Five pirates have 100 gold coins. They have to divide up the loot. In order of seniority, each pirate proposes a distribution of the loot. All the pirates vote, and if at least half accept the proposal, the loot is divided as proposed. If not, the most senior pirate is eliminated, and they start over again with the next senior pirate. What solution does the most senior pirate propose? Assume they are very intelligent, extremely greedy, and interested in surviving.
Hint: Who said anything about fairness?
candidates quickly suggest something like, ‘The most senior pirate takes
half and divides the rest up among the least senior pirates.’ No, that misses
the whole point. Any response without a specific strategy behind it is
invalid. If I ask you why a senior pirate would give anything to a junior
pirate, I don’t want to hear ‘because he’s fair.’ There’s nothing about fairness
in the statement. I want to hear the leap that solves the problem. Once
I hear it, I usually don’t take the time for candidates to finish the puzzle.”
The key insight is that the senior pirate needs to get the cooperation
of the junior pirates (the same thing is required for the success of a development
project). In other words, what’s to stop the rest of the pirates from voting in a bloc against your plan and eliminating you? Then there
would be only four pirates to share 100 coins, not five.
The base situation is one pirate. Obviously he would keep all 100 coins
for himself. How about two pirates?
The most senior pirate knows that he needs to get two other pirates
to vote for his solution in order for him not to be eliminated. So why
would any pirate vote for him? The next most senior pirate would
surely see it in his self-interest for the senior pirate to be eliminated.
Let’s start simplifying. If there were only one pirate, there would
be no puzzle. The pirate would take all the loot, and no one
Now consider the situation with two pirates. Same outcome. The
senior pirate takes all the loot, and the other pirate can’t do a thing
about it as the senior pirate’s vote represents half of the voters.
It gets more complicated for the senior pirate when there are
three pirates. Let’s number the pirates from least to most senior:
1, 2, and 3. With three pirates, pirate 3 has to convince at least one
other pirate to join his collation. Pirate 3 realizes that if his plan
is not adopted, he will be eliminated, and they will be left with two
pirates. All of them know what happens when there are two pirates:
pirate 2 takes all the loot himself and pirate 1 gets nothing. So pirate
3 proposes that he will take 99 gold coins and give 1 coin to pirate
1. If pirate 1 has any self-interest at all, he really has no choice. If
pirate 1 rejects the offer, he gets nothing. So pirate 3’s plan will
pass two to one over pirate 2’s objection.
With four pirates, an even number, the senior pirate needs just one
vote other than his own to impose his will. His question now is which
one of the other pirates’votes can be exchanged for the fewest number
of coins. Pirate 2 recognizes that he is most vulnerable. Therefore, pirate 4 knows that if he gives pirate 2 anything at all, he will vote for it.
A rule is emerging here. In each case, the senor pirate should buy
only the votes he needs, and buy them as cheaply as possible. Apply
the rule to the five-pirate case. Pirate 5 needs two votes plus his own.
The goal is to toss a coin or two to pirate 1 and pirate 3, the two
pirates in the most vulnerable positions. Both will be empty-handed if the senior pirate is eliminated and four pirates remain. So pirate
5 offers one coin to pirate 3 and one coin to pirate 1. Pirates 2 and
4 get nothing.
Solution: The pirate offers one coin each to the pirates in the first and third positions.
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