Thomas Graves Posted August 13, 2015 Share Posted August 13, 2015 Glenn Are you saying the correct answer is E and 7? Robert, Yes, that's what he's saying. The "E" must be checked to make sure that the rule is being followed -- if it has an odd number on the other side, then the rule isn't being followed. The "7" must be checked to make sure it doesn't have a vowel on the other side -- if it has a vowel on the other side, then the rule isn't being followed. The "K" doesn't need to be checked because it doesn't matter whether it has an even or an odd number on the other side -- the rule doesn't say that only a vowel can have an even number on the other side. The "4" doesn't need to be checked because the rule doesn't say that if a card has an even number on one side, it must have a vowel on the other side. --Tommy Link to comment Share on other sites More sharing options...
Kenneth Drew Posted August 13, 2015 Share Posted August 13, 2015 Glenn Are you saying the correct answer is E and 7? That's what he's saying, but that is conditional, in my opinion on whether the answer is based on the first version or the 2nd version. Link to comment Share on other sites More sharing options...
Kenneth Drew Posted August 13, 2015 Share Posted August 13, 2015 (edited) Glenn Are you saying the correct answer is E and 7? Robert, Yes, that's what he's saying. The "E" must be checked to make sure that the rule is being followed -- if it has an odd number on the other side, then the rule isn't being followed. The "7" must be checked to make sure it doesn't have a vowel on the other side -- if it has a vowel on the other side, then the rule isn't being followed. The "K" doesn't need to be checked because it doesn't matter whether it has an even or an odd number on the other side -- the rule doesn't say that only a vowel can have an even number on the other side. The "4" doesn't need to be checked because the rule doesn't say that if a card has an even number on one side, it must have a vowel on the other side. --Tommy The "E" must be checked to make sure that the rule is being followed -- if it has an odd number on the other side, then the rule isn't being followed. I still disagree. It was stated originally that IF there is a vowel there will be an even number. Since there is a vowel, then you already know there is an even number. the 7 does not to be checked because there can't be a vowel there because the rule said if there is a vowel, then there will be an even number. So if the other side had a vowel, then you couldn't have an odd number. This is like saying that it is a fact that if a traffic light is red one direction then it is green for the other direction. then asking if the light is red for your direction can we assume it is green for the other direction, or do we actually have to go look and see? I have never gotten out of my car at a red light to go and see if it was green for the other direction. Edited August 13, 2015 by Kenneth Drew Link to comment Share on other sites More sharing options...
Thomas Graves Posted August 13, 2015 Share Posted August 13, 2015 (edited) Glenn Are you saying the correct answer is E and 7? Robert, Yes, that's what he's saying. The "E" must be checked to make sure that the rule is being followed -- if it has an odd number on the other side, then the rule isn't being followed. The "7" must be checked to make sure it doesn't have a vowel on the other side -- if it has a vowel on the other side, then the rule isn't being followed. The "K" doesn't need to be checked because it doesn't matter whether it has an even or an odd number on the other side -- the rule doesn't say that only a vowel can have an even number on the other side. The "4" doesn't need to be checked because the rule doesn't say that if a card has an even number on one side, it must have a vowel on the other side. --Tommy The "E" must be checked to make sure that the rule is being followed -- if it has an odd number on the other side, then the rule isn't being followed. I still disagree. It was stated originally that IF there is a vowel there will be an even number. Since there is a vowel, then you already know there is an even number. the 7 does not to be checked because there can't be a vowel there because the rule said if there is a vowel, then there will be an even number. So if the other side had a vowel, then you couldn't have an odd number. This is like saying that it is a fact that if a traffic light is red one direction then it is green for the other direction. then asking if the light is red for your direction can we assume it is green for the other direction, or do we actually have to go look and see? I have never gotten out of my car at a red light to go and see if it was green for the other direction. Ken, Try looking at it this way. Assuming that there are four cards lying on a table showing an "E" , a "7" , a "K" , and a "4," and that you've been told by a trustworthy person who has looked at both sides of the cards that each of the cards has a letter on one side and a number on the other, and bearing in mind that the rule states, in so many words, "If there is a vowel on one side of a card, there should / must be an even number on the other side," -- The Question Is: Which of the four cards must you turn over to make sure that the person who created this exercise didn't "space out" and put an odd number on the other side of a vowel by mistake (or a vowel on the other side of an odd number, for that matter), which is the only combination that's prohibited? The Solution: Since you are only concerned with what is prohibited, and since the only prohibited combination is a vowel and an odd number, you are concerned just with what's on the other side of the only vowel showing ("E"), and what's on the other side of the only odd number showing ("7"). Bear in mind that the rule doesn't say that there should / must be a vowel on the other side of every even number, just that there must be an even number on the other side of every vowel !!! Since it's human nature to occasionally make mistakes when writing down letters and numbers, you have to check the "E" card and the "7" card to make sure that the person who wrote the letter or number on the other side of those two cards didn't "space out," but actually followed the rule. Remember: It's okay if there's an even number on the other side of the "K," or a consonant on the other side of the "4" because the rule doesn't prohibit those combinations. The only thing the rule prohibits is the combination of a vowel and and an odd number / an odd number and a vowel. --Tommy Edited August 13, 2015 by Thomas Graves Link to comment Share on other sites More sharing options...
