Highway Mathematicians

Seven mathematicians have found out that it doesn't really pay well to be a mathematician, and decided to become highwaymen instead. To stay true to the ideology of their new line of work they all pledge to make all decisions according to honesty, seniority, and self-interest. That policy becomes a problem when they try to decide how to divide the loot they got, 5,040 coins of gold, as fairness is not a part of their new creed.

Finally, they decide to follow the process of

1. The most senior member group member alive proposes how to divide the loot.
2. Everybody else votes on whether they accept the vote or not.
3a. If aye's win, the division holds.
3b. If nay's win, the member who made the proposal is shot, with his or her share of the gold buried with him or her as the living mathematicians will feel bad about having shot one of their own.

How will the coins end up divided, who lives and who dies if a draw in the vote to be a. a successful division or b. a reason to shoot the person who proposed it?

One more clarification - in case someone is shot, the share to be buried with him or her is an equal share of the total loot, or 720 coins.

I am sure the answer is cleverer than this, but how about the eldest 4 share the total. That gets the majority on side but also the most vulnerable majority. It seems to be the only way they can prevent someone younger getting them shot for the lot.

**Edit I haven't come up with a final split for these 4

The four eldest ones could certainly do that, and it would make the most senior one very happy - he gets to stay alive. Would they split it evenly? And is this the answer to both versions of the puzzle, with the two different interpretations of the a tie vote?

Eldest highwayman takes 5037, giving three others 1 each. If one of the 1's votes to shoot him then they have 3 coins to share amongst 6 and will end up with one or less so there is no gain to shooting him.

My solution -

5033, 0, 1, 2, 0, 1, 3 if tie means BANG
5036, 0, 1, 2, 0, 1, 0 if tie means OK.

Reasoning.

If an even split in the vote means BANG:

If six of the highwaymen die, the last gets 720, the loot after six burials.
So the second to last survives by offering the last a better deal. 719 for him, 721 for the last.
Third to last survives only by two yes-votes. So he sweetens the deal for the other two from his 720. Split 718 - 720 - 722.
Fourth to last needs two yes-votes but saves money by accepting one nay. Split 1440 - 719 - 721 - 0.
Third, or fifth to last needs three votes in favor, and one can be against. Split 2157 - 0 - 720 - 722 - 1.
Second needs three in favor, and two can be against. Split 3596 - 0 - 1 - 721 - 0 - 2.
First needs four in favor, and two can be against. Split 5033 - 0 - 1 - 2 - 0 - 1 - 3.

The tie-means-accepting-the-split works almost the same.

720
719 - 721
1440 - 720 - 0
2158 - 0 - 721 - 1
2875 - 0 - 1 - 722 - 2
4314 - 0 - 1 - 2 - 0 - 3
5036 - 0 - 1 - 2 - 0 - 1 - 0

Ok, lets look at my solution:

5037 - 1 - 1 - 1 - 0 - 0 - 0

Since they are all mathematicians, we can assume they will act logically, and will not vote a division down out of spite, only if they can gain by it.

We start off with 7, so there can't be a tie

would one of the ones vote my answer down? Presumably only if they gain....

if they vote it down can any of the 1's get 2?

say
2-1-0-0-0-0

no they can't, the four zeros can expect a better deal (on average) by voting this down.

So the oldest has to do
1-1-1-0-0-0 (if tie is ok), which goes through or
0,1,1,1,0,0 (if tie means bang), which goes through

None of the 1's gets a better deal by voting the first division down, so it will pass.

I think my answers deviate from yours at the start, they would be:

5040 - 0
5039 - 1 - 0
5039 - 1 - 0 - 0
5038 - 1 - 1 - 0 - 0
5038 - 1 - 1 - 0 - 0 - 0
5037 - 1 - 1 - 1 - 0 - 0 - 0



As guy 2 can't get a better deal by voting this down (regardless of bangy draws)

mine were based on the 720 coin funeral costs.. without those, I would totally agree with you.

Ah, I missed your amendment and didn;t read the problem carefully either!. My answer is only right if the loser is buried with the share he allocated to himself, and if the allocator gets to vote as well as the others.

By the way, did you see my answer to your marble puzzle? Did you want a full derivation?

I did, and I'd be a delight to see fully how you came up with that?

Originally posted by talzamir
Reasoning.

If an even split in the vote means BANG:

If six of the highwaymen die, the last gets 720, the loot after six burials.
So the second to last survives by offering the last a better deal. 719 for him, 721 for the last.
Third to last survives only by two yes-votes. So he sweetens the deal for the other two from his 720. Split 718 - 720 - 722. ...[text shortened]... 58 - 0 - 721 - 1
2875 - 0 - 1 - 722 - 2
4314 - 0 - 1 - 2 - 0 - 3
5036 - 0 - 1 - 2 - 0 - 1 - 0

I remain to be convinced. Surely on that reasoning the split is one that each successive person can do better with, so why wouldnt 6 others vote against it? It is not about spite but if I was numbers 4-1 then I think I can do better than 2-0-1-3 with 2880 to split. This would mean a vote against the proposal and second can then work a better split of the proceeds.

Aye, the solution depends on a number of assumptions - such as, no negotiations take place. But if negotiations DO take place, and it's possible to make deals that are based on pooling votes, bargain etc which would naturally be possible..?

What happens when the eldest mathematician says to the other six, "Look guys, I propose that I get a seventh of the loot and face the risk of standing before the guns. If I'd ask for more than 720, you'd just shoot me and get more on the average as a result so I won't. But I won't settle for less than 720 either as you would just call me an idiot for throwing money away, so I won't make an offer of a division where I get less than 720. Other than that, I'll propose to divide the other 4,320 coins any way you want - so that the two (or three) greediest get to say they want me dead, and they ones who settle for the least shares get what they ask for. For starters, let's say everyone gets an even split of 720, and start the bargaining from there. If you can come up with a split that guarantees I survive, I'll go for that, and if there are multiple such proposals I'll choose that one with the biggest share for me, propose that, you say aye, and most of us will be happy."

Originally posted by talzamir
Aye, the solution depends on a number of assumptions - such as, no negotiations take place. But if negotiations DO take place, and it's possible to make deals that are based on pooling votes, bargain etc which would naturally be possible..?

What happens when the eldest mathematician says to the other six, "Look guys, I propose that I get a seventh of the ...[text shortened]... the biggest share for me, propose that, you say aye, and most of us will be happy."

I think it is the second oldest that has the power, he has a potentially large sum to play with and can offer to give the others really good deals if they vote the current proposal down.

Does he? The sum goes down by 720 if the first proposal is rejected, so if the eldest accepts even a coin less, isn't even better for the other six?