# Five Pirates and 100 coins

+1 vote
762 views

5 pirates of different ages have a treasure of 100 gold coins. On their ship, they decide to split the coins using this scheme:

The oldest pirate proposes how to share the coins, and ALL pirates (including the oldest) vote for or against it.

If 50% or more of the pirates vote for it, then the coins will be shared that way. Otherwise, the pirate proposing the scheme will be thrown overboard, and the process is repeated with the pirates that remain.

As pirates tend to be a bloodthirsty bunch, if a pirate would get the same number of coins if he voted for or against a proposal, he will vote against so that the pirate who proposed the plan will be thrown overboard.

Assuming that all 5 pirates are intelligent, rational, greedy, and do not wish to die, (and are rather good at math for pirates) what will happen?

posted Mar 6, 2014
Looking for solution? Promote on:

Similar Puzzles

There are 5 pirates, they must decide how to distribute 100 gold coins among them. The pirates have seniority levels, the senior-most is A, then B, then C, then D, and finally the junior-most is E.

Rules of distribution are:

1. The most senior pirate proposes a distribution of coins.
2. All pirates vote on whether to accept the distribution.
3. If the distribution is accepted, the coins are disbursed and the game ends.
4. If not, the proposer is thrown and dies, and the next most senior pirate makes a new proposal to begin the system again.
5. In case of a tie vote the proposer can has the casting vote.

Rules every pirates follows:

1. Every pirate wants to survive.
2. Given survival, each pirate wants to maximize the number of gold coins he receives.

What is the maximum number of coins that pirate A might get?

There are two candles. Both will only burn exactly for an hour.
How will you use these two candles to measure forty-five minutes?

+1 vote

An errand boy was collecting boxes of cakes for the summer fair. He collected boxes from various people in his local village and each box was labelled in Roman Numerals with the number of cakes in the box. By the time the errand boy has collected the last box. he was quite hungry, and really needed to eat at least one cake. Luckily the last box was marked with an underlined XI, meaning there was 11 cakes in it. He had the brain wave of turning the box around and underlining the number again to give the impression there were IX, that is 9 cakes. However, after eating the cakes 2 cakes, he was still hungry.
How can he change the number shown on the box again and eat more cakes?

+1 vote

You and a friend play a game where you start with a pile of 1000 stones, and each turn you can either add 2 stones to the pile or remove 10 stones.

The player to remove the last stone wins.

Should you start first or second to win this game?

+1 vote

A Blind-folded man is handed a deck of 52 cards and told that exactly 10 of these cards are facing up. He is asked to divide those cards into two piles, each with the same number of cards facing up.

He can't peek, get help, or damage the cards, but may use any strategy that occurs to him to do so.

How can he do it?

Note: Think out of box