» 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

• Incorrect Answer
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  I agree with your 44 as the best answer. Another way to solve the problem is to realize that the 6 and 9 packs are going to give you every multiple of 3 from 6 upwards and that you are going to have to construct two additional sequences to fill in the gaps. You do this by first adding 20 and then adding 40 to the sequence. When you put all these numbers together in a series and sort it you can see that 44 is the answer. When I did it in my head I came up with 45 (which is better than the answer) but putting it in a spreadsheet revealed 44.

  Posted by GaryScarr on November 11, 2006 at 06:16:46 Pacific Time

• Incorrect Answer
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  Actually, the answer using the methology stated in the answer key is correct. On the last line, number 44 is crossed off, and number 45 is crossed off. The last statement is that anything greater than 46 can also be crossed off. That leaves 43 left on the last line. So anything 44 or greater has been crossed off the list. It just doesn't mention that 44 is the smallest number.
  
  Personally, I solved this puzzle by looking for the set of independent linear equations in the format 6*x + 9*y + 20*z = n, where x,y,z >= 0, that could generate N, N+1, N+2, N+3, N+4, N+5, etc. Once you hit N+6, then the rest of the sequence can be filled in by incrementing x in the x*6 element. This is essentially the same as the methodology in the answer key.
  

  Posted by ericbronnimann on January 31, 2007 at 11:49:00 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

• Thanks!
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  Thanks, Michael. I always enjoy the column.

  Posted by TPIRman on November 07, 2006 at 14:26:10 Pacific Time

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  Michael, I'm not clear on what got added. The problem and solution appear consistent to me although, being British, I never saw the show but was assuming that more was involved in the real show. Seems like any gambler would want $300k to not switch but risk-averse individuals could be swayed by lesser amounts.

  Posted by GaryScarr on November 11, 2006 at 06:32:18 Pacific Time

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  I meant to add - to make it interesting it seems like the contestant would need to know that there was a prize of some description behind one curtain but wouldn't know the value of it and would be offered varying amounts of guaranteed money before the curtains were opened. If not, there must have been many millionaires produced by the show.

  Posted by GaryScarr on November 11, 2006 at 06:42:12 Pacific Time

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  Gary -- Michael is referring to the aforementioned error in the Chicken Nuggets solution. There was no error in the Monty Hall problem/solution (as far as I can tell).

  Posted by TPIRman on November 11, 2006 at 14:03:52 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

• Monty Hall Problem
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  Oops. In the last sentence I should have said "has a 1/3 chance of winning at the moment when he picks a door if he resolves not to trade, has a 100% chance of winning if he makes it to the point where he's offered a trade and declines it, and has a 0% chance of winning if he accepts a trade."

  Posted by Indy_Darren on November 06, 2006 at 18:37:43 Pacific Time

• Monty Hall Problem
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  In the situation you stated -- the contestant picking the correct door from the beginning -- yes, the contestant has a 100% chance of winning. But this scenario occurs only 1/3 of the time. The other 2/3 of the time, the contestant has made the wrong choice initially. Since the contestant has no idea whether he has made the correct choice or not, he can only rely on probability, which demands that he switch.
  
  I don't have the magazine in front of me, but I'm pretty sure that the problem stated that the host reveals a door with a goat behind it. This is the only important behavior pertinent to the host: he always reveals a goat.
  
  The intentions of the host (i.e., whether he is rooting for the contestant) are irrelevant. As long as he reveals a goat, he has made switching a mathematically attractive prospect for the reasons laid out in the answer.
  
  Incidentally, Monty Hall always (jokingly) objected to this being called the "Monty Hall Problem" because this particular exchange never actually played out on the real "Let's Make a Deal." The reason, of course, is that it would be an unbalanced game: one choice would ALWAYS be more sensible than the other. "LMAD" worked by constructing "deals" in which the contestant could depend only on intuition and luck as they tried to make the best choice.

  Posted by TPIRman on November 07, 2006 at 14:24:11 Pacific Time

• Monty Hall Problem
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  The problem in the magazine doesn't say that the host always reveals a goat, only that he reveals a goat on this occasion: "Monty then counters by showing
  you
  one of the doors with a goat behind it. . . " The given solution only works if the contestant can rely on him always revealing a goat after the contestant selects a door. The assumption that the host will
  always
  reveal a goat can't be made from the problem as stated.

  Posted by Indy_Darren on November 18, 2006 at 17:20:18 Pacific Time

• Monty Hall Problem
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  Well put -- that I can agree to.

