Posted by: kalisto
Is this a 'how smart are you'? Or more of a 'how good are you at propositional calculus'?
Smart is a pretty subjective word.
there's no trickery but most people will make certain assumptions that will lead them to the wrong answer. so it's smart in the sense of how you evaluate the rule and then apply that to the cards. there's no need for any special math skill, it's purely based on logical reasoning.
one thing for sure is that you have to turn over more than one card.
I would flip over all the vowels, just like Vanna.
Posted by: kalisto
I would argue you do not. What is the reasoning for more than 1 card?
In order to prove the validity of a rule, you need to examine not only cases that will prove your rule, but also those that could disprove the rule.
Flip em all Syncro, it's the only way to know.
Posted by: chupacabra
Flip em all Syncro, it's the only way to know.
Just had my ah-ha moment on this.
Assuming the 4 cards shown are the 4 Syncro posted (it just says "let's say we have 4 cards so it might be 4 other cards"), I would flip the A, the D and the 7 and leave the 4 unflipped.
Posted by: chupacabra
Posted by: chupacabra
Flip em all Syncro, it's the only way to know.
Just had my ah-ha moment on this.
Assuming the 4 cards shown are the 4 Syncro posted (it just says "let's say we have 4 cards so it might be 4 other cards"), I would flip the A, the D and the 7 and leave the 4 unflipped.
I derped hard too.
this is the right answer. A, D, 7
No need to flip D. I doesn't matter wrt the rule what is on the other side of the D card.
Posted by: syncro
Again don't explain your reasons or give the answer away.
You're not the boss of me!!!
Posted by: KenN
No need to flip D. I doesn't matter wrt the rule what is on the other side of the D card.
What if it's a vowel? ;)
Posted by: kalisto
Posted by: KenN
No need to flip D. I doesn't matter wrt the rule what is on the other side of the D card.
What if it's a vowel? ;)
I'm betting you've got it now, but for the greater good ...
There's only one rule. Humans have a flaw in their logic, where they will fill in a logical void without even stopping to think it out. There's a tendency to assume that "if there is a vowel on one side, there is an even number on the other side" implies another rule - that if there is a consonant on one side, there must be an odd number on the other side. That isn't at all the case. The rule makes no assumptions about the numbered side of the card, nor does it make assumptions about what is on the back side of a consonant card. A consonant could have an even number, an odd number or the Greek letter AEtna for all we care - it still doesn't negate the single rule as stated.
Posted by: KenN
Posted by: kalisto
Posted by: KenN
No need to flip D. I doesn't matter wrt the rule what is on the other side of the D card.
What if it's a vowel? ;)
I'm betting you've got it now, but for the greater good ...
the cards are supposed to obey the following rule: If there is a vowel on one side then there is an even number on the other side
There's only one rule. Humans have a flaw in their logic, where they will fill in a logical void without even stopping to think it out. There's a tendency to assume that "if there is a vowel on one side, there is an even number on the other side" implies another rule - that if there is a consonant on one side, there must be an odd number on the other side. That isn't at all the case. The rule makes no assumptions about the numbered side of the card, nor does it make assumptions about what is on the back side of a consonant card. A consonant could have an even number, an odd number or the Greek letter AEtna for all we care - it still doesn't negate the single rule as stated.
I think we have official proof that KenN, is in fact, human. There is no rule that says the card must have a number on one side and a letter on the other, therefore what kalisto is saying is possible and must be checked. The only card that doesn't require flipping is the 4 because any letter or number on the other side would maintain the one and only rule.
EDIT: Sorry Syncro, but we were tired of waiting for you and went all Leroy Jenkins on this shit.
the right answer is not A, D and 7.
KenN is on the right track with his thinking, people are making incorrect assumptions about the rule.
it's a conditional in the manner of: If X, then Y.
we want to know is if a card obeys the rule of: If there is a vowel on one side then there is an even number on the other side. so in the case of D, it can't violate the rule of: if vowel, then even number. it's a consonant so the rule does not apply to that card at all.
in the case of number 4, if there is a vowel then it's good and if there is anything else then the rule also does not apply.
so the correct answer is A and 7.
here's a link to the same test but using different conditions. try it out and see what happens.
http://www.philosophyexperiments.com/wason/
Posted by: chupacabra
I think we have official proof that KenN, is in fact, human. There is no rule that says the card must have a number on one side and a letter on the other, therefore what kalisto is saying is possible and must be checked. The only card that doesn't require flipping is the 4 because any letter or number on the other side would maintain the one and only rule.
EDIT: Sorry Syncro, but we were tired of waiting for you and went all Leroy Jenkins on this shit.
no worries. what kalisto is saying is possible (there being a vowel on the other side of D), but it doesn't matter if there is a vowel on the other side of D.
the first condition is vowel, and the second condition is even number. in the case of card D, D is not a number so it cannot violate the rule of if vowel then even number even if there is a vowel on the other side. don't fill in anything extra to the rule.
Posted by: kalisto
Here's a fun interview question:
Don't post up how you did it, just the number of drops in worst case scenario. If you want, you can easily google the answer, but that's way less fun.
You're standing in front of a 100 story building with two identical bowling balls. You've been tasked with testing the bowling balls' resilience. The building has a stairwell with a window at each story from which you can (conveniently) drop bowling balls.
To test the bowling balls you need to find the first floor at which they break. It might be the 100th floor or it might be the 50th floor, but if it breaks somewhere in the middle you know it will break at every floor above.
Devise an algorithm which guarantees you'll find the first floor at which one of your bowling balls will break. Post up the number for the max number you would have to do in the worst case.
Should ask "lowest numbered floor" and not "first floor".
Divide and conquer algorithm.
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