A Land Far Away

A classic logic puzzle with the names changed to protect the innocent.

If you *know* the answer from prior experience, please refrain from spoiling it for those trying to work it out;

The Puzzle

5 Lawful Evil adventurers have found 100 bars of gold.

They have a strict order of seniority: A is superior to B, who is superior to C, who is superior to D, who is superior to E. Their rules for sharing the treasure are as follows:

1) The most senior Adventurer proposes a distribution of gold.
2) The Adventurers then vote on whether to accept this distribution:
3) The proposer is able to vote, and has the casting vote in the event of a tie.
4) If the proposed allocation is approved by vote, it happens.
5) If the proposed allcation is rejected, the proposer is murdered by the other adventurers, and the next most senior Adventurer makes a new proposal to begin the system again.

None of them want to die, so the most senior must choose his distribution carefully.
However, every adventurer wants to maximise his amount of gold.
They actively enjoy murdering each other, but only within the rules.
They are all 100% logical and very clever at working out the pros and cons of various suggestions.

What should 'A' propose in order to maximise his share and not get killed?

So what's the question? The best initial proposed distribution?

What are we supposed to figure out?

The last three will settle for 33 coins.
Explanation:

The last three will reject any offer that gives them less than 33 coins each, cause they have a majority.

But, when it comes to 3 adventurers, the last one can not vote against C's offer cause he will be left with no money,since D's offer would be to take all the money, and he breaks the tie.

I forgot one trivial detail: the question is,

What should 'A' propose in order to maximise his share and not get killed?

[Answer temporarily removed so others can solve the problem]

[Methodology removed temporarily to let other folks have a chance to solve it on their own.]

Did you just work that out yourself?

Perhaps you can delete your answers so other people can have a go.

And that, ladies and gentlemen, is how you solve a game via backwards induction.

Yes, I just worked that out myself while talking on the phone to the VA benefits folks - see, dealing with a government bureacracy gave me a +2 synergy bonus to figuring out lawful evil puzzles.

Anyway, I will remove my solution for now.

Mikayla wrote:Yes, I just worked that out myself while talking on the phone to the VA benefits folks - see, dealing with a government bureacracy gave me a +2 synergy bonus to figuring out lawful evil puzzles.

Anyway, I will remove my solution for now.

Thanks.

I found the puzzle and answer intertwined, so didn't get the opportunity to try and solve it. i suspect that I wouldn't have, even though the answer makes perfect sense it retrospect.

We could also try the "black and white hats" puzzle and the "3 way duel" puzzle.

I've heard this one before, though with a piratey theme. I don't remember the answer, so the following may not be correct. In case it is, try not to read it if you're considering the puzzle (the answer isn't written explicitly, so you wont see it while just scrolling over.)





For any number of adventures, call {B,D,F,...} the even adventurers and {C,E,G,...} the odd adventurers.

Assume that each adventurer enjoys stabbing people more than receiving no gold bars, but less than receiving one or more gold bars.

Throughout, n is a natural number less than 201.

P(n) is equiv. to the statement 'with n adventures, A will assign 1 bar to each odd adventurer and the rest to himself.'

Assuming P(n), if A is killed the even adventurers receive at the least 1 bar each and the odd adventurers nothing. So if all odd adventurers are assigned 1 bar they will all accept. As together they form one of the smallest subsets of adventurers required to vote yes it is impossible for A to assign more gold to himself and survive (due to stabbing assumption.) Therefore P(n+1).

An adventurer who is alone will obviously assign himself 100, therefore P(1). As P(n)=>P(n+1), for all n: P(n).