A Hundred Men On A Dead Man's Chest

[alex_bell]

I found this one quite tricky.

100 bloodthirsty pirates have stolen a chest containing 100 gold pieces, and
they need to divide up the loot between them. Being greedy, argumentative, and
just incidentally all perfect logicians, they decide not to do this in the
obvious and fair way, but instead they vote on it.

The pirates form a line. Then the man at the head of the line makes a proposal
as to how the gold should be shared out. All the pirates vote on the proposal
(including the proposer). If at least half of the pirates vote Yes, then the
decision is made, the gold is shared out, and the process ends. But if over
half the pirates vote No, then the man at the head of the line gets his throat
cut, and the pirate next in line takes his turn to make a proposal, and so on.

On the reasonable assumption that every pirate would like to end up with as
much gold as he possibly can, but that every pirate would prefer to be alive
and penniless rather than dead, what happens?

Answer Below (Highlight to read)

Let's number the pirates 1 to 100, in such a way that pirate 100 is the first
one to propose a division of gold, and pirate 1 is the last.

If we were to reach the point where only pirates 1 and 2 remained, then pirate
2 would be able to get _any_ proposal accepted; so he'd simply propose awarding
himself all the gold, and pirate 1 would be unable to do anything about it.

Now suppose 3 pirates remain. Pirate 2 is going to vote against anything pirate
3 suggests, because he knows he can do better if the proposal is rejected.
Pirate 1, on the other hand, knows he will get no gold if pirate 3 gets the
chop, so pirate 3 need only award him 1 gp in order to get his vote. So pirate
3 awards 1 gp to pirate 1, nothing to pirate 2 (because if you can't please him
anyway, why bother trying?) and the remaining 99 to himself. This would be
voted in and the game would end.

If 4 pirates remain, then of course they have all reasoned out what will happen
if pirate 4 gets the chop. So this time pirate 2 will vote for anything that
gives him at least 1 gp; so pirate 4 can get the votes he needs by awarding 1
gp to pirate 2 and the rest to himself.

With 5 pirates: this time pirates 1 and 3 will vote for anything that gives
them a gp, so pirate 5 gives them each 1 gp and keeps the rest. With 6 pirates:
pirates 2 and 4 know they'll get nothing if pirate 5 gets to have a turn, so
pirate 6 awards them a gp each and they vote him in.

By this time we're starting to see a pattern:

Pirate 1 2 3 4 5 6 7 8

Pirate 2's proposal 0 100
Pirate 3's proposal 1 0 99
Pirate 4's proposal 0 1 0 99
Pirate 5's proposal 1 0 1 0 98
Pirate 6's proposal 0 1 0 1 0 98
Pirate 7's proposal 1 0 1 0 1 0 97
Pirate 8's proposal 0 1 0 1 0 1 0 97

Continuing this pattern to the end, we determine that the actual course of the game runs like this:

- Pirate 100 proposes giving 1 gp to each of the other even-numbered pirates,
giving nothing to any of the odd-numbered pirates, and keeping the remaining
51 gp for himself.

- All the odd-numbered pirates vote Against.

- All the even-numbered pirates vote For.

- The motion is passed, and nobody gets killed.

 

[Peter West]

Re: A Hundred Men On A Dead Man's Chest"

On my assumption that most pirates would have been educationally challenged, i would think it would take them so long to work that out that at least 20% of them would have died through other causes. 10% would have emigrated to England to take advantage of the NHS for fixing their peg legs, and 1% (the clever one) would have sneaked away with all the gold while the other pirates werent paying attention.

 

[TheKaiz]

A riddle for a completely hypothetical situation...

Personally I think the smallest pirate that hid in the corner and took the gold after all of the other ones killed each other would get 100 gold.

The meak shall inherit the earth.