Highway Mathematicians

Originally posted by iamatiger
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.

Instinctively I think the first has more power than the second if a majority is required as he needs 4/7 votes not 4/6.

Unless he gets a fair share, the second can say "if you vote this bozo down I'll give you all fair shares and you can shoot me if I don't!"

Originally posted by iamatiger
Unless he gets a fair share, the second can say "if you vote this bozo down I'll give you all fair shares and you can shoot me if I don't!"

what if the 3rd 4th and 5th all have more than their fair share as they have the 2nd, 6th and 7th along with their own?, so 1st 720, 3rd,4th 5th all get 1440?

First claiming 720 and three people supporting him with shares of 720+ each sounds a viable way to pursue a solution. However.. if he suggests a deal of that kind with 2nd, 3rd, and 4th as beneficiaries, wouldn't the 5th pipe up and say,

"hey, I'll vote to keep you alive for a bit cheaper than the three you picked, and you get a few extra coins from my share if you pick me rather than, say, 2nd?"

Perhaps 720 to everybody is in fact the only stable solution.

Originally posted by talzamir
First claiming 720 and three people supporting him with shares of 720+ each sounds a viable way to pursue a solution. However.. if he suggests a deal of that kind with 2nd, 3rd, and 4th as beneficiaries, wouldn't the 5th pipe up and say,

"hey, I'll vote to keep you alive for a bit cheaper than the three you picked, and you get a few extra coins from my s ...[text shortened]... me rather than, say, 2nd?"

Perhaps 720 to everybody is in fact the only stable solution.

The 1st could reply to the 5th's suggestion by telling the 2nd-4th, you will not know if you will survive until afterwards, is it worth taking that risk for a few extra coins?

True. But then, what does anyone have to gain from duplicity? And #5's offer can be made to #2 and #3 instead. "including me in the triplet that says yes and it's worth an extra 50 for each of you. Still a lot better than me ending up with zero."

So.. #1 makes a bid and it seems clear that he can get away with 720 exactly. Any more and he's voted down, gets 720 and a bullet, and the voting resumes.

Some vote yes and get a share of the rest and the rest.. nothing? But if the yes-votes ask for more than 720, one of the nays can say "I'll do it for less." So if yes-votes are auctioned, then it would seem that asking for more than 720 won't fly.

Guess it comes down to information and timing.

Eldest: "everyone, write on a piece of paper the lowest figure for your own share that you accept to say "yes". I'll pay the lowest three figures of the loot, the three greediest get nothing, and I'll keep the rest of the loot, if any. Commit that you follow through with a yes-vote, and I commit to honor that division if it is possible, as long as I get at least 720. If the sum of the three lowest and my own share is over 5040, just shoot me and good luck for the next guy to make the attempt. If there's a tie, I'll choose randomly among those with the same but feasible request."

Is that the same or different problem? There is no information disparity here, as all get the same data at the same time. Since situation is similar to all, it makes sense they all write same number on the paper. They can all write 1,440 for a 50% chance of getting that and a 50% chance of getting nothing for an expected value of 720. But knowing that, it's tempting to write 1,439 for a 100% chance to get that, or any value in the 721-1439 range, as long as it is just barely in the bottom three.

Asking for 720 might actually be the best strategy.

Or. "We all know that we can do this 0-1680-1679-1678-2-1-0. Let's agree that no one gets the same share of the loot, that's nickles and dimes anyhow. And then we go around, and the one with the least share can propose a new way to divide the share of the one with the most loot, increasing his share and mine, as long as all the numbers differ. When the one with the least share refuses to do that, that's how we split the loot."

Even that would diverge to roughly 720 each.

When there are two mathematicians left, the older one will propose 1440 - 0 and the younger one will not be able to do anything about that. So when there are three mathematicians left, the oldest will propose 2160 - 0 - 0. The middle one will vote against, but the youngest one will vote yes (killing the oldest wouldn't help him). Following this readoning, the oldest of seven will propose 5040 - 0 - 0 - 0 - 0 - 0 - 0.

The youngest five will not vote to kill him, because it won't help their case.

Originally posted by Thomaster
When there are two mathematicians left, the older one will propose 1440 - 0 and the younger one will not be able to do anything about that. So when there are three mathematicians left, the oldest will propose 2160 - 0 - 0. The middle one will vote against, but the youngest one will vote yes (killing the oldest wouldn't help him). Following this readoning, ...[text shortened]... - 0 - 0 - 0.

The youngest five will not vote to kill him, because it won't help their case.

If there were two mathematicians left the older one can never propose 1440-0 because the younger one would reject it and be left with 720. So your answer is wrong.

Originally posted by damionhonegan
If there were two mathematicians left the older one can never propose 1440-0 because the younger one would reject it and be left with 720. So your answer is wrong.

But the older one will not reject it. He holds 50% of the votes, so he can't die, can he?

I see what you mean. So when the votes are 50-50, it is rejected...

That will make the reasoning a bit different, but my conclusion is the same. It now goes:

720 - 720 (2 vs 0)
1440 - 720 - 0 (2 vs 1)
2160 - 0 - 720 - 0 (3 vs 1)
2880 - 0 - 0 - 0 - 0 (3 vs 2)
3600 - 0 - 0 - 0 - 0 - 0 (4 vs 1)
etc.

So with five or more mathematicians, the oldest one can give himself all the money (when they're using this system).

Originally posted by Thomaster
But the older one will not reject it. He holds 50% of the votes, so he can't die, can he?

The one who proposes it doesn't vote. If they were only two left only the younger one would vote.

Originally posted by Thomaster
I see what you mean. So when the votes are 50-50, it is rejected...

That will make the reasoning a bit different, but my conclusion is the same. It now goes:

720 - 720 (2 vs 0)
1440 - 720 - 0 (2 vs 1)
2160 - 0 - 720 - 0 (3 vs 1)
2880 - 0 - 0 - 0 - 0 (3 vs 2)
3600 - 0 - 0 - 0 - 0 - 0 (4 vs 1)
etc.

So with five or more mathematicians, the oldest one can give himself all the money (when they're using this system).

Every one would reject that proposal.

Originally posted by damionhonegan
Every one would reject that proposal.

Why? Why would, par example, the youngest one reject it? He can't possibly get any money. When there are three left, the oldest proposes 1440 - 720 - 0 and it will surely be accepted. Therefore, the youngest one will never vote to kill the oldest mathematician. Doing so wouldn't help him to get more money.

Originally posted by Thomaster
Why? Why would, par example, the youngest one reject it? He can't possibly get any money. When there are three left, the oldest proposes 1440 - 720 - 0 and it will surely be accepted. Therefore, the youngest one will never vote to kill the oldest mathematician. Doing so wouldn't help him to get more money.

(Using the assumptions that a tie in votes and the proposer gets killed) If they were three left and they divided it 1440-720-0. The youngest would reject the proposal because he would be able to get atleast 720. Then there would be two left. The second youngest would propose either 720-720 or 719-721. If he proposes that he gets a larger share than the younger one he would die because the youngest would be left with 720 so he has to give the youngest at least 720.

That's why the youngest will propose to kill the oldest remaining mathematician in that scenario.

Originally posted by damionhonegan
[b](Using the assumptions that a tie in votes and the proposer gets killed)

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.