Yeah, my first answer was “One Card, the 8″.

Only two cards are *showing* numbers, the 3 and the 8. The two letter cards aren’t *showing* numbers, so they don’t need to be checked, no?

Or did the person asking the question really mean *The rule is: If the card has an even number on one face, the reverse face must have a vowel.*

Then, given four cards showing E, 3, 8, D, what is the minimum number of card that need to be turned over to verify that all four cards follow this rule? In this case the answer would be two.

Note: there’s no rule that non-even mumbers can’t have a vowel on the reverse. So:

E = Don’t care

3 = Don’t care

8 = Care, the other face **must** be a vowel

D = care, the other face must **not** be a vowel

Think about it, you could get the right answer (“TWO”) and still get the reasoning wrong (the wrong “E and 3″).

I hate these questions which use terms and/or phrasing that are ambiguous – thus giving you multiple ‘correct’ answers – ESPECIALLY when the questions are for logical puzzles .

This habit/trait/quality of reading multiple possible meanings of an imprecise question is the bane of people with Asperger Syndrome, borderline AS, and computer hackers.

See http://catb.org/jargon/html/speech-style.html

But “If not P, then not Q” and “If Q, then P” are not.

Replace the P’s and Q’s for the second statement and you get the first statement.

“If not P, then not Q” and “If Q, then P” (logical equivalents) becomes:

“If not Q, then not P” and “If P, then Q” (logical equivalents **and** your first statement).

This one was the least intuitive one from my Discrete Math class, until I used an everyday example: If it’s Thursday, then it’s payday. If it’s not payday, then it’s not Thursday.

]]>Similar to “Anyone named Debbie is a female” does not imply that all females are named Debbie.

]]>if the number showing is even, then the back must be a vowel

we assume that the backs of the cards, if seen, would follow the rule.

even(side)==>vowel(side)

if p then q.

p therefore q

not q therefore not p

this is the logic.

hence, not[vowel(side)]==>not[even(side)] or even(side)==>vowel(side)

both conclusions must hold. this is why you test D as well as 8.

it seems that you would want to test E, but that is incorrect for this “toy problem” since vowel(side) DOES NOT ==> even(side) and odd(side) DOES NOT ==> consonant(side)

the rule is lame however, since it refers only to what’s face up not to what’s potentially face up. this is important because otherwise, you’d have to check every card.

]]>Psychology often gets a bad rep, and one of the reasons is that its cool and applicable findings are often called by other names. ]]>

So, flip over 8. Make sure it has a vowel.

Flip over D. Make sure that the side is odd.

You don’t care about 3, because it isn’t even.

You don’t care about E, because an odd card could have a vowel or a consonant.

The two statements “if P, then Q” and “if not Q, not P” are logically equivalent.

But “If not P, then not Q” and “If Q, then P” are not.

]]>http://www.mrperezonlinemathtutor.com/A2/11_2_Prob_Comb_Perm.html

As math fans you maybe like it as well!

]]>If instead the rule said “if one side of the card is an even number, then the other side must be a vowel”, then how many cards do you need to check this time? *THREE* of them. That’s because, you see, there could be a “4″ on the other side of that “3″ and so it would be violating the rule. Any of those could be violating the rule, except the first one.

And if instead the rule said “there is a number on one side and a letter on the other side of each card. If the number is even, then the letter must be a vowel. Which cards must you flip to make sure the rule is followed?” – Then you only need to check two cards. 8 and D.

Hope that helps.

]]>And the solution – 2 cards – is not 100% correct.

Quote:

You need to make sure that 8 has a vowel on the back, and you need to make sure that D does NOT have an even number on the back. (End Quote)

You also need to make sure that the “3″ card doesn’t have a even number on the other side. This means you have to flip 3 cards.

The given solution 2 cards assumes that all cards have a number on one side and a letter on the other. The rules don’t actually claim this. So the “3″ card need checking.

]]>You see four persons, one drinking beer, one drinking coke, one very young and one very old but you can’t see what the two last ones are drinking. Who to check? Obviously the beer drinker and also the drink contents of the young one. ]]>

Also, agree with Jere7my.

Looks like the guy really did fuck up the puzzle to the point of un-relevance. ]]>

E need not be flipped because it is vacuously true; it does not satisfy “If the number showing is even” and as such an be a vowel or not yet the statement holds.

3 need not be flipped for the same reason.

8 needs to be flipped to ensure the truth of the second part of the statement.

D needs to be flipped to ensure the contrapositive holds, that is “If the back does not have a vowel, then the front is not even” which is logically equivalent to the original statement.

Basically an “if…then” statement is only false when the “if” is true and the “then” is false.

]]>I’m not sure it’s the social aspect that makes it easier: just rewriting the rule in the card game to “If the number showing is even, then the back of the card MUST NOT have a consonant” makes it much easier. People just understand ‘forbidden’ better than ‘mandatory’.

]]>you got game….theory

lovely n-dimensional curves

20 sided fuzzy dice in the mustang of love

there’s a fraction too much friction

]]>A better analogy would be “You see four people. One is drinking beer, one is drinking coke, one is 22 years old and has an unknown beverage, and one is 19 years old and has an unknown beverage. Whose IDs and/or drinks do you have to inspect?”

]]>Calling it “punk” will make it alluring? The windblown comb-over, maybe?

]]>The puzzle is very old and the results are consistent. When the same logic underlay puzzles in social situations, the subjects tend to ace them. ]]>

In re: the social phrasing: Thanks to Dan Wineman for your correction. That looks logically equivalent, and the cheat-detecting check is “Ignore the cola drinker, ignore the 22-year-old; check only the beer-drinker’s ID and the underage drinker’s beverage.”

In re: the hair: Nothing seems to help~

-Tom Henderson

]]>I would love to not only hear from this intelligent crowd their reactions to that critique, but also a response to a question I have about this task:

It seems like turning over each card offers either a chance for confirmation or disconfirmation (or both or neither) of the rule (correctly stated as indicated in comments above):

“E” – offers only confirmation

“3″ – offers neither

“8″ – offers both

“D” – offers only disconfirmation

My question is: Most people indicate that the subject should turn over “8″ and “D” – is the value of turning over “8″ purely in the chance for disconfirmation? Because, if there is value in confirmation, then couldn’t that be possibly achieved by turning over “E” as well?

Thanks! This was an awesome comment thread to read – I need to come over to boingboing more often!

]]>