Screwy Pirates

Problem

Five pirates looted a chest full of 100 gold coins. Being a bunch of democratic pirates, they agree on the following method to divide the loot.

The most senior pirate will propose a distribution of the coins. All pirates, including the most senior pirate, will then vote. If at least 50% of the pirates accept, the gold is divided as proposed. Otherwise, the senior pirate is fed to the shark...

How will the gold coins be divided in the end?

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Hints

Hint 1

Pirates are perfectly rational: they will do anything to survive, then maximize their gold, and prefer fewer pirates on board if everything else is equal.

Hint 2

First solve what happens with only 1 pirate. Who gets what, and why? And then go for 2,3,4, and 5 pirates.

Hint 3

Work backward from the simplest scenario. What happens to each pirate if the current proposer is thrown to the sharks?

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Solution

Show Solution

Let the pirates be denoted as $P_1, P_2, P_3, P_4, P_5$ with $P_5$ being the most senior and $P_1$ the least senior. The pirates follow these preferences:

1. Stay alive
2. Maximize coins
3. Prefer fewer pirates if indifferent

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Case $n=1$:
Only one pirate:

• $P_1: 100$

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Case $n=2$:
$P_2$ proposes, needing 50% of 2 votes (1 vote).
They vote for themselves:

• $P_2: 100$
• $P_1: 0$

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Case $n=3$:
$P_3$ proposes, needs 2 out of 3 votes.
Looking ahead if $P_3$ is thrown to the sharks, from the $n=2$ case:

• $P_2: 100$
• $P_1: 0$

$P_1$ would get 0, so $P_3$ can offer $P_1$ one coin, which is better than zero:

• $P_3: 99$
• $P_2: 0$
• $P_1: 1$

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Case $n=4$:
$P_4$ proposes, needs 2 out of 4 votes.
If $P_4$ is thrown overboard, then from $n=3$:

• $P_3: 99$
• $P_2: 0$
• $P_1: 1$

$P_2$ gets 0 in that scenario, so $P_4$ can buy their vote with 1 coin:

• $P_4: 99$
• $P_3: 0$
• $P_2: 1$
• $P_1: 0$

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Case $n=5$:
$P_5$ proposes, needs 3 out of 5 votes.
If $P_5$ is thrown overboard, then from $n=4$:

• $P_4: 99$
• $P_3: 0$
• $P_2: 1$
• $P_1: 0$

$P_3$ gets 0, $P_1$ gets 0, so $P_5$ can buy their votes with 1 coin each:

• $P_5: 98$
• $P_4: 0$
• $P_3: 1$
• $P_2: 0$
• $P_1: 1$

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Therefore the final division for $n=5$ is:

$$P_5: 98, \quad P_4: 0, \quad P_3: 1, \quad P_2: 0, \quad P_1: 1$$

Follow-up Question

Rather than 5, what about $n$ pirates?

As observed from the solution above, the pattern is to look at the previous case $(n-1)$, identify which pirates would otherwise get zero coins, and bribe them with one coin to secure their vote.

Therefore, for general $n$, the most senior pirate $P_n$ will keep as much as possible, distributing coins as follows:

$$P_n: 100 - \left( \left\lceil \frac{n}{2} \right\rceil - 1 \right)$$

and then assign 1 coin each to

$$P_{n-2},\ P_{n-4},\ P_{n-6},\ \dots$$

All other pirates receive 0.

And if you want to prove this more rigorously true for all cases, I suggest using induction as a starting point.