Jon G. Tidd Posted August 13, 2015 Share Posted August 13, 2015 Glenn's problem boils down to this testable proposition: If A, then B. One can test the truth of this proposition by examining samples that present either A or B. If a sample presents A, we know the proposition is true if the sample also presents B. If the sample presents X, we know the proposition is false if the sample also presents B. What we don't know explicitly from Glenn is the sample size. Glenn has presented four cards. The E card might have a 4 on the other side, and the proposition would be true so far. The 7 card might have a G or might have a U. A G means the proposition is correct so far. A U means the proposition is incorrect so far. Glenn hasn't told us how to test the truth of his proposition. If truth is established by one instance out of four, that's one thing. If truth requires consistency among among all four instances, that's another thing. If truth requires consistency among the infinite number of possibilities, that's quite another thing. Why an infinite number of possibilities? There is a finite number of vowels: a, e, i, o , u. As for even numbers, there are infinitely many. Link to comment Share on other sites More sharing options...
John Iacoletti Posted August 13, 2015 Share Posted August 13, 2015 (edited) If there is a vowel on the other side of the K card, then the proposition is proved false. WRONG. Please explain how this is wrong. The proposition is "If a card has a vowel on one side, then it has an even number on the other side." If you turn the K over and it has a vowel on the other side, then this card has a vowel on one side and something other than an even number on the other side (a K). The proposition would be proved false. Edit: I see that Glenn corrected himself in a later post. It does change the answer if you stipulate that the cards must have a letter on one side and a number on the other. Edited August 13, 2015 by John Iacoletti Link to comment Share on other sites More sharing options...
Thomas Graves Posted August 13, 2015 Share Posted August 13, 2015 (edited) Glenn's problem boils down to this testable proposition: If A, then B. One can test the truth of this proposition by examining samples that present either A or B. If a sample presents A, we know the proposition is true if the sample also presents B. If the sample presents X, we know the proposition is false if the sample also presents B. What we don't know explicitly from Glenn is the sample size. Glenn has presented four cards. The E card might have a 4 on the other side, and the proposition would be true so far. The 7 card might have a G or might have a U. A G means the proposition is correct so far. A U means the proposition is incorrect so far. Glenn hasn't told us how to test the truth of his proposition. If truth is established by one instance out of four, that's one thing. If truth requires consistency among among all four instances, that's another thing. If truth requires consistency among the infinite number of possibilities, that's quite another thing. Why an infinite number of possibilities? There is a finite number of vowels: a, e, i, o , u. As for even numbers, there are infinitely many. Dear John, We can reasonably assume that the question applied to only the four cards mentioned in the exercise itself. How could / would he have worded the exercise if he'd wanted us to consider all possible cards in the universe that had a rational number on one side and an upper-case letter from the Modern English alphabet on the other? Does the fact that he didn't do this suggest that he wanted us to infer that that was what he had intended? "Which of the infinite number of cards in the universe would you have to turn over...." LOL Or then again, we can unreasonably quibble, split hairs, and apply "if, then" "then, if" logic like so many of us do in our JFK assassination "research" and "analysis." A perfect microcosm (or is it macrocosm?). Solution for an infinite number of cards: Turn over over all of the cards showing a vowel, and turn over all of the cards showing an odd number. If any of the former have an odd number on the other side, or if any of the latter have a vowel on the other side, then God is either messing with you or has made an honest mistake. --Tommy Edited August 14, 2015 by Thomas Graves Link to comment Share on other sites More sharing options...
