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| Jul-01-08 | | Once: Here's one for you:
On the first day of term, a teacher tells his class that he will give them a mystery test on one day that term. It is a mystery test because when they arrive at school on the day of the test, they will not know that the test will happen that day. At this point, the cleverest boy in the class shouts out: "But that means you can never give us the test!" This is the boy's logic:
"You cannot give us the test on the last day of term, because as we arrived at school on the last day, we would know that there were no more days left. So the test had to be on that day. And that means it would not be a mystery test any more." "And you could not give us the test on the next to last day of term for the same reason. If the last day of term cannot be test day, then when we arrive at school on last day minus one, we would know that would be test day. So that doesn't work either." "And by the same logic, you can discount all the other days in the term." Who is right - the boy or the teacher? |
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| Jul-01-08 | | clocked: First day |
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Jul-01-08
 | | al wazir: Since the rest of the crowd seems to have moved on, I will post the answers to the Geoquiz now. 1. Cuba and the Bahamas are less than 75 miles from Florida, but the
correct answer is Russia. The easternmost tip of Siberia is only 56 miles from Alaska across the narrowest part of the Bering Strait. What's more, in the middle of the strait Russia's Big Diomede Island is so close to America's Little Diomede Island that there's barely enough room between them for the International Date Line to slip through. 2. Reno (longitude 119º 48' W) is west of Los Angeles (118º 14'). Washington (77º 2') is west of Lima (77º 0'). But since two minutes of longitude corresponds to about the distance from Dupont Circle to the Capitol (or from the Plaza de Armas de Magdalena del Mar to the Parque Mariscal Castilla, if you will), nobody could fault you for calling it even. (c) Naples (latitude 40º 50' N) is north of New York (40º 45'). Rome (41º 53') is north of Chicago (41º 52'). Nice (43º 42') is north of Toronto (43º 40') -- but each pair is so close that they're practically ties. (d) West, in all three cases, by a small amount. (Look at a map.) If, like <OCF>, you interpret this to mean "In which direction do you have to go?" the answers are NW, S, and N. 5. The answer to (a) is easy: Alaska. The answer to (b) is Alaska too. (Did you say Hawaii? Look at a globe.) The answer to (c) is Alaska again. (Not Maine: The Aleutian Islands extend across the 180th meridian, putting the ones at the end of the chain in the eastern hemisphere.) The answer to (d) *is* Hawaii. |
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Jul-01-08
| | al wazir: <Once>: It comes down to the definition of a "mystery" (or "surprise") test. If I expect the test on day 1, day 2, etc., and the test is not given until the last day (day N), how can I claim that I "knew" when the test was going to be given just because I was right on the Nth day? After all, I was wrong N-1 out of N times. To put it another way, suppose that 1/N of the class guesses each of the N possible days ahead of time. Then (N-1)/N of them will be wrong, and only 1/N will be right. Isn't a test that surprises (N-1)/N of the class pretty mysterious? The boy's reasoning is that if A then B; if B then C, etc. But the reasoning on day N is valid (i.e., certain to be right) only after N-1 days have passed. The reasoning on day N-1 is contingent on the validity of the premise, "The test cannot be given on day N," which is not yet known to be true. And similarly for all the other steps. |
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| Jul-01-08 | | Once: <al wazir> Eloquently argued as always! But the mystery is not just about whether the boys can guess what day the test will be held. The teacher stipulated that they will not know that the test will be held on the day of the test. That changes the logic. So 100% of the class would know that the test cannot be held on the last day of term. And by the same logic, the test could not be held on the penultimate day of term. The boys know that the teacher cannot wait until the last day for the test, because it would no longer be a mystery. So it must be held on the penultimate day. But if the boys know that, it cannot be held then either. And so on.
I don't think that this can be solved by considering probabilities because the probabilities change after every day passes by. One way to consider this is to imagine that the term is just one day long. Then the test is clearly impossible because the boys know the day it will be held. Then imagine a term that is two days long. Again, the test cannot be held. It cannot be held on the last day (not a mystery). Nor can it be held on the first day, because that is the only other choice and therefore not a mystery either. And a three day term? Same problem. Can't be the last day (not a mystery). Can't be the second day, because if we have got to the second day than we are back in a two day term, which we have already dealt with. |
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| Jul-01-08 | | Jim Bartle: Here's a question many of you probably have already heard: Three cards are placed face down, two hearts and a spade. You're asked to pick out the spade. You pick, someone turns over a card and shows it's a heart, and gives you a chance to change your guess. Should you stay with your choice, change it, or doesn't it matter? |
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| Jul-01-08 | | cuendillar: Obviously change it, that's basic probability. The point is that you then have a 66% chance of getting the spade (if that what you want). Basically, if you change you are in fact choosing two cards rather than one. |
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Jul-01-08
| | al wazir: <Once>: I stand by what I wrote. The reasoning on on all days earlier than day N is contingent on the validity of the premise, "The test cannot be given on day N," which is not YET known to be true. |
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| Jul-01-08 | | Once: <jim bartle> Ah, the old Monty Hall problem! Here's the wikipedia entry for it. http://en.wikipedia.org/wiki/Monty_...
