How can the pirates divide the treasure?
Two pirates want to evenly split the bag of gold coins they just acquired. The coins are all of equal value, and they initially figure they can just take turns removing one coin at a time. The problem with this method is that if the number of coins is an odd number, the person who goes first will always get the extra coin.
Without bothering to count the coins, what small modification can be made to this procedure so that if the number of coins is even they again each get the same amount, and if the number is odd, they each have an equal chance of getting the extra coin?
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6 Answers
If there is a coin left over when there is an uneven amount of coins, just flip the last coin to let chance decide who gets the last coin.
[The following response is unhelpful.]
Did passenger ships and trading vessels ever really sail around freely carrying bags and chests full of gold, with no security, or is that all just part of the lore?
The chatbot I asked this question to came up with this solution:
One possible solution is for the first pirate to take one coin, and then for the second pirate to take two coins, and then for the first pirate to take two coins, and so on. This way, if the number of coins is even, each pirate will end up with the same number of coins, and if the number of coins is odd, they will each have an equal chance of getting the extra coin.
No idea if this works.
Yes that is what I had in mind. Did a chatbot really come up with that? I am impressed.
You can see how it works. After each person’s turn, that person is ahead by 1. At the end, if the person can take just 1 coin then the amount is even and they each have the same amount. Otherwise the last person to choose has the extra coin.
Here is a variation of the problem. Suppose there are a lot of coins and they decide to take 10 at a time. If they just alternate taking 10 coins, the first person will be ahead by as many as 10 coins and they will only have the same amount if the number is divisible by 20. If the first person takes 5 coins, and after that they alternate taking 10, the odds are the same for each and neither can get more than 5 coins than the other.
The problem can also be generalized to any number of pirates. The idea is to alter the order of choosing after each round. A simple way of keeping track is to go around in a circle clockwise and have the person who chose last in the current round be the first to choose in the next round. For 3 pirates, number them 1 to 3. The order of choosing for the first 3 round would be (1,2,3),(3,1,2) and (2,3,1) and then things would repeat.
You can have one pirate divide the shares with a stick, or however they like, and then have a pirate who isn’t his best friend pick which share they want. If anyone still thinks the shares aren’t fair, they can take a turn trying to make fair shares, and someone else can pick which to take. Further disputes can be solved by bartering chores, or arm-wrestling, or gambling.
Math “tricks” are unlikely to be satisfying to pirates, unless they’re all mathematicians or otherwise able to be understood and accepted by every pirate – good luck with that!
@LostInParadise No, the AI did not come up with the answer itself. If found an answer that a human had written, and offered it.
@Jeruba There were ships with chests full of gold, but they had locks and armed men on them.
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