds a chest and gives all the loot to himself.

Solution to a crew of 2 pirates:

No matter what pirate 2 (P2) proposes, Pirate 1 (p1) will down vote it. (because he'd automatically get 100 gold this way, garaunteed). So even if P2 says P1 gets 99 gold, and I (P2) get 1 gold, P1 will still down vote it to get all 100 gold. In the case that P2 says I get 0 gold, P1 gets 100 gold. The situation is identical, as in he doesn't care which way he votes, because of this we can choose that he votes to kill.

Solution to a crew of 3 pirates:

P2 knows that if it gets to him he'd die (or it'd be up to a coin toss because the situatinos are identical, and P1 can choose either). Because of this garaunteed death no matter what P3 says, P2 will agree giving P3 an automatic pass. So the solution is P3: 100 gold P2: 0 gold P3: 0 gold

Solution to a crew of 4 pirates:

All players know if they kill P4, they end up with the (crew of three pirate solution), so anything that beats out the crew of three situation (which they'd be forced into, would win). So the optimal solution that beats the crew of three solution is P4: 98 gold P3:0 gold P2: 1 gold P1: 1 gold

Note: The reason he can't give them 0 gold, even though that's identical to the P3 situation, is because that would endanger his life, they could very easily just kill him, since the p3 situation is identical. As you can see the choice to kill him if he gives that scenario does result in the 1 gold.

Solution to a crew of 5 pirates:

The solution here, so the P5 pirate doesn't die is to beat the P4 solution (by the smallest margin). So the solution is P5: 96 gold P4: 0 gold P3: 0 gold P2: 2 gold P1: 2 gold.

The only other solution I can see is if people argue that since identical situations occur if they offer 0's then P5 can end up with all 100g. However I find my solution in bettter keeping with the original conditions of valuing your own life the most.

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The pirate world's rules of distribution are thus: that the most senior pirate should propose a distribution of coins. The pirates, including the proposer, then vote on whether to accept this distribution

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This line has invalidated my solution, I set the condition the proposer couldn't vote on his own plan. Also after noticing the solution to the correct problem I noted an earlier oversight. The proper solution (with the LD conditions is):

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P5: 97 gold P4 : 0 Gold P3: 1 gold P2: 2 gold P1: 0 gold Or P5: 97 gold P4 : 0 Gold P3: 1 gold P2: 0 gold P1: 2 gold

The error arose is because i didn't go over each players view point for each crew# case (after doing do I improved my.

You're still wrong. :o

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Originally Posted by Loki

You're still wrong. :o

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Actually with the condition I assumed, that the proposer of the plan cannot vote (which from Wraith's post was, for me, ambigious). Then I did eventually come to the correct collusion of:

Spoiler:

P5: 97 gold P4 : 0 Gold P3: 1 gold P2: 2 gold P1: 0 gold Or P5: 97 gold P4 : 0 Gold P3: 1 gold P2: 0 gold P1: 2 gold

The cool thing is with the voting assumption, I made, you have to reason it out on your own, since you can't look up the answer online.

On what basis do you think that's the correct conclusion?

Ok my proposed solution, if the proposer can't vote and 50% (not 50%+1) need to vote yes:

Spoiler:

If 1 pirate there wouldn't be a vote, presumably he just takes it all.
If 2 pirates it doesn't matter what P2 proposes, P1 votes no to see P2 die and take it all.
3 pirates, pirate 3 could propose he takes all 100. P2 would have to vote yes knowing that although he gets zero he'd die if he says no.
4 pirates: P4 98, P3 0, P2 and P1 1 (both will get none if it goes to P3).
5 pirates: P5 97, P4 0, P3 1, 2 to either P2 or P1 and zero to the other.

You're thinking about this from a wrong angle. Start with only the last two pirates remaining. Think about what offer the more senior of the top two pirates would make. Then bring in the third pirate, and think about what he'd have to do to get his way. Then the second, and then the first.

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D would offer E nothing, because he could tie him in a vote.
C knows that and would therefore offer E 1 coin, and E would vote with him.
B knows all this and offers 1 coin to D. B & D tie C & E, which allows the deal to go through.
A knows all of the above, and offers 1 coin to C and E.

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No you're wrong if the proposer can't vote.

If just 2 pirates left:
D could offer E all 100 gold. E gets the only vote (D can't) and would still vote no to getting everything as he'd rather D dies and he gets everything.

The proposer does get to vote.

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It's a version of an ultimatum game. You have to think of it game-theoretically, starting from the preferences of the last pirate and going backwards.

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You can solve the game top down, or bottom up. I think bottom up is easier to visualize though. Top down, you would start by saying, suppose I die what would 4-crew case solution be, in order to solve that, you say suppose 4-crew captain dies what would the three case solution be etc... In some sense you can consider this to be still bottom-up.

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Ohh, uh, I dunno, the pirate of the highest rank proposes 33 coins to himself and the two other highest ranking pirates and the other two can go and fuck themselves?

