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The 44th Riddle

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Hi,
 
I tried for several days to solve a puzzle that I found in an android application, without success.
 
This is the riddle number 44 of the android application : "66 Riddles" of which the following statement:


Five bankers are sharing 12 golden ingots. 
They decide to proceed that way : The elder one will suggest an ingots allotment. 
The rest will vote for or against it. 

If the majority accepts, the sharing is ratified.
If not, the elder will be dismissed. 
So, the sharing would be done between the remaining banker with the same rules. Knowing that they are set from left to right in a diminishing order of their ages, how would be the allotment ?


It remains for me than this to unlock the 66th riddle in the app, can anyone help me please?

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Assuming that the golden ingots cannot be cut/ sub-divided and that everyone wants to get fair share while maximizing his/her own, the best outcome is 2 people getting 3 ingots while 3 people get 2 each.

In this case, it is best for the first elder to suggest an allocation of 2, 3, 3, 2, 2. He will then win the votes of the next 2 elders and the allocation will be ratified.

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Assuming a tie vote of the remaining means its ratified

Spoiler

The oldest suggest 9,1,0,2,0

Simply if the it gets down to the last 2 the youngest will get all the gold so the youngest will always vote no. The second youngest knowing this will want to make sure that it doesn't get there so will vote yes for anything 1 or more with only 3 people left. So the third youngest would ideally suggest 11,1,0 split to get the split vote and 11 for himself. This leaves the second oldest with no possible options since the middle is expecting 11 gold and the youngest is expecting 12 so nothing her can do will please a majority of bankers and give him any gold. The oldest knowing all of this just has to make sure that the second and fourth youngest are getting more then they would if the first vote wasn't ratified. knowing that going further the 4th is destined for none and the second is destined for 1, he offers 1 and 2, respectively and can keep the rest for himself.

 

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Spoiler

6 0 1 3 2

 

 

Spoiler

Second possibility: 7 0 1 3 1 - the formulation of the problem is quite ambiguous.

 

Edited by harey

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Hi this is how i think about this problem.

There are 2 possible assumptions. A tie is ratified (1) or a tie is not ratified (2).

Starting with 2 bankers left, (1) 12-0 and (2) 0-12.

3 bankers: (1) 11-0-1, (2) 11-1-0. The banker receiving 1 will accept because when he doesnt he will receive nothing.

4 bankers: (1) 11-0-1-0, (2) 11-0-0-1

5 bankers: (1) 10-0-1-0-1, (2) 10-0-1-1-0

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I tried this two answers, but it's not working,

 

i think that if there are 2 bankers left,  whatever the solution, the younger will not accept, so he will take all the ingots. So the fourth one has to accept everything to not to remain two.

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After getting the answer it seems a tie is not ratified.

Defrak was really close to the answer but he made a small error...

4 bankers: 9-0-2-1 (since 3 is needed for a majority)

Therefore 5 bankers will be: 9-0-1-0-2

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Assuming that a tie is not ratified, the first elder can suggest the allotment as: 9-0-1-1-1, knowing that the second eldest one will never accept the allotment as once the first one is dismissed he is going to get the biggest share, but remaining bankers will accept the allotment knowing that 1 is the maximum share they are going to get. Youngest is always going to decline the share for maximum share but if he is smart he would know the moment the sharing goes to the second eldest he would go to cut him loose.

But if the 4 bankers vote down the first, the second eldest can suggest either of the three allotments: 10-1-1-0 or 10-0-1-1 or 10-1-0-1, all the bankers know that if they vote down this banker, the voting would come to third, third knows whatever he does the voting would go for a tie dismissing him, the forth knows that the sharing is the maximum he/she gets as if allotment comes to him he would not get anything

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