Rational Pirates Revisited
There are
pirates
with the strict order of superiority
.
They find 100 gold coins. They must distribute the coins among themselves according to the following rules:
- The superior-most surviving pirate proposes a distribution.
- The pirates, including the proposer, then vote on whether to accept this distribution.
- In case of a tie vote the proposer has the casting vote.
- If the distribution is accepted, the coins are disbursed and the game ends. If not, the proposer is thrown overboard from the pirate ship and dies (leaving behind pirates and their original ordering), and the next most superior pirate makes a new proposal to begin the game again.
Every pirate knows that every other pirates is perfectly rational , which means:
- They will always be able to deduce something logically deducible.
- Their first preference is that they want to survive.
- Their second preference is that they want to maximize the number of coins they receive.
- Other things remaining same, they prefer to kill a pirate rather than having him alive.
What is the minimum value of such that the superior-most pirate can propose no distribution of coins that will let him survive?
The answer is 203.
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If we take the famous puzzle of 5 pirates, let the 5 pirates be A,B,C,D and E. with superiority order be A>B>C>D>E. I'm using Y for Yes for the proposal by most superior pirate and N for No.
In this puzzle we work backwards (bottom up approach). If we had only D and E, D proposed 100 for himself and 0 for E. As has the deciding vote so he gets all 100. Vote=[Y,N]. Allocation=[D:100 , E:0]
If there are three left, C, D and E. C and E know that D will offer 0 coins to E if C is killed. Thus C offers 1 coin to E and 99 to himself. E also says yes, as it's the only option to get a coin.Vote=[Y,N,Y]. Allocation=[C:99, D:0, E:1]
If B, C, D and E remain. B knows the above allocation, So he decides to give 1 coin to D to get his vote. Thus, he proposes [B:99, C:0, D:1 , E:0]. He could give 1 to E, but E knows he won't get more than 1 even with C, so he could get B killed. B wants to live so... Vote=[Y,N,Y,N]. Allocation=[B:99, C:0, D:1 , E:0]
And since A know all these allocation lists, he will now take the odd numbered pirates with him. He proposes [A: 98, B0, C:1, D:0 , E:1]. C and E will say yes as they know if they get A killed, B takes it all giving them 0 coins each. Vote=[Y,N,Y,N,Y]. Allocation=[A: 98, B0, C:1, D:0 , E:1].
As you can see there is a pattern in the vote casted and thus the allocation list. So if we have say N=100 pirates, then each even numbered pirate would get 2 coins each( Vote=[Y,N,Y,N...] Allocation=[2,0,2,0...]).
And if we had say 200 pirates, then each even numbered pirate would get 1 coin each( Vote=[Y,N,Y,N...] Allocation=[1,0,1,0...])
N=201, if pirate 201 is the most superior, he can stay alive only by offering all the gold to odd-numbered pirates, keeping none for himself.( Vote: 101 Yes > 100 No and allocation: 0 to pirate 201, 0 to 200th, 1 to 199th, 0, 1, 0,...)
N=202, if pirate 202 is the most superior, he can stay alive only by offering all the gold to even-numbered pirates, keeping none for himself.( Vote: 101 Yes = 101 No but 202 has the deciding vote so he lives. and allocation: 0 to pirate 202, 0 to 201th, 1 to 200th, 0, 1, 0,...)
N=203, if pirate 203 is the most superior, there isn't enough gold available to have a majority. He has only 100 coins, so only 100 will say YES + 1 for himself, As 101<102, so he will die. Hence answer is 203.