I was going to write another post on cards (this time on Texas Hold’em) but I can’t find any of my notes from the game theory class I took last semester, so instead I’m going to tell a riddle. It’s probably easier than a lot of “hedge fund interview” questions, but it’s a very good example of iterated logic.

There are five pirates who have just discovered a treasure chest with 100 gold coins in it. Being greedy, they do not want to split it evenly, mainly because each of the men has been working as a pirate for a different amount of time (and the “senior” pirates feel the deserve–probably rightfully–a larger share). In fact, Pirate 1 has been working for one year, Pirate 2 for two years, Pirate 3 for three years, etc. Their proposed method of solving the gold distribution problem is this: The most senior pirate will propose a plan and everyone proceeds to vote “yes” or “no” to the plan (**including the senior pirate**). If there is at least 50% agreement (it could be exactly 50%), the plan goes through. If there is less than 50%, however, the senior pirate has to **walk the plank**. Once he is gone, they would proceed again with the plan, this time with the next most senior pirate proposing.

Assuming all the pirates are rational and know that every other player is also rational (and emotions play no part in their choice), how will the most senior pirate decide to divide the coins so that he can maximize his profit and avoid walking the plank?

(HINT: the exact number of pirates on the boat does not affect the logic of the plan)

** SOLUTION** (

**Don’t read if you still want to work it out! I’m not going to write a “filler paragraph” like David did.):**

Let’s assume the pirates weren’t very smart about their plan proposals and everyone was eliminated but Pirate 1 (the rookie). Obviously, he’s going to get all 100 coins. What if there were two pirates left? Pirate 2 knows that all he needs is a 50% vote to go through with his plan, so he would propose that he gets all of it (since his vote alone counts 50%).

It gets a little trickier when Pirate 3 is still around, though. He needs one more vote besides his to get his proposal through, and he knows that anything he says Pirate 2 will veto (because Pirate 2 is walking out with all the booty if Pirate 3 goes overboard). He also knows that Pirate 1 will get nothing unless Pirate 2 dies, too—so he offers Pirate 1 a single gold coin (which is better than nothing), assuring his proposal goes through, and takes 99 coins for himself.

When there are four pirates left, Pirate 4 only needs one other vote to guarantee his plan. He **could **simply give Pirate 1 two coins instead of one (which would be better than the situation in which he dies), but he realizes with his greed that, if he dies, Pirate 2 will get nothing, so all Pirate 4 needs to do is offer Pirate 2 **one** coin to secure his vote, which is what he does.

With five pirates left, Pirate 5 needs two more votes to get his booty, and all he is going to consider is the situation in which he dies (getting repetitive at this point). If he’s gone, Pirate 4 is getting 99 coins and Pirate 2 is getting 1, so Pirate 5 just needs to offer one coin each to Pirates 1 and 3 (who are getting nothing if he dies) for the situation to be better off for them.

Hence, Pirate 5 is going to make out with 98 coins, and Pirates 1 and 3 will get one each. This problem can be easily iterated for higher numbers of pirates. If there were 15 pirates, Pirate 15 would simply offer his one-coin bribe to pirates 13, 11, 9, 7, 5, 3, and 1, hoarding 93 coins for himself. That wasn’t so harrrrrd, now, was it?

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