Read the post below first if you don't want to have the answers given away.

Census-taker brainteaser

The key is spotting an unusual piece of data. At least for a brain teaser. It's clear there are 3 kids, the product is 36, the sum is the house number, and the oldest likes bananas. There is another piece of information in there that's hard to identify. I'll give you a clue. The typical response to this question is "don't I need to know the house number?"

Well the census taker knows what the house number is. She can look up at the number. Isn't it odd that she still cannot solve the problem? This is the piece of data we are missing. If she cannot solve it, then it follows that all possible solutions for the product = 36 must lead to 2 sets that add up to the same number. Those are 2,2,9 and 1,6,6. All other product sets add up to a unique number. Since the oldest likes bananas only the first can be a solution. There are two 'oldest' in 1,6,6.

I love this teaser because it illustrates a great talent in problem solving - understanding the data you have and what it does and doesn't say.

Pirate brainteaser

This one requires a framework to solve correctly. In this case the easiest way to do this is to simplify the problem. Make it 2 pirates. Since we know how they will behave, we can determine how 3 will behave. We can do this because if they decide to vote #3 off the plank, the scenario with 2 pirates is what they'll be left to deal with. It will influence their decision with 3 pirates present.

With 2 the answer is easy. Pirate #2 (the 2nd to last pirate) will propose he gets 0 coins and the last gets 500. Anything else and Pirate #1 will vote him down (he has a better economic position if the #2 is gone and he gets everything) and there will be no majority - #2 walks the plank. It's pure self-preservation. Solution here is 0-500 (bolded votes 'yes'; unbolded votes 'no').

Add pirate #3. He needs 2 votes to get his proposal accepted. His is a guaranteed vote so he just needs to influence one of the other pirates. To get #2 to vote yes he simply has to give him 0 coins. He's indifferent to this economic outcome and the previous one (and probably doesn't want his pirate friend to die) so he will vote yes. Solution here is 500-0-0 (bolded votes 'yes'; unbolded votes 'no').

Add pirate #4. He needs 3 votes to get his proposal accepted. Again he just has to lure the pirates with the least booty in the next round - namely #1 & #2. So he offers 500-0-0-0 (bolded votes 'yes'; unbolded votes 'no').

Add pirate #5. It's clear what will happen here. He offers 500-0-0-0-0 (bolded votes 'yes'; unbolded votes 'no'). So the answer is the first pirate proposes he keeps everything and 4 vote yes. No one dies.

What a completely unintuitive solution. It's clear this probably wouldn't work in real life. I could see pirates #1-#4 teaming together and screwing #5. In the heat of the moment it would seem a smart thing to do. But once #5 is gone, #1-#3 would team together and screw #4. They'd quickly realize that would be a disastrous move.

Even more interesting is if the pirates are more bloodthirsty (spiteful?) and need an economic benefit to not vote their pirate friends off. In other words if a pirate can get x coins now and x coins in the next round, he will vote the proposer off the plank. The answer is 497-0-1-0-2.

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