Kenneth Drew Posted August 11, 2015 Share Posted August 11, 2015 hmmm... ahh but which is one side and which is the other? The puzzle said there was an E a K a 4 and a 7 those are the only sides you can look at and if one of them answers if, then you get to look at the other side. Link to comment Share on other sites More sharing options...
Kenneth Drew Posted August 11, 2015 Share Posted August 11, 2015 I don't mean to annoy anyone - i do feel that this is an appropriate (and fun) topic in each of us understanding more what's involved in the conclusion forming process which can either lead to errors in deduction or to progress in our pursuit of accuracy and truth in the solution. one of the real reasons i'm into this thing so much is my passion for 'problem solving,' and i'm sure that's the case for many of ya'll. so these kinds of things are fun, and good for our brains (which we need to solve this thing!) i'm just pasting in this little bit of text and this quick test i found (that I failed) without the answer. if any of you have seen it already, which is very likely, please don't publish the answer, or cheat. so, check it out: If...then... Conditional reasoning is based on an 'if A then B' construct that posits B to be true if A is true. Note that this leaves open the question of what happens when A is false, which means that in this case, B can logically be either true or false. Conditional traps A couple of definitions: In the statement 'If A then B', A is the antecedent and B is the consequent. You can affirm or deny either the antecedent or consequent, which may lead to error. Denying the consequent Denying the consequent means going backwards, saying 'If B is false, then A must also be false.' Thus if you say 'If it is raining, I will get wet', then the trap is to assume that if I am not getting wet then it is not raining. Denying the antecedent Denying the antecedent is making assumptions about what will happen if A is false. Thus if you say 'If it is raining, I will get wet' and is not raining, I might assume that I will not get wet. But then I could fall in the lake. Affirming the consequent This is making assumptions about A if B is shown to be true. Thus if I make the statement 'If it is raining, I will get wet', then if I am getting wet it does not mean that it is raining. The card trap A classic trap was created some years ago; Four cards are laid out as below: The condition is now established (true): 'If a card has a vowel on one side, then it has an even number on the other side.' The problem is to decide which are the minimum cards that need to be turned over to prove that the conditional statement is true. How many and which card(s)? Discuss it among yourselves... Sounds as if you're now changing the original intent: So let's start again: here is the statement: 'If a card has a vowel on one side, then it has an even number on the other side.' You can not rearrange the conditions, it clearly says 'if a card has a vowel on one side', then.... So you can only look to see if a card has a vowel on one side,, if you find one that satisfies the if. you find a vowel, then: then you can see if the other condition is met' then it has an even number on the other side. You can't rearrange it to say if there is an even number on one side then it has a vowel on the other side. if there's a vowel, then.......... No vowel, no 'then'. Link to comment Share on other sites More sharing options...
Glenn Nall Posted August 11, 2015 Author Share Posted August 11, 2015 Ken, I don't understand why you think i'm trying to twist things. I haven't changed a thing, i've tried to clarify the directions because you misunderstood the 'original intent,' i was thinking perhaps my words were not clear. but really, they are not my words - i copied these STRAIGHT from the website from whence they came. and i won't let myself succumb to offering help. I can see where the mistakes are made - IF they're being made. and i'm using the plural 'mistakes' so as to not imply any one person. perhaps someone else will read this and point out what they think is illogic... Link to comment Share on other sites More sharing options...
Kenneth Drew Posted August 11, 2015 Share Posted August 11, 2015 Ken, I don't understand why you think i'm trying to twist things. I haven't changed a thing, i've tried to clarify the directions because you misunderstood the 'original intent,' i was thinking perhaps my words were not clear. but really, they are not my words - i copied these STRAIGHT from the website from whence they came. and i won't let myself succumb to offering help. I can see where the mistakes are made - IF they're being made. and i'm using the plural 'mistakes' so as to not imply any one person. perhaps someone else will read this and point out what they think is illogic... Ok, then if the answer 1 is not correct, then the hypothesis can not be as stated and there will have to be a twist that changes conditions. If it stays as was originally stated If, then. you can't get to the 'then' until you've satisfied the If. I won't say that you are changing the statement, but your coaching implies that there is a twist to it and in this type problem, a twist is NOT allowed. you can not jump over to then it's an even number so it has to be a vowel on the other side. You can't see what on the other side of an even number, because you can only look on the other side of a vowel. If vowel, then........ not even number so maybe vowel.....no no Link to comment Share on other sites More sharing options...
