Save the Robber
Five robbers steal 1000 gold coins and devise the following method to divide them:
The youngest robber has to present a plan to divide them. The plan is accepted only if majority of robbers accept it. (Two out of four won’t do.) If the plan is rejected the presenter of plan is shot and the next older robber gets a chance. So what plan should youngest robber present to save his life?
The youngest robber has to present a plan to divide them. The plan is accepted only if majority of robbers accept it. (Two out of four won’t do.) If the plan is rejected the presenter of plan is shot and the next older robber gets a chance. So what plan should youngest robber present to save his life?
Labels: logic
posted by Rajesh Lal at
7:49 AM
30 Comments:
Because two of the oldest realise, that the most profite for them is divide coins just between each other, they will vote against any time. From the condition of the problem I see, that if they do it, the youngest robber suggestion will be rejected. If so, his sole proper desision is to give all money to the oldest two persons 500/500.
quantense, you're forgetting the last step: if the oldest two are left over, the oldest can reject the second oldest's division, shoot him and keep everything for himself. So he would possibly reject the youngest's plan to give the oldest two equal parts, hoping for more. I think it would be wiser to bribe the 2nd and 3rd youngest with something like 200/800 respectively. the 2nd youngest will be happy to get anything at all, since he'd have to give everything away to save his life when it's his turn.
600, 300, 99, & 1 oldest to youngest
Hi mo. Yeah, I think your objection makes sense. The thing is that for the second youngest person the most profit thing is to vote against any time except the youngest gives him all coins. If they kill the youngest after that, the 2nd youngest will vote for all money for himself. 3rd and 4th won't vote against him, because if they do it, among the rest 3 robbers the oldest will vote anytime against, kill everuone and take all coins.
The youngest should give 500 or more to the second oldest, and divide the remainder up between the rest except for the oldest. The second oldest is guaranteed at least 500 regardless of him rejecting or accepting, so he might as well accept. the 3rd and 4th oldest must accept. so theres a 3 to 1 vote.
I know im probably missing something here...???
Wasn't there a problem just like this a few months ago, but with pirates?
The youngest should say, "Ive figured this out. Let me show you how I am going to pass these coins out." He then puts his hand in the bag, and draws a gun, and shoots the other four guys.
Very fun, Ragknot. Have you thought about scenario writing?
Ok, i'm a little bit drunk, but lets think this though. we presume that all robbers are perfectly rational and have full information, which means that everyone knows that everyone else is perfectly rational and have full information. furthermore we should presume that their foremost priority is to survive, their second priority is to get as much cash as they can. as a third-rate priority and providing it does not affect their first two priorities, they could i) be completely indifferent towards the fate of the others, meaning that it would be a fifty-fifty chance that they let someone live or die, or ii) rather let someone live (for ethical or nostalgic reasons, or because bullets cost money, too), or iii) rather have others killed , 'cause they're sadistic swines.
Situation D: If only the oldest (Mr. 1) and second oldest (Mr. 2) were left, then Mr. 2's only hope of surviving would be to give Mr. 1 all the money (and in the case of iii even then he would die).
Situation C: That means that if only Mr. 1, Mr. 2 and the third oldest (Mr. 3) were left, then Mr. 2 will agree on any division Mr. 3 proposes (in the the case of i and iii to secure his survival, in the case of ii because he would rather save Mr. 3). That puts Mr. 3 in the happy position to be able to propose that he gets 1000 gold coins and Mr. 1 and Mr. 2 get nothing.
Situation B: That means that the forth oldest (Mr. 4) knows that unless he gives everything to Mr. 3 (and in the case of iii even then), Mr. 3 will vote against his proposal. He also knows that Mr. 1 and Mr. 2 have no hope of getting anything unless he proposes so and his proposal is accepted. so he can buy their approval for 1 gold coin each and keep the rest.
Situation A: Now we come to the youngest: Mr. Smith. In the case of ii, Mr. Smith can offer Mr. 1 OR Mr. 2 one gold coin and keep 999 coins. Mr. 3 and whoever he gives the coin will vote for his proposal to save his life and because it is no disadvantage to them compared with Situation B. In the cases of i and iii, he would have to offer Mr. 1 or Mr. 2 two gold coins and Mr. 3 one coin to be sure of their approval. That means he gets to keep 997 gold coins.
That much to game theory, in reality if he proposed to keep 997 coins and hand out 3, he would almost certainly be shot, due to certain psychological factors. So in practice I would sick to my first suggestion: give Mr. 4 200 and Mr. 3 800, of the four remaining robbers they have the strongest incentive to have the situation resolved.
mo, very nice answer, even so, I'm going to nit-pick. In situation C, 3 would give 2 $1 to absolutely guarantee his vote. Otherwise 2 might veto just for the fun of it. 1 still gets $0.
