Yesterday's puzzle got a couple correct answers in the comments of both the blog and on the Facebook notes thread. (I should really search harder for a solution which would allow combining both sources of comments...) There was also some good discussion about what would happen if one of the pirates wasn't a "perfectly logical being" which is actually where I was planning on heading with tomorrow's post. So I'm going to essentially ignore that for now but will be returning soon. Today I'm going to provide a tangentially related anecdote, the game theory "game" that sparked my remembrance of that puzzle, and the solution for those who may not read comments.
Last month was Monopoly at McDonalds. I was out eating lunch there with Andrew and he pulled Park Place off of his drink. If you collect both Park Place and Boardwalk you win a cool million dollars. The conversation turned to what would happen if I had Boardwalk on my drink. Andrew seemed to assume it was a given that we'd just split it and take $500k each. But part of me is thinking that Boardwalk is more likely to be the rare one so my expected value in terms of sheer money would be to reject that offer and just buy some more drinks in the hopes of getting a Park Place of my own. (Or to haggle him down to a better split I guess.) But in reality the marginal utility of the extra $100k between $500k and $600k isn't all that much and certainly isn't worth any strife that might be caused by making such demands from a friend. So if I'd had Boardwalk I'm sure we would have just taken $500k each and been ecstatic.
I wanted to continue the thought experiment though, so I asked Andrew what he'd do if a random dude had walked up to him and offered to buy Park Place off of him. What values would he accept? (Clearly he'd sell it to me for $500k, so what about the random dude?) The conclusion Andrew came to was he simply wouldn't sell it to a random dude no matter what the price was. The thing is we don't know which piece is actually worth the million bucks. One of Boardwalk and Park Place is worth almost a million dollars and the other is practically worthless. (It has some value to the owner of the real piece but is worth nothing to everyone else.) So if some guy showed up and offered a large amount of money then the odds are he's trying to scam Andrew out of the real piece. $50k sounds great but there's no way a stranger with the real piece is paying that much for the common one was his argument. After all, the stranger could just go buy 30 drinks, get 60 pieces, and almost certainly get his million bucks. If the stranger was offering a low amount of money (say, $50) then Andrew still wouldn't take it. Even though the stranger is probably just trying to save himself from buying 30 drinks there's a small chance he'd be trying to pull a fast one and it wouldn't be worth the risk.
Personally I like the idea of getting $50 for my worthless piece of paper, though I think what I'd do is try to find a second person in the restaurant with a Park Place and try to haggle the stranger to buy them both for $100. I don't think the stranger is scamming me at all but I will admit it would be a huge blow to my sanity if I sold it to him and it turned out to be the real piece and I probably wouldn't be willing to take that risk either. (Mostly I don't think it would ever happen at all. I know if I thought I had the winner I wouldn't advertise it in a room full of strangers. I'd quietly go home, hide the winner, and then just eat at McDonalds for a while.)
Which brings us to the 'game' I only just found out about earlier this week in my reading... The Ultimatum Game. The basic idea of the game is a random dude (probably in a top hat with a handlebar moustache) is going around making the following proposal to two strangers (in this case referred to as Alice and Bob.):
Alice is given $100 dollars and is told to split that money up into two piles. One pile for herself and the other pile for Bob. Then Bob gets to vote on the split. He can either vote for the split in which case Alice and Bob walk away with some cash. Or he can vote against the split in which case our moustachioed hero takes his money back. Either way the game is over. Alice and Bob can't negotiate over the split or communicate in any way. Alice makes a split and then Bob decides whether to take it or not.
So the question is, if you're Alice what split do you propose? If you're Bob what splits do you accept?
Now, the solution to yesterday's puzzle.
The trick to this puzzle is to find the base case and figure out what happens. Then work your way backwards until you end up at the current case. Since you only need half of the votes to pass a proposal the base case is when there are two pirates left since whatever is proposed is guaranteed to pass. (Obviously you vote for your own split since otherwise you're dead.) I'm going to break form from how everyone else labeled the pirates and assert that the pirates are, in reverse order of seniority, E-D-C-B-A.
Two pirates alive (E & D): D knows that whatever he proposes will pass and she's greedy so she offers a 0-100 split. It passes and D is rich!
Three pirates alive (E, D, & C): C knows that if his proposal gets shot down that the split will be 0-100. He also knows that he just needs one other vote in order to have his proposal pass. He needs to bribe E or D to vote with him. Clearly he can't bribe D to vote with him since D is already going to get all the money if C loses. On the other hand E is going to get absolutely nothing if this vote fails. One coin is better than no coins, after all, so C proposes a 1-0-99 split. E is logical and greedy and knows it isn't getting any better so he has to accept.
Four pirates alive (E, D, C, & B): B knows that if his proposal gets shot down that the split will be 1-0-99. He also knows he needs just one vote in order to win. The big loser is the last split was D. She has visions of coming home with all the coins way back in the base case but she now knows that's a pipe dream. She's going to get nothing if B dies. B knows that D knows this and once again one coin is better than no coins. His proposed split is 0-1-0-99 and it passes.
Five pirates alive (E, D, C, B, & A): I hope the pattern is becoming clear. A knows she needs to get 2 more votes and she knows that if she dies the split will be 0-1-0-99. The two easiest people for her to bribe and going to be E and C since they'll be getting nothing if A dies. So A proposes a split of 1-0-1-0-98 and walks away with most of the loot.
As Snuggles might say, "It's good to be captain!"
