Pirate Game Theory Puzzle

Tee Jet Whaw

Five pirates of different heights have a treasure of 100 gold coins, which they plan to share amongst themselves. They decide to split the coins using the following scheme:

The tallest pirate proposes how to share the coins, and ALL pirates (including the tallest) 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 is thrown overboard, and the process is repeated with the pirates that remain.

Since pirates are generally known to be bloodthirsty, if a pirate will get the same number of coins if voting for or against a proposal, they will choose to vote against so that the proposer is thrown overboard.

Assuming that all 5 pirates are intelligent, rational, greedy, and do not wish to die, what will happen?

Scroll down to reveal the solution

.

.

.

.

.

.

.

.

.

.

Solution

We start by numbering the pirates from 1 (tallest) to 5 (shortest) and work backwards through the problem.

If pirates 4 and 5 are the only one remaining, pirate 4 can simply propose that they have all 100 coins and given only one vote is required for the plan to triumph, they can vote in favour and receive all of the treasure. Pirate 5 can do nothing to prevent this.

Outcome for two pirates (4, 5) = (100, 0)

If there are three pirates remaining, pirate 3 needs only give pirate 5 a single coin and they will already be better off. Pirate 4 knows that they will receive 100 coins in the next round of voting and so will vote against any proposal made here. Therefore, pirate 3 needs to only sacrifice 1 coin to pirate 5 to ensure a winning vote and does not need to give anything to pirate 3.

Outcome for three pirates (3, 4, 5) = (99, 0, 1)

With four pirates remaining, the key vote lies with pirate 4. Since they are intelligent and rational, they know that if this proposal fails and only three pirates remain they will end up with nothing (and most importantly can do nothing to stop this). Therefore, pirate 2 only needs to give 1 coin to pirate 4 to ensure their vote and therefore a winning proposal.

Outcome for four pirates (2, 3, 4, 5) = (99, 0, 1, 0)

We have finally worked our way back up to the original problem. We can see from the above scenario with four pirates that pirates 3 and 5 can be most easily bribed. If we allocate a single coin to each of them, they will vote in favour of our proposal (since the alternative sees them receive zero coins without the ability to change anything). The final outcome is therefore:

Pirate 1: 98 coins
Pirate 2: 0 coins
Pirate 3: 1 coin
Pirate 4: 0 coins
Pirate 5: 1 coin

As an extension problem, can you say something about the general case of n pirates? And if so, can you prove it by induction?

Scroll down for some hints (if needed)

.

.

.

.

.

.

.

.

.

.

Hint 1: Write out the above solutions in a table and try to look for a pattern. You may find it helpful to add more rows.

Hint 2: Formulate a proposition with the pattern found.

Hint 3: Try proving the proposition by induction. If you find your proposition to be wrong, can you modify it so that it works?

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s