Highway Mathematicians

Originally posted by Thomaster
Ok, I'll use that assumption now. The one who proposes cannot vote, and a tie means death (for some reason).

My backwards induction now is as follows:
720 (no voting)
720 - 720 (1 vs 0)
720 - 720 -720 (2 vs 0)
1440 - 720 - 720 - 0 (2 vs 1)
2160 - 0 - 720 -720 - 0 (3 vs 1)
3600 - 0 - 0 - 720 - 0 - 0 (3 vs 2)
5040 - 0 - 0 - 0 - 0 - 0 - 0 (4 vs 2)

So the oldest one can give himself all the money.

Well, it says this in the puzzle "a. a successful division or b. a reason to shoot the person who proposed it?"

So there are actually two assumptions: one for a tie being an agreement and another for tie being death of the proposer. So you can have one or two different answers.

(Using the assumption that a tie is an agreement) I have this:
1260-1260-1260-1260-0-0-0

(Using the assumption that a tie means death) I have this:
0-0-0-0-0-0-720
0-0-0-0-0-719-721
0-0-0-0-718-720-722
0-0-0-717-719-721-723
0-0-716-718-720-722-724
0-715-717-719-721-723-725
714-716-718-720-722-724-726

Actually, the first answer I gave could work for both assumptions.
1260-1260-1260-1260-0-0-0

Ah, I must have misread the OP. I'm sorry for that.

I disagree with your solution for two reasons. You increase the money the mathematicians get with every step, but that's not needed. If a mathematician knows that he will get 720 dollar no matter wether the oldest will be shot or not, he will surely not vote for a death. Also, in the fourth step you say the youngest one will be given 723 dollar. The oldest one doesn't need to give him that money, because 1440-719-721-0 would be accepted as well (with two votes).

Originally posted by talzamir
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 the ...[text shortened]... n the vote to be a. a successful division or b. a reason to shoot the person who proposed it?

Lets work backwards to solve for b.

If a tie kills the one who proposes the division then the least senior should always vote no unless he is given 720 coins because this is the highest amount he can get if everyone is shot anyway.

Knowing that the least senior will vote no unless given 720 coins the second least senior math wiz should vote no if least senior will vote no and yes if he is given 720 coins.

etc etc.

So the Most senior in his first proposal should be 2160 0 0 720 720 720 720 and if everybody is acting in their honesty, seniority, and self-interest this should be the only way he can guarantee that he won't be shot with the most money.

Example: with 2160 720 0 720 720 720 0 it would be a tie since the second senior math wiz would vote no because he can get more than 720 if most senior is shot and thus would vote no along with third most senior and last senior math wiz.


I haven't looked at it for a. which would seem to be a completely different answer.

Originally posted by tomtom232
Lets work backwards to solve for b.

If a tie kills the one who proposes the division then the least senior should always vote no unless he is given 720 coins because this is the highest amount he can get if everyone is shot anyway.

Knowing that the least senior will vote no unless given 720 coins the second least senior math wiz should vote no if lea ...[text shortened]... wiz.


I haven't looked at it for a. which would seem to be a completely different answer.

Well, for a. you could give the oldest mathematician 2880 and three of the other six mathematicians 720 and the remaining three mathematicians 0.

Originally posted by talzamir
My solution -

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

This is probably right.