So 4 should keep $997 and give 3 $0, 2 $2 and 1 $1, 4 would be rich and definitely alive, because 1 and 2 will definitely vote in 4's favour. Their blood lust can be bought for an incredibly low price.
PS I've updated "A flipping fortune" after seeing your post. Greetz.
the youngest should hide the money then tell the other 4 that if they don't agree with splitting the cash $200 each they if they shoot him no one will see a dime
i think he shud take the money and RUN RUN RUN !!
At each round the presenter of his plan will be shot, regardless of what he proposes, so that there will be one less robber to share the loot.
The oldest will then end up with the full 1000.
Wiz - I hope you're joking or drunk.
I've gone mad. I've only just noticed that there were 5 robbers. I thought there were only 4. So the youngest has two options: 995,0,0,3,2 or 997,0,1,0,2; the youngest would choose the latter.
A table of results, 0 padded for layout reasons:
000,000,000,0,1000 robbers 5+4
000,000,999,1,0000 robbers 5+4+3
000,997,000,2,0001 robbers 5+4+3+2
997,000,001,0,0002 robbers 5+4+3+2+1
The credit goes to mo, for getting me started. The problem is easy when you do it backwards.
If there had been 6 robbers (so 3 others need to agree),
then would have got 996,0,1,2,1,0
Chris i think you should read the question again. I dont think the youngest has a vote, it is a vote among the remaining 4. If there is a 2:2 split decision the youngest gets shot.
The only hope for the youngest is to offer the 2nd oldest an amount over 500, and 3, 4 and himself any amount over 0, and the oldest nothing.
this way 2,3, and 4 will all vote for this proposal, the oldest will vote no, but that doesnt matter.
Hi Larry. I read it as the proposer automatically votes in favour of his own proposal. I think you've done the same as I did and misinterpreted the '2 out of 4 isn't a majority' guideline. I'll give you that the question doesn't make it clear that the proposer can or can't vote for himself.
If you are right, then with three robbers, 3 would have to offer 5 $1000 to get his vote, but then 4 would probably vote against you (just for the hoot of it). So 3 would die. So that certainly makes the problem interesting, because number 3 won't want to vote in such a way that only three robbers would be left - that would be suicide.
I can't spare any more time on this for about the next 6 hours. I think that what you suggest may be soluble. I've already been too long in this reply.
Can you prove that your suggestion is logical (with a clear majority vote and with the proposer unable to vote)?
Since the youngest needs a majority rule in order to live, he should divide the money between the older 3 in order to spare his own life. That way they will agree and he will live.
... no, they won't agree to that.
The last Anonymous has suggested that the proposer may not be allowed to vote. So just for the heck of it, here's what would happen:
000 000 000 000 1000 4+5
000 000 DIE XXX XXXX 3+4+5
000 999 000 001 0000 2+3+4+5
996 000 001 002 0001 1+2+3+4+5
4+5 only as before, 4 has to give all the loot to 5. With 3+4+5, 3 has had it, he can't win two votes, as 5 wants the whole $1000. With 2+3+4+5, 2 doesn't have to pay 3 (as 3 isn't suicidal). 4 will be happy to get the $1 as it's better than $0. So that's 2 of 3 votes. With 1+2+3+4+5, 3, 4and 5 will vote for 1 as they'll all be $1 better off than if they voted aginst him. So that's 3 of 4 votes.
... sorry, it was Larry who originated the suggestion about excluding the proposer from voting.
Chris, I really am convinced that in situation C Mr. 3 can keep all the loot. Mr. 1 will vote against him, but Mr. 2 HAS TO vote for him because otherwise i) there is a 50% chance of Mr. 1 shooting him just for the heck of it, or ii) he's not going to get any money anyway and might as well let Mr. 3 live, or iii) there is a 100% chance of Mr. 1 shooting him just for the heck of it! (i, ii and iii as described in my preceding post.)
I've just reread my post and found that I made a mistake in situation B in the case of ii (the generous "if having someone killed isn't to my advantage, I might as well let him live" case). In that case, Mr. 4 doesn't even have to give anything to Mr. 1 or Mr. 2, they'll vote for his proposal just to save his life. That means that in situation A, Mr Smith (ehr.., Mr. 5) can keep ALL the loot -- but only in the case of ii, in the cases i and iii he still has to hand out three gold coins.
Interesting, that a tiny bit of soft-heartiness on behalf of 5's follow robbers would allow him to claim the whole prize.
