Idealogian

At the end of the day ,I believe there are 2 pirates left. Each with 250 gold coins.

Nope.

(Trying to decide whether or not to point out why that isn't correct, but I think I'll leave it for now.)

(Also - the usual disclaimer of "no funny stuff" [e.g., a pirate who's been voted out can't try to stand up against those remaining], and, of course, all the pirates are intelligent enough to figure out what is best for them.)

definitions

first guy to propose = 5th pirate
last guy to propose = 1st pirate. (assuming they all would propose)

Let's assume there are 2 pirates.

the only way the proposal works is if the other pirate agrees. Of course he'd disagree (if he doesn't get all of it) as he'd just be the last one left.

so if 2 pirates. if proposer wants to live, he proposes to give it all to the other pirate. Then it's up to the other pirate to determine if he's willing to let the guy live. Since money makes him happy, he lets him live.

Similiar with 3 pirates.

here #3 needs one more vote besides his own. But since we already know the 2 person case, he can keep 499 for himself and give 1 to the 2nd guy. Since that 1 is better than what would happen otherwise to that guy, you get 2 yes votes. and it passes. guy #1 looses out totally. Though one could argue that since guy #2 totally looses out in next case, one could give him zero, but operating under assumption that if he's going to get zero, he's going to let people die, as lack of money makes him unhappy. so need to give him 1 coin to keep him happy.

now let's expand to 4. We know the 3 person case, so for a guy to vote yes, it has to be better than what would happen in a 3 person case.

the only way to improve #3's position is to give him it all, but that won't work. so #3 gets 0 this time around. #4 keeps 497 for himself, give 2 to #2 and 1 to #1. Since that's better than what they'd get in 3 person case, they're happy and vote yes.

now let's expand with same logic to 5 person case.

#5 keeps 495 for himself, gives 0 to #4, gives 0 to #3, gives 3 to #2 and 2 to #1.

since that's better for #2 and #1 than the 4 person case (where they would get 2 and 1 respectively) they vote yes, so we have 3 vs. 2 and they all live.

or did I overthink this?

small error.

5th guy can save more for himself.

since in 4 person case, #3 gets 0, he'd be happy with 1 coin.

so if give #3 1 coin and #5 2 coins, he can keep 497 for himself.

First 3 pirates split it evenly (as best you can divide 500/3). They win (as majority), and the last 2 can't do anything about it.

Good correction, Shaya (though I think in the last line you mean "#1" where you say "#5," given your backwards definition).

But you still lose three points for reading comprehension. Why three? Because that's how many coins you made #5 (the first pirate to suggest) lose unnecessarily. You can earn those points back by telling me where you were wrong.

Revision: #3, #4, and #5 vote against the first plan, then the second, so there are 3 guys left. #3 proposes splitting it evenly with #4, and they win. #5 can be assumed to always vote no, because he has nothing to gain/lose ... there will always be an alliance against him.

well, considering I said he can get 497, I'm not sure how he could get another 3 coins (that be all).

I guess you could save one more coin by just giving #1 1 coin, the same as he'd get in the 4 person case, but my logic is that you have to improve each time, otherwise the guys will just let the proposer be killed for being rude to them (no skin off their back, so to speak)

so unsure how the first proposer could do better than 497 (and especially by 3)

One more point ... whatever the final scheme is (and assuming it doesn't involve any coins for at least the first proposer), they could still all live, provided that #1 proposes that theory, knowing it is the eventual outcome. Especially in the scenario I proposed, #1, #3, and #4 could agree that #3 and #4 split it, saving everyone's life, and crediting the two who have the power to contain the cash.

"Though the pirates have been friends for years, business is business. A pirate won't hesitate to reject a plan if he knows he can get a better deal."

The clear implication (not in a symbolic logic sense, but certainly in a common-sense sense) is that a pirate will not vote against his friend unless he can get a better deal.

Thus, going through your thought process again, if there were only two pirates left, #2 would have to give everything to #1 in order to avoid getting killed.

Therefore, if there were three pirates left, #3 could keep everything for himself, and #2 would go along with it because he has nothing to gain, and thus no reason to vote against his friend, #3.

Therefore, if there were four pirates left, #4 could keep everything for himself, and #2 and #1 would go along with him because they have nothing to gain.

Therefore, when there are still all five pirates left, #5 can keep everything for himself, and #3, #2 and #1 will go along with him, because they have nothing to gain by voting against him.

(By the way, I've seen the question phrased differently, such that you would be correct [with the word "bloodthirsty" thrown in a few times to indicate a general proclivity towards death / negative votes], but I like my version better because the result is more surprising.)

Reuven, with your emphasis on friends first, and the implication that the end result is 1-4 will get zilch, they could team up and get 1/4 each (or 1-3 could get 1/3 each), etc. Especially with your emphasis on their friendship, #5 is going to get nothing.