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| Jul-01-08 | | Jim Bartle: Thanks for the reference. Yes, it's true. Sticking with original choice gives you 33% chance; changing gives you 50%. |
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| Jul-01-08 | | SetNoEscapeOn: <Jim Bartle>
I've heard it referred to as the three doors problem, and yes, you should change your choice. This problem about probability has a high probability of inducing arguments , for whatever reason. I have seen people whom I consider to be otherwise intelligent argue for literally <hours on end> that it doesn't matter which choice you make, because the chance is simply 1/2... |
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| Jul-01-08 | | Shams: <<Once>: I stand by what I wrote. The reasoning on on all days earlier than day N is contingent on the validity of the premise, "The test cannot be given on day N," which is not YET known to be true.> <al wazir>, given this, do you reject the "winning solution" for the problem of the six pirates dividing 100 gold coins? |
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| Jul-01-08 | | SetNoEscapeOn: And changing gives you 66%, not 50% |
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| Jul-01-08 | | SetNoEscapeOn: or 66.67%... |
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| Jul-01-08 | | SetNoEscapeOn: <Shams>
Please state the Pirate problem.
<al wazir>
I think that
<The reasoning on on all days earlier than day N is contingent on the validity of the premise, "The test cannot be given on day N," which is not YET known to be true.> is incorrect because by the definition of the test they are to be given, we do know that the test can never be given on N. What you said would be true for tests that don't meet the condition <that they will not know that the test will be held on the day of the test.> but since we are only concerned with tests that meet the definition, the statment is invalid. |
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| Jul-01-08 | | Once: <al wazir> I think we may need to mark down the mystery test conundrum as one where we agree to disagree. As far as I am aware, there is no solution to the problem. The difficulty is in disproving the boy's logic. I have never managed to do this, despite many attempts. I'll have one last go. To put it into your terminology: The test cannot be given on day N, because the boys would know that the test was going to happen on that day. If the teacher really wants the test to be a mystery, then he cannot wait until the last day of term to spring it. That logically means that the test cannot be given on day N-1 either. The boys know that the teacher cannot wait until day N, because the test would not be a mystery. So having arrived at school on day N-1, they know that the test has to happen that day. The teacher cannot wait until the last day. But that means that the test is not a mystery. Repeat for each preceding day.
In terms of logic, I am with you. It must be possible to have a mystery test at some point in the term without the boys knowing when it will be. But how do we demolish the boy's logic that it is impossible to hold the test? |
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| Jul-01-08 | | Shams: I'm fairly sure the pirate problem has been posted here before, but here it is: Five pirates have obtained 100 gold coins and have to divide up the loot. The pirates are all extremely intelligent, treacherous and selfish (especially the captain). The captain always proposes a distribution of the loot. All pirates vote on the proposal, and if half the crew or more go "Aye", the loot is divided as proposed, as no pirate would be willing to take on the captain without superior force on their side. If the captain fails to obtain support of at least half his crew (which includes himself), he faces a mutiny, and all pirates will turn against him and make him walk the plank. The pirates start over again with the next senior pirate as captain. What is the maximum number of coins the captain can keep without risking his life? |
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Jul-01-08
| | al wazir: <Once>: The test can obviously be given on any day the teacher wants, or twice, or not at all. The question is, will it or will it not be a surprise when it is given? That self-evidently depends on what you mean by a surprise. Your logical argument is incomplete if you don't define "surprise." |
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| Jul-02-08 | | Once: <shams> Do the other pirates confer before making a decision? If so, they could all decide to make him walk the plank whatever distribution he chooses. Then the gold has to be divided between four rather than five, giving everyone a higher share. |
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| Jul-02-08 | | Once: <al wazir> The test is a surprise in that when the boys arrive at school that morning they do not know (or deduce with 100% certainty) that the test will be that day. But I suspect that we may have flogged this one to death. Let's start another. What is the shortest word in the english language which contains all the vowels, once and once only and in the correct order? There are two answers here, depending on whether you count "y" as a vowel, but the root word is the same. |
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| Jul-02-08 | | kellmano: <once> My first contribution, the horrible word 'abstemious'. |
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| Jul-02-08 | | pawn to QB4: 'facetious' is one shorter, but what's this stuff about "y" as a vowel? |
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| Jul-02-08 | | Once: <pawn to QB4> Great handle, BTW, and the right answer. I mentioned "y" because some people argue that it is an honorary vowel, or functions as a vowel in words such as "sky" and "why". When I have used the puzzle in the past, some pedants have claimed that "facetious" does not qualify because it excludes "y". So to pre-empt arguments like this, I also allow "facetiously". |
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| Jul-02-08 | | pawn to QB4: Thanks a lot. Only the second stumper I've "won". I still have nobody topping this: Opening Explorer (hope that came through) as the position where the largest number of masters went unimaginatively for the same move. So in tribute to your own handle: everyone else picks one move; a lone hero picks another and wins - what's the largest number of sheep successfully shown a winning method in our database? |
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Jul-02-08
 | | OhioChessFan: Believe it or not, the first time I heard that, with the y included, I got abstemiously and facetiously in an instant. |
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