Saturday, May 23, 2015

Sailor's Delight

10 pirates are ranked in order, first to last. After finding a treasure chest of 100 gold coins, they are discussing how to divide up the booty. They allow the lowest ranked sailor to divide up the coins and then vote on his idea. If the number of pirates who like the division is equal to or greater than the others who don't like it, then the boss will say, "Make it So." (The proposer of the idea also has a vote.)

Otherwise... well, being pirates their simple solution is to dump the unfortunate sailor into the deep blue sea and let the next pirate in line decide how to divide up the spoils.

pirates and coins game

How many pirates will be thrown into the sea?

 

Notes:
  • Pirates are smart, want money, and love life, especially their own!
  • This one is harder than average. If you are stuck, think of fewer pirates...
  • Why would #1 ever vote for any schemes?
  • Why would #2 ever vote for any schemes?
  • ... hmmmm

Sailor's Delight Puzzle Solution

We asked you readers to send us a solution to this puzzle, and we kept an open mind about it.
The first person with a simple, elegant, and, in our opinion, valid solution, is Saurabh Gupta, a faithful reader and avid puzzler. Here is Saurabh's solution.

How many pirates will be thrown into the sea? None. And the correct distribution is:
Pirate
with rank
Number
of coins
1 0
2 1
3 0
4 1
5 0
6 1
7 0
8 1
9 0
10 96

Pirate 10 divided the coins. He will get the votes of pirates 2, 4, 6, 8, and himself. This is taking into consideration what each of the pirates will get if this plan is not passed.

Starting with a situation when there is only pirate 1. He keeps all the 100 coins for himself and live happily by passing the division with his only vote.

In case that there are two pirates, pirate 2 divides and he keeps 100 coins for himself while giving none to pirate 1. He still gets the division passed with his vote and live happily ever after.
In case there are three pirates, pirate 3 divides and gives pirate 1 a single coin and keeps the other 99 coins for himself. Pirate 1 would now vote in his favor because if he votes against, then pirate 2 would get a chance to divide and would keep all the loot for himself.

If four pirates are present, pirate 4 divides and now gives pirate 2 a single coin to gain his vote (who otherwise gets nothing if pirate 3 has a chance to divide the coins). In this case, pirates 1 and 3 get nothing.

Therefore, in a similar manner, the distribution when there is an extra pirate is achieved as follows:

With 5 pirates,
pirate 5 divides:
Pirate
with rank
Number
of coins
1 1
2 0
3 1
4 0
5 98
With 6 pirates,
pirate 6 divides:
Pirate
with rank
Number
of coins
1 0
2 1
3 0
4 1
5 0
6 98
With 7 pirates,
pirate 7 divides:
Pirate
with rank
Number
of coins
1 1
2 0
3 1
4 0
5 1
6 0
7 97
With 8 pirates,
pirate 8 divides:
Pirate
with rank
Number
of coins
1 0
2 1
3 0
4 1
5 0
6 1
7 0
8 97
With 9 pirates,
pirate 9 divides:
Pirate
with rank
Number
of coins
1 1
2 0
3 1
4 0
5 1
6 0
7 1
8 0
9 96
With 10 pirates,
pirate 10 divides:
Pirate
with rank
Number
of coins
1 0
2 1
3 0
4 1
5 0
6 1
7 0
8 1
9 0
10 96
This solution holds because all pirates are rational and think through the situation. They understand that they don't get anything if the next ranked pirate gets a chance to make the division.

There is, however, still some doubt about this solution. After all, it was said that pirates are smart. With this solution, based on a method of dividing the booty supposedly approved by all, it is shown that only one (the last one) is very smart, while 4 of them get only a coin, and 5 of them get nothing. This seems to be in conflict with the statement that pirates are smart. Even if the method was approved only by half of them (the ones who get at least a coin), 4 of these voters don't seem very smart if they get only one coin. Certainly, it couldn't have been a dictatorship decision taken by the leader, as he ends up empty-handed.

An alternative, to make the pirates look brighter, is that the dividing pirate actually shares the loot evenly between the pirates likely to give him a positive vote. In that case, the pirates who previously got only one coin, end up getting just as much as the lowest ranked, while the others get nothing again. This would entail the following solution:
Pirate
with rank
Number
of coins
1 0
2 20
3 0
4 20
5 0
6 20
7 0
8 20
9 0
10 20

This alternative solution, however, defies a bit the mechanism of the division logic and process explained in the main solution. Perhaps, it would be fair to make a re-wording of the puzzle, stating that, although pirates are smart, the method of dividing the booty is forced on them by some external entity, and no pirate, not even the leader, can oppose to it. However, a rewording of the puzzle will NOT take place, as this is how it was presented on Brent's page, and we are going to preserve it.

Also, it might be possible that this is not the only acceptable solution, and another alternative solution might exist, that doesn't raise any doubts with the puzzle premises.

So, we are still asking you: send us your solution!