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This is the bit from the text which explains the correct answer. Quote:
I must look at the card with the vowel showing to find out what is on the other side because it could be an odd number and thus would show me that the statement is false. I must also look at the card with the odd number to find out what is on the other side because it could be a vowel and thus would show me that the statement is false. I don't need to look at the card with the consonant because the statement I am testing has nothing to do with consonants. Nor do I need to look at the card with the even number showing because whether the other side has a vowel or a consonant will not help me determine whether the statement is false.
In other words: A and 7.
I think this is a pretty tough problem! The solution depends on a very literal-minded interpretation of "Which card(s) must you turn over to determine whether the following statement is *false*?" which statement, to answer the problem correctly, should not be equated with proving that the statement is true.
In other words, it is not a "true or false?" problem; but a "prove this false" problem. Considering this fact, that "whether" in the statement is very unkind. It tempts you into thinking "whether or not" where "not" means "true" and therefore resolve the statement in your mind into the "true or false" dichotomy you are meant to avoid to solve the ******* problem.
Apparently, the problem is made difficult by the "confirmation bias": The fact that the brain prefers to try to prove something is true (and therefore not false) than false (and therefore not true).
Anway, it got me.
(Well done Faldage)
Last edited by Hydra; 10/26/06 02:30 PM.
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But again, [not to sound pompous or campy,] the answer should be A or 7, or possibly both, depending on the interim result. Disproving the statement may be possible with a single card. 
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The instructions are "Which card(s) must you turn over to determine whether the following statement is false?". Your solution is to "Which card(s) MIGHT you turn over...".
Turning over A and 7 must determine the outcome. Turning over either A or 7 might determine the outcome.
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But even turning over A and 7 may not prove the statement false, depending on what is revealed. For that matter it was not even stipulated that this was the complete card set vice a sampling from a larger pack [not an original observation at this point]. Given the latter possibility, 'might' is all there is to work with. So "A or 7" is correct if one or both of those cards falsifies the statement. The answer does not suggest being committed to selecting a particular card; only that one of them (if any) will be sufficient, despite any undue melodrama, to dispatch the villain. 
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Four cards are presented: A, D, 4, and 7. There is a letter on one side of each card and a number on the other side. Which card(s) must you turn over to determine whether the following statement is false? "If a card has a vowel on one side, then it has an even number on the other side."
question: why would you interpret this to suggest that there are more than four cards?
suggested answer: to further muddy the situation, and to further muddy the situation only.
Last edited by tsuwm; 10/27/06 07:19 PM.
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Quote:
Four cards are presented: A, D, 4, and 7. There is a letter on one side of each card and a number on the other side. Which card(s) must you turn over to determine whether the following statement is false? "If a card has a vowel on one side, then it has an even number on the other side."
question: why would you interpret this to suggest that there are more than four cards?
suggested answer: to further muddy the situation, and to further muddy the situation only.
Why would you assume that there are only four cards?
But irregardless of whether or not there are or aren't only four cards or not, only by the turning over of A and 7 do you have a hope of disproving the hypothesis.
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it's a classic philosophical/psychological test which seems to always have four "cards", as here and here .
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Quote:
it's a classic philosophical/psychological test which seems to always have four "cards", as here and here .
Or here?
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The whole thing is a long-winded approach to seeing if the examinee understands that the assumption "if a then b" implies the statement "if not b then then not a." And, the red herring "not a" (which does not prove "not b") is there to tempt the examinee.
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