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5 Bank Robbers have pulled off a daring heist and escaped with 100 gold bars.
Safe in their hideout they decide to split the loot using this scheme:
The eldest robber proposes how to share out the gold.
It is put to the vote and if 50% or more vote for it then they agree, otherwise he is shot in the face and the process is repeated with the next eldest and remaining robbers.
The eldest of the 5 quickly realises that this doesn't look good for him. Assuming these are the intelligent, greedy but rational type of robbers, what shares should he propose to avoid being shot in the face?
(importantly, the robber proposing the scheme doesn't get a vote!)
does someone else give them a his camel?
Eh? 20 each shoorely, or have I missed some crucial part of the puzzle?
I suspect the shares will look like this:
100% to the robber who proposed the crazy scheme, who is able to sneak out the back with all the gold whilst the other four are arguing about percentages and how old they all are.
If he offers 33.3% to each of two of the other robbers and takes the other third himself, two of the other robbers will vote yes, giving him his 50% of the vote and ensuring his survival and cut.
Unless we are dealing with particularly generous fair playing robbers...
I'd say the split goes like this:-
Oldest gets 16.66666 bars
Second oldest gets 33.333333 bars
Third oldest gets 50 bars...
The youngest two get nothing..
1st' guy offers a split 96/1/1/1/1
that's if they all have to have something
[i]Eh? 20 each shoorely, or have I missed some crucial part of the pu
zzle?
[/i]
Game theory omnitel. With that suggestion they all vote no and then only have to divide between four.
The robbers will always vote against anyone older unless they think they'll get shot. The youngest always votes no as he'll never get shot. So presumably the answer is to propose as many bars as possible to the second and third eldest, since they need to vote yes for the first guy to live....
CharlieMungus is closest so far
Nah, the second guy will take pretty much anything he is offered, if he doesn't then he will get shot next, and so on.
Omnitel should be 'innit' - silly phone.
you mean 100/0/0/0/0
Ah, just read it again and I understand now (well, sort of).
No idea about the answer though!
Hint: the eldest robber not only avoids being shot in the face, but he does pretty well out of his proposed scheme.
He shoots all the others and takes the lot? 🙂
vote [b]for[/b] it
So 50% or more have to vote 'yes', or just have to agree?
Cos if the oldest says he's keeping it all, they will all say no, but will all have agreed on 'no', so he gets to keep it.
If they have to vote yes then I don't know
Working youngest to eldest I'd reckon
90/4/3/2/1
or
0/0/40/40/20
It's not a trick question, just game logic.
The other robbers just vote whether they accept the proposed shares.
They are all remarkably adept at thinking through the options and make the most logical choice that leaves them with the most gold and a non-shot face.
97/2/0/1/0
The way I see it
When there are 2 left, the 4th guy is stuffed, he's gonna get shot unless he offers the 5th guy everything. So, the most he's gonna make here is zero.
So when there are 3 guys, he's going to vote for anything in which he gets more than 0, i.e. 1.
Which means that the 3rd guy can offer him 1 and the last guy nothing. so, he stands to get 99 bars.
So when there are 4 guys, Unless he gets 99 bars he's not going to go for it. Which means the 4th has to offer him 99, and keep 1, which means the 2nd guy will be happy with anything more than 1, so, the 1st guy offers him 2 and the 4th guy 1, keeps 97 an everyone else can whistle
It all depends if the bus falls off the cliff or not.
I see.
As long as there are people younger than you, they will always in theory vote no to reduce the number of players and therefore in theory increase each remaining player's share.
So you don't ever want to be the oldest. So in order to keep your head on your shoulders you have to vote yes to whatever the first guy says otherwise you'll get shot.
Based on the assumption that your life is worth more than 100 gold bars. The proposer is the one in the firing line remember. So he is going to do whichever one ensures he will not get shot and NOT result in an iterative elimination of players.
Charlie.. I was just going along similar lines, but..
98/1/0/1/0
For the 2nd guy to survive, he'd also have to give away all the gold (99 to 3rd guy and 1 to 4th guy), so he'd accept anything greater than 0...
Yeah, Charlie has it - start at the end game and work back.
Are we back?
We're back:
I think you've more or less got it Charlie. The answer I have in front of me is:
From youngest to oldest:
Robber1: 2
Robber2: 0
Robber3: 1
Robber4: 0
Robber5: 97
See http://www.mathsisfun.com/puzzles/5-pirates-solution.html
But how will the plane get off the conveyor carrying all that gold?
Shoot the other four leaving one with 100% share.
Simple!
😈
but if they were genuinely intelligent they would work together to maximise their gain, not selfishly.
yeah, i sent a correction, but it disappeared into the STW fermata, i was working on needs more than 50%,
Is there one gun or are they all 'carrying'? How many rounds do they have?