Flip over this one: |4|

Since any one of them can prove the rule wrong, then you have to turn them all over to prove it right.

Unless your missing some assumptions.

Ah wait ... Haha, tricky. I like it, the only part of the question that matters is "if its a vowel" you only have to turn over the E card.

Let's save the explanation for now. I would be interested to see if there are any other answers (I going to withhold if anyone is right or wrong for now). Once we have had a few more answer we'll discuss.

Wait again, if the 4 has a vowel, then it disproved the rule, so you have to flip it as well. So flip both 4 and E

Oh, ok. Sorry. Obviously my explanations are even confusing me though, so hopefully I'm not giving anything away. :-)

Yeah, philcore. After thinking about it more, I don't see how either of E and 4 can be left unflipped and you are still certain the rule is correct.

|E| and |4|.

Waaiiit... maybe we want to flip neither of E and 4? That must be it. He wants us to DISprove the rule, not verify it.

E 4

Oh, I misread the question. I thought it was asking "which is the one card you can flip over to verify or refute the claim", but I don't think you can do it with one card. So yes, the answer is flip E and 4.

I would turn the cars [E] and [4] up. The cars [C] and [5] are irrelevant.

Car [5] is only relevant for the case:
""If and ONLY IF a card has a vowel on one side, then it has an odd number on the other side."

Are they on a glass table? Or is there another way of determining what is on the reverse without flipping them?

No they are not on a glass table. Lol.

It is of course cards E and 4.
This test, which is very famous, is called the Wason Selection task. The key is to attempt to disprove the hypothesis by turning over the 4 card (if there is a vowel the statement must be false) rather than confirm the hypothesis (by turning over the 5 to see if there is a vowel).
When this test was first put to university students 90% got it wrong (and this has been replicated many times). The answer most people are immediately drawn to is turn over cards E and 5. This illustrates the human tendency towards confirmation bias.

I feel nervous about my answers as I'm the first person to give it :P

E, 5, 4

I hope I haven't made a horrific embarrassing mistake :P But It seems to make sense to me at least.

Okay, the answer has been posted but I still don't see why 5 is incorrect.

"If a card has a vowel on one side, then it has an odd number on the other side."

5 is odd, on the other side it could have A or B. A would confirm, B would prove wrong the hypothesis. So Surely you have to flip it over?

''disprove the hypothesis by turning over the 4 card (if there is a vowel the statement must be false) rather than confirm the hypothesis (by turning over the 5 to see if there is a vowel).''

But doesn't turning over 5 also mean potentially disproving the hypothesis if there is an B there for example.

I'm going with E, 5 and 4 too - you need to check all three to verify the statement - C is the only one that is irrelevant.

...er... 10 minutes after the answer was posted!

E and 4 Are obvious. I got those instantly

C is irrelevant as it isn't odd, so it can have whatever the hell it likes and it doesn't matter. Can anyone please explain in detail while I go have a shower. Thanks :)

Well I fit into the 90% of students who got it wrong, so at least I don't feel to bad, but I still want to know why I'm wrong.

Fat lot of good all those years studying logic at university did me. Sigh.

People want to turn over the vowel card to see if there is an odd number... so far so good. Then it seems natural to turn over the odd number, card 5, to see if there is a vowel, which would appear to confirm the proposition. But in fact if there were a consonant instead, this does not actually invalidate the proposition. Likewise with card C... the proposition makes no comments about consonants, thus it is irrelevant. But card 4 if there is a vowel on the other side then the proposition must be false.
Hence the two cards needed are E and 4.

Okay but we have no idea what the other cards have, and surely they are relevant.

"If a card has a vowel on one side, then it has an odd number on the other side."

As we have no idea what's on the other side of the cards, they can be anything so I will layout a plausible scenario.

We have

E
C
5
4

You turn over E and 4

E1
C
5
4B

You think great it's true then.
E has a vowel on one side, and an odd on the other. = Fit
4 Is an even, B is a consonant. This makes it irrelevant to the statement

"If a card has a vowel on one side, then it has an odd number on the other side."

As it has neither a vowel or an odd number on the other side it seems that the statement is true. But you can't yet assume that the statement is true. just by turning over those 2 cards, You need more information

You decide to turn over the rest of the cards just for arguments sake and they end up being

E1
C6
5
4B

The statement is no longer true, While A confirms the statement, C and 4 Are irrelevant, 5 now proves it wrong. As on the other side of the odd, the is a consonant not a vowel.

So after initially turning over 2 cards you could say with the information you had that the statement is true, but by turning over the rest it got proven false. As stated earlier, C is just irrelevant, Simply turning over 5 would have proved this. As this is an entirely plausible scenario, how is the answer E, 5, 4 Incorrect considering this could easily be the scenario, and by simply turning over E and 4 you wouldn't of had enough information.

I don't mind being wrong, I just don't see how I am... And that does annoy me.

Mintyboy, after your shower, please explain "C is irrelevant as it isn't odd" ?

Tip for you Mintyboy:
"If A then B" =/= "if B then A"

Minty - The statement is in effect "all vowels are odd" not "all odd numbers are vowels". So as long as the one even number shows a consonant and the one vowel shows an odd number, the statement stands as true for the limited set in question. it does *not* however prove it definitevely because there may be an even vowel in the unexplored part of the deck.

@Mujus - Wow yeah, that was just me not having woken up yet properly.

@Draug - "If a card has a vowel on one side, then it has an odd number on the other side."

But surely this then means this applies to all cards, not just necessarily to the 4 cards on the table. The cards on the table are just a sample. If it would say, 'If a card on the table has a vowel on one side, then it has an odd number on the other side." This would be a different question. But for now there is nothing stopping me from assuming it applies to all cards, and as there are many unchecked cards, it's impossibly to determine whether the statement is true of false just by turning over two.
And as you said ''it does *not* however prove it definitevely because there may be an even vowel in the unexplored part of the deck. ''
My point precisely, we don't have enough information. The statement isn't specific enough to the 4 cards on the table. So by turning over the cards E and 4, we can't determine if for all cards.

Also for the point ''The statement is in effect "all vowels are odd" not "all odd numbers are vowels"

What's the difference? If I would say 1+2=3 How is this different to saying 3=2+1
It's exactly the same information just in another format. It shouldn't effect the question.

What other cards? The scenario only mentions four cards.

You don't understand basic statement logic. "All dogs are animals" isn't the same information as "All animals are dogs".