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Interestingly enough, this solution actually wouldn't work (in either voting assumption) If the proposer can vote, he still needs two other votes to get greater than 50%, and if the proposer can't vote then he still needs to get two out of the 4 voters. The reason it wouldn't work is because P5 would vote for himself. P4 would decline becaues he knows if P5 dies, he can force more than 33 gold for himself. P3 would vote yes, since he'll get zero under the P4 case. P2 will vote no, and P1 will vote no. So you would have 2/5.

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Ok my proposed solution, if the proposer can't vote and 50% (not 50%+1) need to vote yes:

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That's correct I got the same answer.

No, you cannot solve game theory problems top down. You need to figure out the payoffs first, which you can't do without starting from the bottom.

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Originally Posted by Loki

The proposer does get to vote.

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In the original question, but we already have that solution (as does Google). Leb challenged what would be the solution if the proposer doesn't get to vote, I took up that challenge and therefore got my logic.

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Originally Posted by Lebanese Dragon

Actually with the condition I assumed, that the proposer of the plan cannot vote (which from Wraith's post was, for me, ambigious). Then I did eventually come to the correct collusion of:

Spoiler:

P5: 97 gold P4 : 0 Gold P3: 1 gold P2: 2 gold P1: 0 gold Or P5: 97 gold P4 : 0 Gold P3: 1 gold P2: 0 gold P1: 2 gold

The cool thing is with the voting assumption, I made, you have to reason it out on your own, since you can't look up the answer online.

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Originally Posted by Loki

On what basis do you think that's the correct conclusion?

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Originally Posted by RandBlade

Ok my proposed solution, if the proposer can't vote and 50% (not 50%+1) need to vote yes:

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If 1 pirate there wouldn't be a vote, presumably he just takes it all.
If 2 pirates it doesn't matter what P2 proposes, P1 votes no to see P2 die and take it all.
3 pirates, pirate 3 could propose he takes all 100. P2 would have to vote yes knowing that although he gets zero he'd die if he says no.
4 pirates: P4 98, P3 0, P2 and P1 1 (both will get none if it goes to P3).
5 pirates: P5 97, P4 0, P3 1, 2 to either P2 or P1 and zero to the other.

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Sorry, must have missed that. I believe the correct solution would then be:

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If the proposer can't vote, then E kills D regardless of the offer.
C gives 0 coin to D and E. D presumably prefers to live, even if he gets nothing.
B gives 1 coin to D and 1 coin to E.
A gives 1 coin to C and 2 coins to D or E.

Edit: changed solution; had D's payoff wrong.

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Why would C give D anything?

C claims all 100 for himself. Only D and E vote and D votes yes knowing he'll die at the next vote if he votes no now. C knows that so takes everything.

See edit above. :o

So you, Leb and I have all reached same solution :)

Just for that, I'm going to change the solution. :o

While if the proposer can vote 6+ pirates can be worked out simply, it's interesting to think about 6+ if the proposer can't. Some assumptions about risk required then.

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Originally Posted by RandBlade

While if the proposer can vote 6+ pirates can be worked out simply, it's interesting to think about 6+ if the proposer can't. Some assumptions about risk required then.

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I haven't thought about a general case (with n-pirates and k gold to distribute), A computer algorithm wouldn't be hard to conjure to solve it though.
For p6 with proposer can't vote: 3 possible solutions:

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P6: 95, P5:0, P4:1 then give 2 gold to any 2 out of P3, P2, P1 (they should vote yes with only 2 gold because statistically they only have a 50% to get 1 gold and a 50% to get 2, so if they gamble on average they'll get 1.5 gold which is less than 2 gold garaunteed.

Actually your maths is wrong.

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50% 0, 50% 2 = Average of 1

Anyway I can think of 3 solutions:

P6 97 P5 0 P4 1 P3 0 P1 and P2 1 [if risk adverse they might both vote yes prefering 1 guaranteed over a risky average of 1, even though it means one of them gets less than they would otherwise]
P6 95 P5 0 P4 1 and 2 to any 2 of P3-1 [but they can possibly still get 2 and see a pirate die if they vote no so might even turn this one down]
P6 94 P5 0 P4 1 P3 2 and 3 to either P1 or P2 [this offer has to be accepted]

From what we know though, I think its possible to work out which of these solutions is right.

Loki using the information from the original post [but with the sole change to proposer can't vote] and Game Theory, which solution would you think is right? Or do you have an alternative solution?

I should have looked at the p5 case again, it was 0 and 2 to the other (was going off memory).

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Your P6 95 P5 0 P4 1 and 2 to any 2 of P3-1, which was the answer I had, would pass though because it's a simple question of (would you take 2 gold garuanteed or a chance of getting 2 gold). You'd always take 2 gold garaunteed.

The only other option is what you suggested
P6: 97 p5: 0, P4: 1, P3: 0, P2: 1, P1: 1 I think there is argument to be made here, but I can very easily see this solution accepted, because it's a stable 1. I know I would take it.

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"Your P6 95 P5 0 P4 1 and 2 to any 2 of P3-1, which was the answer I had, would pass though because it's a simple question of (would you take 2 gold garuanteed or a chance of getting 2 gold). You'd always take 2 gold garaunteed."