Mark Knight Posted August 11, 2015 Share Posted August 11, 2015 OK...here's my shot at this. We are given the statement, 'If a card has a vowel on one side, then it has an even number on the other side.' If the point of the exercise is to prove the statement, and ONLY to prove the statement, then we only need to turn over the card with the E. My logic is...since E is the only vowel, it is the only card we can turn that can prove the statement. I conclude that because K is not a vowel, so turning that card will neither prove nor disprove the statement. Since the statement does not say that ONLY cards with even numbers on one side have vowels on the other side, the results of turning over the card with the 7 on one side are inconsequential, relating to the statement we are given. And since the statement does NOT say that cards with even numbers on one side will ONLY have vowels on the opposite side, turning over the card with the 4 on one side is also inconsequential in relation to the statement. Therefore, turning only one card--the card with the E on one side--will prove [or disprove] the statement. Am I even still on topic here? Or did I drift off on a tangent? Link to comment Share on other sites More sharing options...
Glenn Nall Posted August 11, 2015 Author Share Posted August 11, 2015 you're reading way too much into this. let me try to say this without giving up the answer. you may not like it, but you're not an idiot and if you have not answered the problem then perhaps there IS a challenge to it, and i'm hoping that at least some OTHERS will find the fun in it. OK - there IS NO twist. it's the common error this doctor says many people make in jumping to an errant conclusion from a given set of circumstances. the statement is: If a card has a vowel on one side, then it has an even number on the other side. If a vowel is visible on a side of the card, then the rule states that an EVEN number will be found on the other side of the card when it's turned over. it does NOT say "If a vowel is seen on a card 'then you can see if'" anything... The premise is: EVEN NUMBERS will ALWAYS be found on the other side of a VOWEL. that's all it says. NOW read the challenge... Link to comment Share on other sites More sharing options...
Mark Knight Posted August 11, 2015 Share Posted August 11, 2015 you're reading way too much into this. let me try to say this without giving up the answer. you may not like it, but you're not an idiot and if you have not answered the problem then perhaps there IS a challenge to it, and i'm hoping that at least some OTHERS will find the fun in it. OK - there IS NO twist. it's the common error this doctor says many people make in jumping to an errant conclusion from a given set of circumstances. the statement is: If a card has a vowel on one side, then it has an even number on the other side. If a vowel is visible on a side of the card, then the rule states that an EVEN number will be found on the other side of the card when it's turned over. it does NOT say "If a vowel is seen on a card 'then you can see if'" anything... The premise is: EVEN NUMBERS will ALWAYS be found on the other side of a VOWEL. that's all it says. NOW read the challenge... Was this addressed towards MY post? Link to comment Share on other sites More sharing options...
Pat Speer Posted August 11, 2015 Share Posted August 11, 2015 (edited) We have to look at all four, because all four could have a vowel on the other side. Actually, I take that back. We don't need to look at the 4. If it has a vowel on the other side, it is consistent. If it does not have a vowel on the other side, it is still consistent. It was never said, after all, that there can not be an even number unless there is a vowel on the other side. Edited August 11, 2015 by Pat Speer Link to comment Share on other sites More sharing options...
Glenn Nall Posted August 11, 2015 Author Share Posted August 11, 2015 (edited) OK...here's my shot at this. We are given the statement, 'If a card has a vowel on one side, then it has an even number on the other side.' If the point of the exercise is to prove the statement, and ONLY to prove the statement, then we only need to turn over the card with the E. My logic is...since E is the only vowel, it is the only card we can turn that can prove the statement. I conclude that because K is not a vowel, so turning that card will neither prove nor disprove the statement. Since the statement does not say that ONLY cards with even numbers on one side have vowels on the other side, the results of turning over the card with the 7 on one side are inconsequential, relating to the statement we are given. And since the statement does NOT say that cards with even numbers on one side will ONLY have vowels on the opposite side, turning over the card with the 4 on one side is also inconsequential in relation to the statement. Therefore, turning only one card--the card with the E on one side--will prove [or disprove] the statement. Am I even still on topic here? Or did I drift off on a tangent? thanks Mark. If the point of the exercise is to prove the statement - right -> "WHAT are the minimum cards that need to be turned over to prove that the conditional statement is true" what we KNOW is: If a card has a vowel on one side, then it has an even number on the other side. --OR-- if VOWEL - then EVEN - if CONSONANT - then ... any of the four (vowel, consonant, even, odd)... right? none are excluded...? if EVEN, then ... any of the four... right? are any excluded? if ODD, then ... any of the four? are any excluded...? you're not off track. it's not as easy as we think. this is why it's easy - for some - to say, "if there are shells on the floor, and the gun is registered to BOB, then Bob HAD TO have shot the gun." Edited August 11, 2015 by Glenn Nall Link to comment Share on other sites More sharing options...
