arunshamli said:
Let me ask a question this time.
5 people have 100 gold coin to split among themsleves. They came up with a solution to divide the coins. the most senior memeber will propose a divison and everyone votes on it. if 50% or more member votes in favor of the proposal they will divide the coin that way, if not they kill the senior member and start all over agian and the most senior surviving member will have to propose the solution and they vote by the same rules, eithe divide the coins or kill the senior memeber,as the case may be. this process continues util one plan is accepted.
If You are the senior member ,What plan do you propose? your own vote will also count. everybody in the gang is extremly logical, greedy and they all want to live.)
Follwing is my solution scheme for ur problem.Please check it and tell me if i m right or wrong:
Lets number the persons from N1 to N5----N1 being the juniormost and N5 being the seniormost....
In the riddle,its given that these people are very very greedy and logical too......
So what the N5 can do.....Here is my answer.........
Lets apply induction.
Case1: only one person.
If there is only one person,then he will have all the share.
Case2: two persons
let it be N1 and N2.In this case N2 would want to take all the coins,and he has 50% votes (his own vote) for
doing this. This way N1 won't be getting anything.
Case3: three persons
let it be N1,N2, and N3.So in this case N3 needs support of one more member.N2 wont support him,as he knows that if he is able to get rid of N3,then only two persons will be left and he will be at gain(see case2).So N3 would lure N1 for 1 gold coin.since each of them is extremely greedy and logical,so N1 would think that 1 coin is better than nothing,so he would accept his offer.In this way,N3 will have 99 coins,N2 will have none and N1 will have 1.
Case4: four persons
Let it be N4,N3,N2,and N1. then N4 needs support of at least one member. But he wont get it from N3 (from case3) and recursively from N1. So the only person left is N2. Again, N4 will have 99coins,and N2 will have 1.
Case5: five persons
N5 needs at least 2 more people on his side,to get his plan proposed......If we think logically,like them, we know that N4 would want to kill N5,since he is the next seniormost.......So he will lure N3 with one coin.N3 knows that if N4 comes,he won't get anything,so he accepts that deal...Next comes N2...Now he also would want to kill N5,since he would get advantage then...So the one that will be left is N1.The N5 person can share 1 more coin with N1.
So the solution scheme can be:
Person# No. of coins
N5 98
N4 0
N3 1
N2 0
N1 1
Similarly, there can be more solutions,(if N5 is less greedy),then
Person# No. of coins OR
N5 97 97
N4 0 0
N3 2 1
N2 0 0
N1 1 2
