Tuesday, July 10, 2007
5 Pirates and the gold problem
Five pirates have 100 gold coins. they have to divide up the loot. In order of seniority (suppose pirate 5 is most senior, pirate 1 is least senior), the most senior pirate proposes a distribution of the loot. they vote and if at least 50% accept the proposal, the loot is divided as proposed. Otherwise the most senior pirate is executed, and they start over again with the next senior pirate. What solution does the most senior pirate propose? Assume they are very intelligent and extremely greedy (and that they would prefer not to die).
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19 comments:
first pirate will distribute it as 98 1 0 1 0 ..
No da its not..
32 32 32 2 2. is that right ??
no..
31,30,28,18,13..is it??
nope :)
The distribution will be 30,25,20,15,10.. The multipla of 5 plus 5 coins...
Is it right?
no its not..
it will be 34,0,0,33,33
i think anupam is almost right
but the order should be different
34,33,33,0,0.
no tats wrong..
k man .. i will tell my logic
you point out the error in it...
The last pirate always has a possibility of getting 100 coins so he is never gonna rise his hand ..however if a situation arises where the p2(second last) has to make a decision he wud hav to give 100 coins to p1 to make him rise his hand ..so p2 wont get a single coin.. now if you consider p3..when he is deciding he need to give p2 only 1 coin so he ll get 99 coins and unless he get 99 he wont lift his hand.. now p4 has to give 99 to p3 aand 1 to p2 to make them rise the hand ..so he will get only 0 coins
knowing all these the most senior pirate realizes that he need to give p4 and p2 only 1 coin each to make them support him and he takes the remaining 98..
dont disclose the answer completely .. just tell where my fault is
PLS gimme time to think..dont give the logic of answer
i think division will take place as
40,30,20,10,0
no its not..
There's no way to get the last 2 pirates in the favour of any proposal.
2 senior most pirates will have to die.
and then, the division will take place as 50, 50, 0
no.. tats not it... :)
When there are two pirates left, 50% of the vote = 1 which means the last pirate will get nothing. The last pirate knows this and will try to avoid this situation. If the situation ever gets boiled down to 3 pirates, the 3rd pirate knowing this as well, will distribute 99, 0, 1 thus swaying the last pirate's vote who will otherwise be left with nothing if he votes out the 3rd guy.
Argue it up to 4 pirates, the 4th pirate will distribute 99, 0, 1, 0 because he knows the 2nd guy being as smart as every one else would take this rather than getting stuck with nothing.
Come to 5 pirates, the 5th pirate knows the 3rd and 4th guys wont vote for him as they stand a chance to potential 99 so he distributes 98, 0, 0, 1, 1. The 2nd pirate knows this is the best he can get and will take it. The 5th pirate knows he will get atleast this no matter what. Being the pirate he is, he could chose to not vote for the 5th pirate so to make sure he gets a better deal, the 5th guy divides 97, 0, 0, 1, 2 thus giving a better deal to the last guy than anybody else would so that seems to work.
Hi,Can you please post the correct solution please?
Correct answer would be 1 0 1 0 98 from the youngest to the oldest pirate.
If you want an explanation let me know and I will post it.
Cheers.
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