Jon G. Tidd Posted August 13, 2015 Share Posted August 13, 2015 Thomas Graves, If the four cards Glenn presents is the entire universe of possibilities, that's an if, then only two cards need to be turned over to test the proposition that in all cases of the universe of possibilities, a card bearing a vowel on one side has an even number on the other side. Those two cards are the E and the 7 card. All cases within the universe, Thomas. Link to comment Share on other sites More sharing options...
Thomas Graves Posted August 13, 2015 Share Posted August 13, 2015 (edited) Thomas Graves, If the four cards Glenn presents is the entire universe of possibilities, that's an if, then only two cards need to be turned over to test the proposition that in all cases of the universe of possibilities, a card bearing a vowel on one side has an even number on the other side. Those two cards are the E and the 7 card. All cases within the universe, Thomas. Jon, Correcto mundo. Mas pravdu. "You are correct, sir." --Tommy Edited August 14, 2015 by Thomas Graves Link to comment Share on other sites More sharing options...
Paul Hailes Posted August 14, 2015 Share Posted August 14, 2015 And the point you discuss immediately above is salient. At no point in the description of the problem are any other cards other than the 4 here mentioned, nor implied. That is one of the many assumptions that many seem to make in looking at this problem. In order to find the correct solution it is imperative to pare down the problem to its minimum. Check all your assumptions at the door before entering. And it is clearly very difficult to do. Link to comment Share on other sites More sharing options...
Thomas Graves Posted August 14, 2015 Share Posted August 14, 2015 (edited) Glenn's problem boils down to this testable proposition: If A, then B. One can test the truth of this proposition by examining samples that present either A or B. If a sample presents A, we know the proposition is true if the sample also presents B. If the sample presents X, we know the proposition is false if the sample also presents B. What we don't know explicitly from Glenn is the sample size. Glenn has presented four cards. The E card might have a 4 on the other side, and the proposition would be true so far. The 7 card might have a G or might have a U. A G means the proposition is correct so far. A U means the proposition is incorrect so far. Glenn hasn't told us how to test the truth of his proposition. If truth is established by one instance out of four, that's one thing. If truth requires consistency among among all four instances, that's another thing. If truth requires consistency among the infinite number of possibilities, that's quite another thing. Why an infinite number of possibilities? There is a finite number of vowels: a, e, i, o , u. As for even numbers, there are infinitely many. Dear John, We can reasonably assume that the question applied to only the four cards mentioned in the exercise itself. How could / would he have worded the exercise if he'd wanted us to consider all possible cards in the universe that had a rational number on one side and an upper-case letter from the Modern English alphabet on the other? Does the fact that he didn't do this suggest that he wanted us to infer that that was what he had intended? "Which of the infinite number of cards in the universe would you have to turn over...." LOL Or then again, we can unreasonably quibble, split hairs, and apply "if, then" "then, if" logic like so many of us do in our JFK assassination "research" and "analysis." A perfect microcosm (or is it macrocosm?). Solution for an infinite number of cards: Turn over all of the cards showing a vowel, and turn over all of the cards showing an odd number. If any of the former have an odd number on the other side, or if any of the latter have a vowel on the other side, then God is either messing with you or has made an honest mistake. --Tommy edited and bumped Edited August 14, 2015 by Thomas Graves Link to comment Share on other sites More sharing options...
Glenn Nall Posted August 14, 2015 Author Share Posted August 14, 2015 If there is a vowel on the other side of the K card, then the proposition is proved false. WRONG. Please explain how this is wrong. The proposition is "If a card has a vowel on one side, then it has an even number on the other side." If you turn the K over and it has a vowel on the other side, then this card has a vowel on one side and something other than an even number on the other side (a K). The proposition would be proved false. Edit: I see that Glenn corrected himself in a later post. It does change the answer if you stipulate that the cards must have a letter on one side and a number on the other. yes, i did. thanks, Link to comment Share on other sites More sharing options...
