FREE MUNNY II

[Beaver Dude]

There's a riddle. But, uh, nope no cash prize. And you guys know it cuz' I'm new and with my funds I can offer a grand prize of 30 munny. But I'm not going to.

(Unless no one answers. In which case I will attempt to look pitiable and pathetic and offer my 30 munny)

Well, we were were riddling away in the Cbox and Rocket said: we should make a Riddle thread. Rhea and I agreed heartily. And then Rocket skipped town and I sorta-kinda hoped that Rhea might make it but then she good nighted the Cbox too- so I guess it's up to me. I suck at this type of thing so if someone wants to make a *real* riddle thread feel free. I kinda made this thread temporary by nature so, mweh.

Alrighty, rules are simple:

-Max of Two Riddles (first come first served. I'll ask the first, anyone else can ask the second)
-Answerer must specify whose riddle s/he is answering
-Once the Riddles are answered we move on to FREE MUNNY III and start a new thread

Riddle:

There are four pirates sailing on a pirate ship. They've just looted a funky looking luxury thing complete with ducks and dogs and little kids carrying oversized keys so they've got some munny to spend. Well, to be precise, exactly 10,000 munny.

Now, unfortunately, these pirates are cursed. To be nameless, perfectly logical *and* democratic. So that means they have to divide up the spoils through a process of voting. It also means that we shall refer to them as nameless pirate 1,2,3, and 4.

Here's how the voting works:

In order, Pirates 1,2,3 and 4 will offer a deal. For each deal the pirates will vote and if the number of pirates voting against the deal is GREATER[/b] than the number of pirates voting for the deal, the unlucky pirate walks the plank. That means that if an equal number of pirates vote for the deal the pirate survives and that deal is accepted.

For example:

If Pirate 1 offers a deal that is unaccepted (three vote against), he walks the plank. Then Pirate 2 offers a deal and so on and so forth. If Pirate 1 offers a deal that IS accepted then everything stops and the money is distributed.

And no, the pirates will not shoot each other for the munny before the vote begins. It's part of the curse.

Now keeping in mind that these pirates are perfectly logical, how does Pirate 1 keep the largest amount of Munny and what is that number of Munny?

Happy Riddling!

[Mimetic Memories]

What kind of riddle is this?..

It's more of a mathematician thing. 10000/4= 2500. Theoretically, if all 4 agree to this, which is perfectly logical because it's fair they each get a 4th of it, then 2500 is the largest pirate 1 can get.

[prythian]

Actually, if they were perfectly logical, they would vote against any deal they could, so as to get rid of the other members. That way, the maximum would be 5000, because they'd all vote together against any deal until there was only 2 of them left, and then whoever made the next deal couldn't be voted down.

[Mimetic Memories]

They have to be democratic as well, remember?

[prythian]

Yes, and democratically they vote down the other plans.

Democratic doesn't mean 'fair'. Perfectly logical means they will split the munny evenly, but...

. . .Woah, I just realized. If they take my plan, that leaves two left, which means that the next deal to be offered automatically wins.

Which MEANS that the person offers the deal of them getting ALL of the munny, while the other gets none, and they can't do anything about it.

So: all of it.

[Mimetic Memories]

Right, since I'm pretty sure that you just solved it..

..I'm going to make my own and post it sometime. I won't be able to offer money, though, for obvious reasons.

[Rocket!]

Well, no... the last two wouldn't have Pirate 1, would it?

[prythian]

Ah, good point. I missed the part that said they went in order.

In which case it IS only 2500.

Which means this isn't really a riddle so much as a math problem.

[Beaver Dude]

Prythian is REALLY close to the answer. He's just not applying it to Pirate 1's situation.

Hint: Pirate 1 can get a LOT of the Munny.

[prythian]

. . .Oh. DUH.

Pirate 1 offers a 3-way split, excluding one of the pirates. The other two will naturally vote that way. So the total would be either 3333 or 3334, I believe.

[Beaver Dude]

If we were playing warmer/colder I'd say you're getting colder now. =3

Wanna 'nother hint?

Well, to be fair, this is a game theory question. When I say logical I mean: HOGOD I AM SO LOGICAL I AM STUPID AND IF THE WORLD REALLY RAN LIKE THIS WE'D ALL BE DEAD.

I feel like someone is going to hit me after I reveal the answer. I shoulda gone with the gunmen question. XD

[raphe]

Pirate 1 makes a deal with pirate number 4 saying that he will split the treasure 50/50 with him he votes with him on all deals. Pirate 4 agrees knowing that a better deal couldn't be offred. Of course pirates 2 and 3 don't like this deal so they vote agains. The votes 2/2 deal isn't made but Pirate 1 doesn't have to walk.

Pirate 2 tried offering the same deal to Pirate 1 but it doesn't work because Pirate 1 knows that if he sticks with the plan he gets half the munny. Pirate 4 votes no for same reason and Pirate 3 votes no because he gets no munny. Pirate 2 walks the plank

Three Pirates left. Pirate 3 tried offering the same 50/50 deal to Pirate 1 but he says no because he knows that if he just sticks with the plan he splits it with Pirate 4. Pirate 4 votes no for the same reason. Pirate 3 walks the plank.

Pirate 4's turn. Two left. He offers a 50/50 deal with Pirate 1. Pirate 1 agrees and they split it 50/50 each one getting 5000 munny.

I think that's it if not oh well.