Glenn Nall Posted August 11, 2015 Author Share Posted August 11, 2015 (edited) you're reading way too much into this. let me try to say this without giving up the answer. you may not like it, but you're not an idiot and if you have not answered the problem then perhaps there IS a challenge to it, and i'm hoping that at least some OTHERS will find the fun in it. OK - there IS NO twist. it's the common error this doctor says many people make in jumping to an errant conclusion from a given set of circumstances. the statement is: If a card has a vowel on one side, then it has an even number on the other side. If a vowel is visible on a side of the card, then the rule states that an EVEN number will be found on the other side of the card when it's turned over. it does NOT say "If a vowel is seen on a card 'then you can see if'" anything... The premise is: EVEN NUMBERS will ALWAYS be found on the other side of a VOWEL. that's all it says. NOW read the challenge... Was this addressed towards MY post? no no, sorry - i wrote that before you had posted, when it was only Ken complaining, i mean playing... that was for him. Edited August 11, 2015 by Glenn Nall Link to comment Share on other sites More sharing options...
Glenn Nall Posted August 11, 2015 Author Share Posted August 11, 2015 (edited) the challenge, in my opinion, is reading the evidence that is given ONLY for what it is, and not adding to it, which apparently is what humans are prone to do, since so few people got this right. reading the evidence that is given ONLY for what it is - an example, (and I'm sorry, Ken, but it works as an example) - Ken took from the GIVEN to mean - not that there will be an even number on the other side of a vowel - but that he could only turn over a card with a vowel on it. I'm not faulting him, i'm fascinated that one person reads that so differently than another. going ONLY where the evidence takes you, and no further... and now Pat has come up with something quite unique... Edited August 11, 2015 by Glenn Nall Link to comment Share on other sites More sharing options...
Glenn Nall Posted August 11, 2015 Author Share Posted August 11, 2015 you're reading way too much into this. let me try to say this without giving up the answer. you may not like it, but you're not an idiot and if you have not answered the problem then perhaps there IS a challenge to it, and i'm hoping that at least some OTHERS will find the fun in it. OK - there IS NO twist. it's the common error this doctor says many people make in jumping to an errant conclusion from a given set of circumstances. the statement is: If a card has a vowel on one side, then it has an even number on the other side. If a vowel is visible on a side of the card, then the rule states that an EVEN number will be found on the other side of the card when it's turned over. it does NOT say "If a vowel is seen on a card 'then you can see if'" anything... The premise is: EVEN NUMBERS will ALWAYS be found on the other side of a VOWEL. that's all it says. NOW read the challenge... Was this addressed towards MY post? rereading your initial offering, you almost nailed it, but blew past a possibility without seeing it. i pledged not to help, but Ken was bitching that the test is flawed, so i was too weak not to respond. the test is not flawed. it's really simple. it's just not easy... i'll provide the link to the original study once it's done... Link to comment Share on other sites More sharing options...
Mark Knight Posted August 11, 2015 Share Posted August 11, 2015 The reason I only went with the card with E is because that is the only one that met the criteria of the original statement. To expect the converse to be true--that if a card has an even number, it will have a vowel on the other side--well, that's NOT within the known parameters of the question. It may or may not have a vowel on the opposite side. And since our statement does not mention odd numbers or consonants, we cannot decide anything for certain. Either of those cards might have ANY number, or ANY letter, on the opposite side. After all, there was no statement that said that only letters had numbers on the opposite side of the card, or that only numbers had letters on the back side of the card. All we know is that cards with vowels on one side have even numbers on the opposite side. And the ONLY card with a vowel on it, of those we are given, is the E. Link to comment Share on other sites More sharing options...
Glenn Nall Posted August 11, 2015 Author Share Posted August 11, 2015 (edited) The reason I only went with the card with E is because that is the only one that met the criteria of the original statement. To expect the converse to be true--that if a card has an even number, it will have a vowel on the other side--well, that's NOT within the known parameters of the question. It may or may not have a vowel on the opposite side. And since our statement does not mention odd numbers or consonants, we cannot decide anything for certain. Either of those cards might have ANY number, or ANY letter, on the opposite side. After all, there was no statement that said that only letters had numbers on the opposite side of the card, or that only numbers had letters on the back side of the card. All we know is that cards with vowels on one side have even numbers on the opposite side. And the ONLY card with a vowel on it, of those we are given, is the E. you are exactly right. and you're missing what i missed. in fact, you're even closer than i was to it... (wow.) right. that's what i meant by "all four" - i think of the possibilities as ONLY 4 different entities - vowels, cons., evens and odds. (since we can see a K and a 7, and by implying that a vowel or an even is ONLY a possibility, then we know that consonants and odds are in play) Edited August 11, 2015 by Glenn Nall Link to comment Share on other sites More sharing options...
Glenn Nall Posted August 11, 2015 Author Share Posted August 11, 2015 so, right - if you turn E over and behold there's a VOWEL, so far the statement is true... and that does not solve the problem... Link to comment Share on other sites More sharing options...
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