Glenn Nall Posted August 14, 2015 Author Share Posted August 14, 2015 And the point you discuss immediately above is salient. At no point in the description of the problem are any other cards other than the 4 here mentioned, nor implied. That is one of the many assumptions that many seem to make in looking at this problem. In order to find the correct solution it is imperative to pare down the problem to its minimum. Check all your assumptions at the door before entering. And it is clearly very difficult to do. right. this is i think the crux of the exercise, the practice of entering into a thing (a religion, a date, a murder investigation, a clue in the investigation) with no presupposition - and the awareness of the human tendency to do so... I am ABSOLUTELY amazed at all that i've learned in what i thought was going to be a brief and boring and easily guessed little ol' test. i've always been fascinated with the human mind, and in a very Nonjudgmental way I've seen INTO the many ways different people can see the same thing. I have had to send my 'main squeeze' - my dutiful and loyal laptop - to the shop for a week and I'm stuck with this sadistic PC with its merciless keyboard; it's like spreading ice cold butter on fresh bread trying to type a sentence with this damn thing. when i can get in front of a keyboard that was manufactured in the twentieth century I look forward to adding my comments - in keeping with the JFK theme and our different approaches to a single piece of evidence. I'd encourage everyone, whether they contributed to the test or not, to read through the discourse and observe all that can be observed. it's pretty illuminating in how problems get solved, individually, as groups (the DPD, the HSCA, etc); and how they don't get solved. wow. too much fun. i had no idea... Link to comment Share on other sites More sharing options...
Glenn Nall Posted August 14, 2015 Author Share Posted August 14, 2015 and guys (and ladies) i couldn't be more grateful for the participation - all of it. utterly dynamic when i expected so much less. huge thanks, there's so much to learn about learning (which is what we're doing, investigating a crime, right?). Link to comment Share on other sites More sharing options...
Glenn Nall Posted August 14, 2015 Author Share Posted August 14, 2015 Glenn's problem boils down to this testable proposition: If A, then B. One can test the truth of this proposition by examining samples that present either A or B. If a sample presents A, we know the proposition is true if the sample also presents B. If the sample presents X, we know the proposition is false if the sample also presents B. What we don't know explicitly from Glenn is the sample size. Glenn has presented four cards. The E card might have a 4 on the other side, and the proposition would be true so far. The 7 card might have a G or might have a U. A G means the proposition is correct so far. A U means the proposition is incorrect so far. Glenn hasn't told us how to test the truth of his proposition. If truth is established by one instance out of four, that's one thing. If truth requires consistency among among all four instances, that's another thing. If truth requires consistency among the infinite number of possibilities, that's quite another thing. Why an infinite number of possibilities? There is a finite number of vowels: a, e, i, o , u. As for even numbers, there are infinitely many. Dear John, We can reasonably assume that the question applied to only the four cards mentioned in the exercise itself. How could / would he have worded the exercise if he'd wanted us to consider all possible cards in the universe that had a rational number on one side and an upper-case letter from the Modern English alphabet on the other? Does the fact that he didn't do this suggest that he wanted us to infer that that was what he had intended? "Which of the infinite number of cards in the universe would you have to turn over...." LOL Or then again, we can unreasonably quibble, split hairs, and apply "if, then" "then, if" logic like so many of us do in our JFK assassination "research" and "analysis." A perfect microcosm (or is it macrocosm?). Solution for an infinite number of cards: Turn over all of the cards showing a vowel, and turn over all of the cards showing an odd number. If any of the former have an odd number on the other side, or if any of the latter have a vowel on the other side, then God is either messing with you or has made an honest mistake. --Tommy edited and bumped Solution for an infinite number of cards (OR the four-cards corollary): Turn over all of the cards showing a vowel, and turn over all of the cards showing an odd number. If any of the former have an odd number on the other side, or if any of the latter have a vowel on the other side, then God is either messing with you or has made an honest mistake. For an infinite number of cards, the options change. the falseness of the antecedent can be proved, but its accuracy can never be. This is why the solution is stated to be the "least" number of turns that need to be made in order to prove true or false. with an infinite number of cards, as soon as a vowel is turned with no even #, or as soon as an even # is turned with no vowel, falseness occurs. and to the contrary, the trueness of the problem can never be proved. Link to comment Share on other sites More sharing options...
Recommended Posts
Please sign in to comment
You will be able to leave a comment after signing in
Sign In Now