Dividing up the loot… again!
Last week, we met 5 pirates with 76 gold coins. Here’s an interesting tweak on the same puzzle…
Now there are 32 pirates, still strictly ordered from most senior to most junior, but this time they have uncovered a stash of only 10 indivisible gold coins that they must divide amongst themselves.
Exactly as in the previous puzzle, these pirates are rational and want to survive, know that the other pirates are rational too, and each wants as much gold for themselves as possible. Also, being pirates and not very nice, they will always prefer a fellow pirate die if they still get as much gold.
Also as in the previous puzzle, the pirates have agreed to divide the coins using the following method:
- The most senior pirate alive proposes a split, and all the pirates vote on it
- If 50% or more of the pirates vote for the proposal then the gold is split as proposed
- If less than 50% of the pirates vote for the proposal then woe betide the most senior pirate who proposed the split, because he must walk the plank and the newly most-senior pirate begins the process again
- The pirates will abide by this agreement
- The pirates will make no other agreements
Which is the most senior pirate that will survive, what will he propose and why will it work?
(Hint: The most senior pirate of all the pirates that can survive is a lucky man – there are several possible proposals that he can make which will keep him alive)