Logic problem: How will the robbers divide the loot?

It’s a logic problem! Logic problems were never designed to be realistic to the real world in any way. It is meant to be a mental exercise that tests a person’s deductive reasoning skills.

How is one to solve this if we do not have the slightest idea how each robber will vote and how big a share they feel acceptable for themselves?
Here is the thing: You say each robber wants to maximise their own share. Couple that with the option to kill one of the competitors each round, each of the robbers is going to want everything, because the murder option makes it possible to remove all competitors.
So, the “robbers murder each other until only 1 is left” is the logical solution.

Well, the most important thing is the democratic system, which will be followed correctly becuase I say so.

Here’s a hint: in the two robber version, the highest robber will always be killed by the lower one. This means that when there are more robbers ranking above him, he will vote yes becuase he wants to avoid being the one to divide the money.

It took seven robbers to steal $1000? They should just start doing real work.

If I was the lead robber, I would split it so the four highest ranks get $100 and the three lowest ranks get $200. Only getting $100 is better than being dead, and the lowest ranks getting the most money will probably dissuade them from voting for the death of the lead robber since they may not see the lion’s share of the loot from anyone else.

Ha! ⅔ of 6. You need 4 to agree on the split. Divide 1000 by 5, they get $200 each and the others get screwed.

One thing that I should have mentioned is that all of the robbers are completely rational in their decision making: their votes are based on what will get them the most money. I know this goes against the whole “robber” side plot

@PhiNotPi So basically, the lowest ranked robber will always vote to kill since they can possibly get more money each go around?

@amujinx Actually, my analysis shows that that is not actually always true. I could link you to a whole magazine article on a different version of this problem, if you want me to.

Well, in my scenario, the head robber has $1000, 6 bullets and good aim.

If he isn’t confident, he gets his ⅔ majority, they all get $200 and I’ll bet they have 2 bullets between them to take care of the others.

7 * ⅔ ≈ 4.6

So 5 of the 7 have to agree. Looks like a variation of the Pirate Game which I coded up a solution to once. Wish I had time tonight to figure it out; looks interesting.

Let the robbers be called labeled with the numbers 1 -7, with 1 representing the lowest rank and 7 representing the highest rank. Given the rules set out by the OP and the additional assumption that each robber would choose staying alive and getting no money over dying and getting no money, then the money will be distributed as follows:

Robber 7 will get three bills,
Robber 6 will get zero bills,
Robber 5 will get one bill,
Robber 4 will get two bills,
Robber 3 will get three bills,
Robber 2 will get one bill, and
Robber 1 will get zero bills.

Why? Since only six robbers get to vote, Robber 7 needs only four of his companions to vote in favor of his solution. This leaves him free to give nothing to two of the robbers just so long as the other four are better off than they would be in a scenario with only six robbers. As it turns out, the distribution above is better for Robbers 2, 3, 4, and 5 than it would be if there were only six robbers because under a six robber scenario:

Robber 6 will get four bills,
Robber 5 will get zero bills,
Robber 4 will get one bill,
Robber 3 will get two bills,
Robber 2 will get zero bills, and
Robber 1 will get three bills.

Robbers 1, 3, and 4 will agree to this solution in a six robber scenario because it leaves them better off than they would be in a five robber scenario in which:

Robber 5 will get four bills,
Robber 4 will get zero bills,
Robber 3 will get one bill,
Robber 2 will get three bills, and
Robber 1 will get two bills.

Robbers 1, 2, and 3 will agree to this solution in a five robber scenario because it leaves them better off than they would in a four robber scenario in which:

Robber 4 will get seven bills,
Robber 3 will get zero bills,
Robber 2 will get two bills, and
Robber 1 will get one bill.

Robbers 1 and 2 will agree to this solution in a four robber scenario because it leaves them better off than they would be in a three robber scenario in which:

Robber 3 will get nine bills,
Robber 2 will get one bill, and
Robber 1 will get zero bills.

Robber 2 will agree to this solution in a three robber scenario because the alternative is to be the top-ranked robber in a two robber scenario. The top-ranked robber in a two robber scenario is always killed (as it allows the bottom-ranked robber to get all of the money and to kill the other robber). Therefore, no one wants to be the top-ranked robber in a two robber scenario.

However:

This solution assumes that the three robber scenario has different rules than every other scenario. If we stuck to the assumption that the solution must have at least two-thirds of the voting group to pass, then the three robber scenario would look very much like the two robber scenario. In that case, Robber 1 gets to kill both Robber 2 and Robber 3 while keeping the money for himself. This leads to a different analysis of what a four, five, six, and seven robber scenario would look like.

Why make the assumption that the rules for a three robber scenario are different? Two reasons. The first reason is that the hint we were given implies—though does not actually state—that only a two robber scenario is supposed to lead to the deaths of everyone but the bottom-ranked robber. The second reason is that there is no elegant solution if the rules are not different for a three robber scenario. Since this is supposed to be a logic problem, and since logic problems are supposed to have elegant solutions, it seems that we are warranted in assuming that the rules are different for a three robber scenario.

What, then, would the problem look like if we did not get to assume that the rules are different for a three robber scenario? A two robber scenario would proceed in the same way as above. A three robber scenario would also proceed in that way, however, meaning that Robbers 2 and 3 would both wind up dead and Robber 1 would get all the money.

This makes a four robber scenario much nicer to the top-ranked robber who can then keep all 10 bills. Robbers 2 and 3 will agree to this solution despite it leaving them with no money because the alternative is death. Thus a five robber solution allows Robber 5 to keep seven bills, while giving one each to Robbers 1, 2, and 3.