True, they could team up, but business is still business. 1-4 could team up to get 5 out of the picture, but then 1-3 could team up to get 4 out of the picture, and 1 & 2 could team up to get 3 out of the picture, and then 2 could vote against 1. I said they were friends; I never said they were honest. That's the clear distinction I make between their friendship and their business "savvy."

(Note: this should not be taken as any indication of how I treat my friends. Nor should it be taken as an indication of how honest I am in matters of business. At most, it's just an indication of how I would treat my friends and how honest I would be, if I were a pirate. There's a reason the riddle isn't about five rabbis.)

The flaw, as I see it, of both your case and Shaya's is the assumption of a base case of 2 pirates. In that case, of course #1 would give up money to save his life. He has no power over #2. But once there is a majority that can
1) save their own lives and
2) get themselves money
they will! And they'll do it by working together to keep it out of the hands of #5 (and possibly others).

David -
(Note: I'm not sure why Shaya did the numbering the way he did - seems backwards to me. So in this explanation, the first to suggest will be #1; the last to suggest, if it gets that far, will be #5.)

Try thinking about it this way:

#1 first has his chance to make a suggestion. Obviously, he wants to keep it all, and, if he could, he would suggest that. He thinks to himself:

"What would happen if I did suggest that? Well, it could either get approved or rejected. If it got approved, great; otherwise, I'd be dead, and #2 would have his chance. What would #2 do? Well, if he could, he'd suggest that he get to keep it all. He'd think to himself:

'What would happen if I did suggest that? Well, it could either get approved or rejected. If it got approved, great; otherwise, I'd be dead, and #3 would have his chance. What would #3 do? Well, if he could, he'd suggest that he get to keep it all. He'd think to himself:

"What would happen if I did suggest that? Well, it could either get approved or rejected. If it got approved, great; otherwise, I'd be dead, and #4 would have his chance. What would #4 do? Actually, #4 couldn't keep anything for himself, or else #5 would vote against him and take it all. That means that if my plan to keep it all isn't approved, #4 would get nothing. Therefore, I, #3, can keep it all for myself, #4 will vote for that plan (having nothing to gain), and my suggestion will be approved!"

[Back to #2] That means that if my plan to keep it all isn't approved, #3 would get it all, and #4 and #5 would get nothing. Therefore, I, #2, can keep it all for myself, #4 and #5 will vote for that plan (having nothing to gain), and my suggestion will be approved!'

[Back to #1] That means that if my plan to keep it all isn't approved, #2 would get it all, and #3, #4 and #5 would get nothing. Therefore, I, #1, can keep it all for myself, #3, #4 and #5 will vote for that plan (having nothing to gain), and my suggestion will be approved!"

Thus, each pirate would take advantage of all those who would be losers if his plan isn't approved, in order to ensure that it is approved.

Or, to directly address your point:

You claim that if #1 suggested that he keep it all, #2-#5 would team up against him, vote him down and split it 4 ways. Let's imagine that. #1 suggests he keeps everything. #2-#5 vote him down. Now #2 suggests: "OK, I now suggest we split it four ways, as we planned, right?" Well, of course #3-#5 would vote him down. Now #3 suggests, "Three ways?" #4 and #5 will vote him down. We now come down to just #4 and #5, and that means #5 gets to keep everything. So #4 would never team up with #5 against #3, because he has nothing to gain. Which means that #4 and #5 would never team up with #3 against #2, because they have nothing to gain. Which means that #3-#5 would never team up with #2 against #1, because they have nothing to gain.

you have to understand the way I think.

i started off with a 1 person case

i.e. pirate #1

if we expand that to 2 person case, we have pirate #1 and pirate #2. Now, to remain consistent, pirate #1 should always be the same.

so on and so forth.

yes, these weird things actually make sense in this messed up head of mine.

Fatal flaw in Reuven's and Shaya's theories ... the "rules" state, "Following each proposal, all the pirates will vote on it." As there is no states resolution to a tie, when there are two pirates (or potentially any even number), a stalemate is all but guaranteed. (#1 has no reason to vote yes to any suggestion of #2, other than #2 giving all gold to #1. But #2 has no reason to give any gold to #1, as #1 alone can't oust #2).

Shaya -
I guess that sort of makes sense. Maybe.

David -
The rules also (carefully) state, "If it is accepted by a majority vote, the suggestion will be carried out; if not, the pirate who suggested it will take a long walk of a short plank, and the next will have his chance to proffer a plan."

If #2 proposed that he keep all the gold and #1 voted against that, the plan would not have been accepted by a majority vote, and #2 would have to walk the plank. (As to #1's ability to enforce the rules on #2, that's why I added the "usual disclaimer" in my first comment.)