Hi mo. I'm happy to try the problem with any rules. I certainly don't insist that I'm right, as you can see from my last post(s). So for completeness I'll present the other two tables. (I wish I'd done the tables in the other order).
So if the proposer won't be voted against by someone who has nothing to gain,
If the proposer can vote:
____ ____ ____ 0000 1000 4+5
____ ____ 1000 0000 0000 3+4+5 (4 being nice)
____ 1000 0000 0000 0000 2+3+4+5 (4 and 5 being nice)
1000 0000 0000 0000 0000 1+2+3+4+5 (3, 4and 5 being nice)
If the proposer can't vote:
____ ____ ____ 0000 1000 4+5
____ ____ 0000 0000 1000 3+4+5 (4 being nice let's 3 live)
____ 1000 0000 0000 0000 2+3+4+5 (3 and 4 being nice)
1000 0000 0000 0000 0000 1+2+3+4+5 (3, 4and 5 being nice)
I hadn't expected that. I also think that completes all variations on the puzzle due the different interpretations of the ambiguities.
I don't necessarily with the solutions given. We don't know the personalities of these guys. If anyone gets less than 1/5 he can say he's not happy and will not vote in favor of the proposition.
If one guys buddy gets less the he thinks he should have got, he can all it fair.
Let's say that one guy gets 100%, he could possibily realize that it's not fair, and kill the youngest guy so they all get 1/4 instead of 1/5.
There's just too much unknown to think that just a few coins difference will change things.
Hi Ragknot. If they behaved like that, I don't think the puzzle could be solveded in a unique way - it'd be very much a matter of opinion as to the right answer.
The robbers were certainly "as thick as thieves". Sorry :<
Assume each robber wants to maximise own personal haul
Assume no robber wants to die
oldest robber = no. 1
youngest robber = no. 5
Working backwards from the case where only 1 robber is left
Case 1
1 robber left
no. 1 gets 1000
Case 2
2 robbers left
no. 2 knows what will happen in case 1
number 2 presents this plan knowing 2 votes will give majority
no. 1 gets 1000
no. 2 gets 0
both vote for it.
any other and 1 will reject and 2 dies
Case 3
3 robbers
no. 3 know what will happen in case 2
no. 3 presents this plan knowing two votes will give majority
no. 1 gets 0
no. 2 gets 1
no. 3 gets 9999
no. 2 is better off than in case 2 so he will vote for it
no.3 is better off than in case 2 so he will vote for it giving a majority
Case 4
4 robbers
no. 4 presents plan knowing what will happen in case 3
he knows that 3 votes are needed to get majority and avoid case 3
no. 1 gets 1
no. 2 gets 2
no. 3 gets 0
no. 4 gets 997
1,2 are better off then case 3 and so will vote for it
4 will obviously vote for it as presenter (wants to maximise gain)
Case 5
5 robbers
no. 5 presents plan knowing what will happen in case 4
he knows that 3 votes are needed to get majority and avoid case 4
no. 1 gets 2
no. 2 gets 0
no. 3 gets 1
no. 4 gets 0
no. 5 gets 997
nos. 1&3 are better off than in case 4 so will vote for it
no. 5 will obviously vote for it giving a majority.
I think this is right. However it assumes that all robbers are master logicians.
In which case they probably wouldn't be robbers!
Sorry if this is a few days late, but just found the riddle.
There are a few things the little guy can do to survive. It's highly likely that nobody will agree with these, but hey.
1.
Put out a phrase like "Hey, were all in this together right? Lets just split it evenly between us all, and just go steal more!"
The others could either keep the successful group of robbers together, and maybe have a long successful career ahead of them. Or shoot him.
2.
He could suddenly realize that, with his life flashing before his eyes, he shouldnt have been a robber. This in itself probably wont help, but it would make him change his ways if he escaped.
3.
He could ask to sleep on it for some ideas. Im sure these kind robbers that shoot each other when they dont have a majority vote will agree, and when hes asleep he can steal the money, get plastic surgery, a name change, a sex change might not go amiss either...
4.
He could request to use a portion of the money to buy booze for everyone, stay sober, and convince them to play Russian Roulette or run away while theyre drunk.
5.
Throw some of his own money into bribing the others into voting in his favor.
6.
Petition to change the shooting idea.
7.
Vote to stop picking on the youngest and have someone else go first. However, this would probably not sit well with everyone else and he would end up shot.
Thats everything ive come up with. You dont actually need to use super intelligent mathematical formulas or logic to solve some of these problems.
Different strokes for different folks. Peace.
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