» MAKE: NOISE — Discuss this article

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• Incorrect Answer
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  The logic given for the answer to "chicken nuggets" is sound, but the conclusion -- N=46 -- is not. The puzzle asks, "Is there such a number N, so that for all numbers bigger than or equal to N, you can buy that exact number of chicken nuggets using 6, 9, and 20 packs?" While 46 is technically a valid value for N, the lowest possible number that satisfies the requirements for N is 44, so that is obviously the best response.
  
  At least, 44 was my answer. I am baffled as to how 46 could be considered correct.

  Posted by TPIRman on November 06, 2006 at 16:27:52 Pacific Time

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  You are entirely correct. Looks like it got added in the editorial process somewhere. I'll contact the publisher.

  Posted by MichaelPryor on November 06, 2006 at 16:33:34 Pacific Time

• Monty Hall Problem
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  In the Monty Hall puzzle, the supposedly "correct answer" depends on an assumption about the behavior of the game show host that isn't stated in the problem, and can't be safely inferred from what is stated.
  
  The assumption is that the host will behave like this: "First, have the contestant pick a door, then, open a door with a goat behind it, then, offer the contestant a chance to trade the door they chose originally with the remaining unknown door." If that IS the way the host behaves, then it's true that the best strategy for the contestant is to trade the door they chose originally for the unknown door remaining after the first goat is revealed.
  
  But what if the host operates under a different strategy? To assume that the host behaves as described above, one must also assume that the host (a) wants to help the contestant win the money, and (b) will open a door to reveal a goat no matter what door the contestant chooses first.
  
  There are any number of other plausible strategies for the host that COULD lead to the factual setup in the problem.
  
  Suppose, for example, that the host simply opens a door at random after the contestant chooses. One third of the time, when the host does this, the money will be revealed and presumably the contestant will lose immediately. The other two thirds of the time a goat will be revealed (just as in the problem) and the contestant's chances of winning will jump to 50%, whether he trades doors or not.
  
  Suppose that the host doesn't want the contestant to have it. If the contestant chooses a door with a goat, the host immediately reveals which door hides the money and announces "Sorry, you guessed wrong, loser." If the contestant picks the door with the money, then the host tries to lure him to another door by revealing one of the goats and offering the prospective sucker a chance to trade for the remaining door (which the host knows hides another goat). From the contestant's point of view, this is the same decision as in the problem, but in this case, the contestant has a 1/3 chance of winning if he does not trade, and is guaranteed to lose if he does.

  Posted by Indy_Darren on November 06, 2006 at 18:39:42 Pacific Time

• to be precise
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  To be precise, the answer to the chicken nugget problem is "Yes."

  
  

  Posted by bknittel on November 17, 2006 at 13:00:03 Pacific Time

• Gold Chain
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  This may seem silly but can't you make a single cut down the middle of the chain length-wise? It'll only take one cut and you can pay the worker in half-links everyday.

  Posted by RogueTemplar on January 01, 2007 at 08:19:22 Pacific Time

• Gold chain
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  The question asks what is the fewest number of cuts _the man_ must make to pay the laborer. If you make the laborer cut off one link each day, as part of his labor, then the man doesn't have to make ANY cuts. Hence: zero cuts is the minimum.

  Posted by hernan43 on February 18, 2007 at 11:58:40 Pacific Time

• Monty Hall and Red Herrings
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  I just revisited this old issue looking for brain teasers for my 8th grade math class. The published answer is plainly wrong, at least by any reasonable reading of the problem.
  
  The whole first choice and door unveiling cycle is a red herring. The real choice happens after the first goat is revealed. At that point, there are two unknown doors - money and goat.
  
  The published answer says I will always be better off (by a 2/3 to 1/3 chance) switching from my current choice. Balderdash! That's confusing my chances of winning with a particular choice with the probability that I will have to switch to win.
  
  What's the difference? Very simple: The difference is in the knowledge I have when I make my final choice.
  
  
  
  • If Monty Hall reveals my initial choice as a goat, then I know I must switch choices to win. There are two doors remaining; one is money, one is a goat. When I switch, I have a 50/50 chance of selecting the money.
  
  
  • If Monty Hall reveals a goat that was NOT my initial choice, then there are still two doors remaining: one with money; one with a goat. I have a 50/50 chance that my current choice is the money. Switching gains me nothing.
  
  
  
  So at the time I must make my choice to switch (or not), I am down to 50/50 odds either way. Switching doesn't improve or reduce those odds.
  
  What we have is (1) a situation in which I know I should switch, and (2) a situation in which I don't know whether I should switch. In case (1), switching improves my odds of winning from zero to 1/2. In case (2), switching does not improve my odds at all, which remain at 1/2.
  
  The published answer combines these cases into an assertion of an overall odds improvement if I switch. That is not very helpful, since I will be able to distinguish between the two cases at the time I need to make my choice.
  

  Posted by Mr. Carl on February 27, 2009 at 06:13:40 Pacific Time