After this is where it gets inelegant. A six robber scenario would be disjunctive: Robber 6 would keep five bills for himself and give one bill to Robber 4, but he would then have to give two bills to two of Robbers 1, 2, and 3 and zero bills to the remainder. This makes the seven robber scenario even more disjunctive. Robber 7 would keep three bills for himself, give one bill to Robber 5, and two bills to Robber 4 just like in the original solution.

Robbers 1, 2, and 3, however, would have to get some distribution of three bills, one bill, and zero bills. What distribution, though? Given the disjunctive element of the six robber scenario, there would be no way for Robbers 1, 2, and 3 in the seven robber scenario to know whether they are doing better or worse than they would in a six robber scenario. As such, it would get very tricky to win their votes. Robber 6, who gets nothing in a seven robber scenario, would be smart to exploit this for his own gain. Thus perhaps Robber 7 would have to give two of them three bills each, leaving only one for himself.

I still say most robbers would say “fuck your elegant solution”. And if they were worth anything as a robber, the point was solved with bullets.

not that I think it’s worth shooting 2 – 6 people over $1000, but we are talking outlaws and shit gets real out there.

@Blueroses The elegance constraint comes from this being a logic problem, not from the nature of the robbers themselves. You are letting yourself get distracted by the wording of the problem. It could just as well be bankers dividing up profits and firing the leader (leaving him with nothing) if they don’t find the solution acceptable.

Oh. I see.
You’re assuming there’s a difference between bankers and robbers.

@Blueroses Okay, you got me there!

I have found an interesting problem for the version that does not assume the rules are different for a three robber scenario. If what I say above is correct, then the final distribution when all is said and done looks something like this (where the arbitrary disjunctive element is determined by giving preference to rank):

Robber 7 gets one bill,
Robber 6 gets zero bills,
Robber 5 gets one bill,
Robber 4 gets two bills,
Robber 3 gets three bills,
Robber 2 gets three bills, and
Robber 1 gets zero bills.

If this is true, then Robber 7 really should prefer the solution given by @Blueroses. Consider the following distribution:

Robber 7 gets two bills,
Robber 6 gets two bills,
Robber 5 gets two bills,
Robber 4 gets two bills,
Robber 3 gets two bills,
Robber 2 gets zero bills, and
Robber 1 gets zero bills.

It is better for Robber 7 than the alternative seven robber solution. Robber 5 should vote for this distribution as it is better for him than both the alternative seven robber solution and and the six robber solution. Robber 4 should vote for it because it is just as good for him as the alternative seven robber solution, and better for him than the six robber solution.

Robber 3 should vote for it because who among the bottom three robbers gets any money at all is decided arbitrarily in both the alternative seven robber scenario and the six robber scenario. The best outcome for him if he votes against the even division solution is that he gets to kill someone and leave with three bills. The worst outcome for him is that he gets to kill someone and leaves with nothing.

The even division solution eliminates the opportunity to kill anyone, but it guarantees him two bills. Given that the money is a higher priority for him, a guarantee of two bills should be more valuable than “either get three bills and get to kill someone or get zero bills and get to kill someone.” The risk of losing the money should not be worth the possibility of getting to kill someone given that he wants to kill only if all other things are equal.

This leaves Robber 6, who exposes one of the paradoxes of egoism. Should he agree to the even division solution or not? If he refuses to be the crucial fourth vote, he gets to kill someone and more than double his money. As such, he would be irrational to accept the even division solution. Once Robber 7 proposed it, he would have no reason to accept it.

Robber 7 knows this, though, and thus will not suggest the even division solution. This leaves Robber 6 worse off than he would have been had he agreed to the even division solution, however, meaning that he would be irrational to reject it. He should consider the relative contrast class to be the alternative seven robber solution. Yet once the even division solution has been proposed, he has no reason to vote for it given the psychology that has been stipulated for him.

Do we assume that the robbers place a value on their lives? I am trying to work this out for the case where there are 4 of them. If the robbers prefer being alive and not paid to being dead, then robber 4 would be able to keep all the money, getting the votes of 2 and 3. The reason for this is that if there are fewer than four robbers, robber 1 gets to prevent a ⅔ majority each time, resulting in the deaths of robbers 2 and 3, and ends up with all the money.

If the robbers are only concerned about how much money they get, dead or alive, then in the four robber case, 4 would get $800, and 2 and 3 would each get $100.

Since you insist on making up the rules, I am guessing that you came up with this problem on your own. It is a GQ in either case.

The problem has to be solved in reverse order, starting with 2 robbers, then moving successively to 3, 4, 5, 6 and 7. At each stage, the money divider gives $100 more to the ⅔ from the previous round who got the least, keeping the rest.

Going on the assumption that none of the robbers place a value on their lives (see previous post), I get the following:
7 -200
6 – 200
5 – 100
4 – 200
3 – 0
2 – 0
1 – 300

Here is my reasoning:
For cases 1 to 3, 1 ends up with everything

4 – 800
3 – 100
2 – 100
1 – 0

5 – 500
4 – 0
3 – 200
2 – 200
1 – 100

6 – 100
5 – 0
4 – 100
3 – 300
2 – 300
1 – 200

@LostInParadise Actually, this problem is a modified form of this other problem. Warning: although the change of rules renders the article’s solution wrong, the link tells you exactly how to solve it.

I started reading the article. My overall strategy is in agreement, though I may have messed up on the details.

50/50 – $500 each