6 comments:
I actually won one of these situations once. When coke was doing their 'spell cokacola' contest I collected all the letters other than k (which is trivial to do) and then a cousin of mine walked up to me, handed me a k and asked if I needed that one. I immediately went and read a can of coke and discovered that I had missed the contest date and was out of luck - no million dollars for me. I pitched it out and thought no more about it and didn't even tell my parents until years later.
In these situations I can't see anyone running around asking to buy the common piece for $50 unless the contest date was over pretty much immediately. I would check the expiry date and unless it was today or tomorrow I would turn down any offer. If it was today or tomorrow things become much more muddled and then I probably try to rely on my read of the person to decide whether to take the money or start buying product furiously to try to complete the set myself.
Well, either that or try to consult the internet, which obviously knows the answer. I assume that isn't kosher for answering this question though since it makes the question irrelevant.
Yeah, it was a thought experiment more than anything else. Though I did just search the internet and it turns out Boardwalk is the rare one. And has been the rare one every time (at least back until 2003, the chart on Wikipedia doesn't go any further back). The rare one in each set is the last alphabetically except for Boardwalk which is just awesome.
Also interesting, in 2010 Boardwalk was actually paired with Park Place (you get 2 pieces per item). I wonder if that was a coincidence or if it was intentional. I'd think intentional.
Most interesting of all, from 1995-2000 all the good prizes were won by friends of the chief of security at the promotion company. He was able to steal all the winners and give them to people. The first million he even gave to a hospital...
Check the length of this thread:
http://contests.about.com/b/2011/09/27/rare-mcdonalds-monopoly-pieces-for-2011-which-2011-monopoly-pieces-are-rare.htm
The great majority of it is people with Park Place trying to con some idiot who got Boardwalk into going halfsies with them. Hope springs eternal I guess.
Cut off because of length, so two posts...
You guys better hope I'm not pirate five. Suppose pirate five votes against the initial offer. Now the other pirates have to factor into their logic that pirate five is not thinking the same way they are. The other pirates know pirate five is still logical and greedy, so they can rule out that he is determining his vote at random or that he is voting no to everything. After all, if it gets down to the two pirate situation then his gambit, whatever it is, will have failed.
So clearly pirate five has his eye on more than one coin, and he is willing to kill someone if he doesn't get it. How many coins does he want? Who knows.
Now if pirate two goes ahead and offers 99-0-1-0 then pirate two is not in the situation described in your logical reasoning. After all, pirate four does not *have* to accept this offer. He could instead vote against and then hope that pirate five rejects pirate three's offer, in which case he gets 100.
You may say that pirate five will never reject pirate three's offer, but that depends on his solution to Newcomb's paradox. In case you aren't familiar, a super-intelligent alien says it has a machine that can predict your actions by scanning your mind. In box A it places $10k. In box B it either places $1M or nothing. You get to choose whether to take box B or take *both* box A and box B. The trick is that if the alien predicts you are going to take just box B then it places the million in there, otherwise it places nothing. So you can argue that you should always take both since you'll get more money, but assuming the alien can actually tell what you are going to do, you'll get more money by taking just box B - after all that, only way that you are *going* to take just box B is if you actually do take it.
The other pirates are predicting pirate five's actions with logic rather than with super-technology, but the point is the same. Newcomb's paradox doesn't really have a solution, and games where your personal outcome is determined by how well someone else is able to predict how you'll act are very complicated. At any rate, it doesn't matter what the solution to this problem is, it matters what everyone else thinks pirate five's solution is. The fact that he would vote no to the first offer gives them that answer.
So when pirate three offers a split, he has to face the possibility that pirate five has a predetermined course of action. We know that if pirate five is willing to accept 1 coin then he always gets one coin and that not voting for the first offer was the wrong choice. But what if pirate five isn't even making choices, what if before the game began he decided that he would accept any offer of X coins or more and reject anything less. This leaves the others to guess at risk rather than have a clear path.
After pirate five votes no, pirate four isn't sure whether five will vote yes or not to the split offered by three. Given that he wants to maximize his coins, he should vote yes to pirate four's offer if the number of coins he is offered is greater than 100 * the odds that five rejects three. Of course he doesn't know the odds that five rejects three, but if two only offers him one coin then he isn't risking much to find out.
So my question is, what should pirate two do? Well, he has the goal of getting one other pirate to vote for him. There are a lot of ways he could approach this. If he offers a Y-Z-0-0 split then he is saying to pirate three, "Look, this way you get a very good share, and the other way you have to deal with that maniac at the end of the line." Of course pirate three might still be being "logical" in which case he gets killed and pirate three thinks he is getting 99. He also has to contend with what pirate three thinks pirate five will take. If pirate three thinks that pirate five will take a 50/50 split, then he won't take less than 50 from pirate two. He could go Y-0-Z-0 and hope that pirate two is unaffected by pirate five's gambit, which might work, especially if he decides to make Z more than one. Or he could go Y-0-0-Z, dealing directly with the maniac. After all, he can reasonably assume that there is an amount of money that will assuage pirate five.
Which one of these ways of thinking is correct? I don't think there is a clear answer. The question for pirate five, when concocting a strategy, is whether he should guarantee himself one coin or introduce a large element of risk in the hopes of getting more. Since one of one hundred coins is not a good take, I think risk is the way to go. We don't know what the odds of getting more are, but it is silly to assume that because we don't know what they are they must be zero.
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