Can you be 100% certain of that, remember from the original rules that 2 gold + a pirates death > 2 gold nobody dies. So might they not vote no in order to see someone die knowing they can still get 2 gold in the next round?

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I think they only care if a pirate dies if the situation is equivalent, in the situation we have, it's they live, maybe 2 gold, and a pirate dies, or they live, garaunteed 2 gold, and no pirate dies. If you go to the original post that was the order of their priority: their life, maximize money (which they statistically are), and then check if they can kill a pirate.

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Yes I agree, the chance of killing a pirate comes below the priority of getting gold, so therefore they'll accept 2 even though they could have got 2+ a death as another's death is below their own gold as priority.

In which case that rules out option 3 so just 2 options left. Which one is the solution and why?

I haven't done any problems with incomplete information for a while, so I could be wrong. But here we go:

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E is killed by F.
D offers nothing to E or F.
C offers 1 coin to E and F.
B offers 1 coin to D, and 2 coins to E with n probability and 2 coins to F 1-n probability (expected value for E is 2*n, expected value for F is 2*(1-n)).
A offers 1 coin to C, 2 coins to D, and 2 to E if n<.5 or 2 to F if n>.5) (keeps 95)

It gets messier if there's a lack of perfect information (some or all of the pirates don't know what n is).

Edit: made a minor mistake of my own.

You generally shouldn't add your own assumptions. We know nothing about whether the pirate are risk-adverse or risk-acceptant, and should therefore assume they're risk-neutral.

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P6 97 P5 0 P4 1 P3 0 P1 and P2 1 [if risk adverse they might both vote yes prefering 1 guaranteed over a risky average of 1, even though it means one of them gets less than they would otherwise] See point above about assumptions; in the strict sense of the concept, being risk-acceptant or averse is irrational anyway
P6 95 P5 0 P4 1 and 2 to any 2 of P3-1 [but they can possibly still get 2 and see a pirate die if they vote no so might even turn this one down] More or less
P6 94 P5 0 P4 1 P3 2 and 3 to either P1 or P2 [this offer has to be accepted] Irrational for P6 because he could have gotten more money with method above

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Originally Posted by Lebanese Dragon

I should have looked at the p5 case again, it was 0 and 2 to the other (was going off memory).

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Your P6 95 P5 0 P4 1 and 2 to any 2 of P3-1, which was the answer I had, would pass though because it's a simple question of (would you take 2 gold garuanteed or a chance of getting 2 gold). You'd always take 2 gold garaunteed.

The only other option is what you suggested
P6: 97 p5: 0, P4: 1, P3: 0, P2: 1, P1: 1 I think there is argument to be made here, but I can very easily see this solution accepted, because it's a stable 1. I know I would take it.

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P4, P2 and P1 get the same expected monetary return from accepting and rejecting, but they both gain more utility from refusing and having P6 killed. They therefore won't accept.

As I told Rand, don't add your own assumptions, especially when they are premised on irrationality.

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Actually I think a rational solution is possible from the information we have. Specifically we're told this:

"These particular pirates are all perfect logicians, and want to stay alive, get as much gold as possible, and see as many of the other pirates die as possible, in that order of priority. They'll never sacrifice from one of the higher priorities for a lesser priority."

So of the 3 solutions I provided:

Solution 1: This means the lead pirate risking his life that neither of the other pirates would risk their gold for more gold.
Solution 2: This means the lead pirate risking his life that neither of the other pirates would risk their gold for a chance to kill
Solution 3: This means the lead pirate accepting less gold to avoid risking his life as the other pirates are guaranteed to accept.

We know the priorities. A pirate won't risk his life to get more gold, so solution 3 could be the obvious answer. But a pirate won't risk his gold for killing, so solution 2 would have to be accepted. Therefore the lead pirate is happy to risk his life safe in the knowledge that the lesser pirates won't risk their gold.

But what's the solution from the remaining 2? Well we know that the pirates won't risk their life for gold, or gold for another's life but we don't know if they'll risk their gold for more gold. That means option 1 has to be ruled out - it is too risky on the lead's pirate life. His own life takes priority over any amount of gold so he'll save his life. So that makes solution 2 the ultimate solution.

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I guess we can think of it this way. Let's say 6 people were getting to play the 6 crew scenario over and over. Suppose they play it once per second for every second that passes for a year. if you could have a 50/50 chance of getting 2 dollars every second or to get 1 dollar garaunteed every second, what would you pick. They're statistically identical in the total money earned. So we live eitherway, we statistically make the same money eitherway, so let's go for the kill. That's one way to think about it... However, they're only equal for you in the long run, so since we're playing this game once (you don't constantly find chests of 100 gold), I think you're better off taking the 1. My ultimate vote is, the solution you pointed out, after doing the correct maths of 50%(0) + 50%(2), is the correct one. So I think P6: 97 p5: 0, P4: 1, P3: 0, P2: 1, P1: 1 is correct. Again, I know I would take it.