  Posted by TPIRman on December 13, 2006 at 10:59:55 Pacific Time

• Monty Hall Problem
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  The "always versus this time" issue is absolutely crucial and the problem does not have an absolute answer if the rule isn't specified that Monty always shows a goat. The classic formulation of this problem is intended to astound folks with one of those counterintuitive probability experiences that starts with the simple idea that with k equal choices and one prize you have 1/k probability of selecting the right one. What seems tricky is that when non-winning options are subsequently and deliberately revealed the 1/k probability of your choice remains constant, whereas the remaining 1-(1/k) probability is then distributed over the remaining non-winning options you didn't select.
  It actually clears up a little when presented as 100 doors with one prize. You select one and are told 98 non-winning doors will be opened and the oprion to switch will then be offered. Ask yourself whether it seems more likely that your original selection holds the prize or that the last remaining door does?
  When the rules change, however, the answer also changes. If the selection of the doors to open is random, and it is possible that a winning selection could have been opened first, then no preference for the non-selected door accrues and either choice is equally valid. If you know that Monty hates you and you will only be offered a choice if you originally selected the right door then, umm, don't switch.

  Posted by StillScaredofDinosaurs on November 23, 2006 at 20:45:18 Pacific Time

• Monty Hall Problem
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  The "classic" Monty Hall problem is as posed in the TV Show: Monty ALWAYS shows the contestant a goat after the contestant picks a door, which means that Monty MUST know where the prize is (his luck at picking Goats CAN'T be that perfect). At that point, there are two unopened doors, one with a goat and one with a prize. It doesn't matter whether the contestant changes doors or not, his/her chances are 50% of winning at that point. No one cares about the 33% of picking the correct door in the first round because you don't open it then - only after Monty opens one - and Monty is playing with sure foreknowledge of the prize's location. Same story with 100 doors if Monty keeps opening doors after offering you a chance to change your choice. As long as the rule is that Monty will show you a goat, you will ALWAYS end up with 2 doors and 98 goats and a choice to make - a 50% chance of winning even with 100 doors, whether you switched 98 times or 0 times because Monty will always open a Goat! Pretty good odds if the game is played as on the TV show. The only reason that the show had 3 doors was to add excitement and suspense.

  Posted by Walrus on November 27, 2006 at 10:40:21 Pacific Time

• Monty Hall Problem
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  Walrus,
  You have managed to get a bunch of things wrong.
  1) As Monty has stated, this was not actually a common bit on the show & there were no "rules" defined there.
  2) IF, however, the rule is as you state (always show non-selected goats) then the odds after 98 doors aren't 50-50 but 99-1. That's what makes this puzzle interesting.
  3) Requiring outside knowledge to answer a puzzle would be one of the dumbest moves I've ever seen. The simple fact is, if the rule isn't correctly defined IN the puzzle, you haven't done your job.
  Goo goo ga Choob.

  Posted by StillScaredofDinosaurs on November 29, 2006 at 20:22:06 Pacific Time

• Monty Hall Problem
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  The problem makes it clear that Monty always shows you a goat, no matter what you initially choose. Also, he never opens the door you select -- if you initially choose a goat, he shows you the other goat. (The problem statement is unfortunately vague on this point, but if he could show you your own goat, the language about having a choice between keeping your door or taking the other one would not make sense, so you have to infer that Monty never opens your chosen door).
  
  So, as the game was set up in the problem statement, Monty's actions are constrained, and his motives and behavior can't factor into the game at all.
  

  Posted by bknittel on November 17, 2006 at 13:11:01 Pacific Time

• Monty Hall Problem
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  For an exhaustive discussion of this problem, see Cecil Adams' "Straight Dope":
  http://www.straightdope.com/classics/a3_189.html.

  Posted by HFTobeason on December 13, 2006 at 17:23:58 Pacific Time

• Monty Hall Problem
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  For an exhaustive discussion of this problem, see Cecil Adams' "Straight Dope": www.straightdope.com/classics/a3_189.html.

  Posted by HFTobeason on December 13, 2006 at 17:24:12 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

• Monty Hall and Red Herrings
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  Oops! Sorry! You misunderstood the problem. Monty always opens *another* door with a goat behind it, not yours.
  
  But if you still don't agree with the correct answer, you're not alone (but still wrong). See here:
  http://www.marilynvossavant.com/articles/gameshow.html
  
  When Marilyn first posted her response, whe got LOTS of mail from all kinds of learned people telling her the answer was incorrect.
  
  Most people run a computer simulation of the problem and then they see the answer is 2/3.

  Posted by MichaelPryor on February 27, 2009 at 08:27:54 Pacific Time