Wednesday, December 22, 2004

Math puzzles for the holidays

I've always been a fan of clever math puzzles, so here's one for the holidays (after which I'll get back to slagging President Smirky and his band of idiots.)

There's an old story of two pirates who chance upon a cache of mixed valuables (gems, gold, that sort of thing) and want to divide it evenly, so that no one's feelings get hurt. The puzzle is, of course, how to divide the goodies so that both pirates feel that the division was equitable. Note the careful wording here -- the division doesn't have to be exactly in half; it simply needs to leave each pirate feeling satisfied that it was fair. The solution isn't hard to figure out -- one of the pirates divides the swag into two piles, after which the other gets to pick which pile he prefers. If you think about it, it's the obvious answer -- neither pirate can cheat the other or bend the rules to his advantage.

But what if you add more people to the mix? Say you have (God help us all) a Christmas fruitcake, and an arbitrary number of people. With just a sharp knife, how can you slice the fruitcake so, once again, every single person in the group is satisfied with their slice and doesn't feel cheated compared to anyone else?

Now it's not so easy, is it?


Dave said...

Sure it is! For three, the first pirate makes three piles. The second pirate has the option of making any modifications. The third pirate gets to choose. Then it's back down to two pirates again.

CC said...

Nope. Let's say the first pirate divides up the loot as carefully and as equitably as possible. Then the second pirate is totally clueless and makes a mess of things so that one pile is clearly more valuable than the others. At that point, the third pirate would see what happened and grab that larger pile.

The first pirate could only look on and see that he's getting shafted, but there's nothing he could do about it, is there? The puzzle is to make it so that every single participant ends up feeling that he got his fair share. In your scenario, the second and third pirates could effectively conspire against the first pirate to screw him thoroughly.

Try again. :-)