You are working with a deck of four cards that have a letter on one side and a number on the other. They are lying on the table with different sides facing up, as shown:
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One theorem about the deck states: If a card has a vowel on one side, then it has an even number on the other side.
Which cards do you have to flip over to verify that the theorem is not false?
Do not read on until you think you have the answer.
The answer is that you must flip over the first and last card. Almost everyone correctly identifies the A. Many people make the mistake of identifying the 4 as well. Few people correctly identify the 7.
If your answer was correct, your grasp of basic logic is exceptional. The problem is straightforward, but it happens to be counterintuitive to humans.
You can read the full explanation at the Skeptic's Dictionary's Critical Thinking mini-lesson 3.
19 comments:
Curiosly, the first time I took this test, I picked 7 but not A. I don't know what this means, but it's a fun test.
So out of A, B, 4, and 7, I said you need to flip over A, B, and 7. The answer said only A and 7, but I think's that's wrong. To verify that something is not false, you have to check anything that could potentially give you proof that it is false. 4 the card doesn't need to be flipped, since 4 is even so if there's a vowel on the other side, then the theorem is true and if there's a non-vowel, then the theorem doensn't apply period. However, B should also be included, because if a vowel is on the other side such as A, then this card's existence would falsify the claim that a vowel always has an even number on its back. Likewise, for A and 7. In fact, the B card is just like the 7 card...it's precisely because neither of those are even numbers, that you have to check the back to make sure the back doesn't have a vowel on it in which case these cards would falsify these "theorem." In other words, that guy's explanation is wrong, unless he's making another implicit assumption that somehow every card always has a letter on one side and a number on the other.
"You are working with a deck of four cards that have a letter on one side and a number on the other."
It's not an implicit assumption, it's explicitly stated.
Yeah, whenever a problem has bolded and italicized text, I tend not to read the normal text. This has gotten me into plenty of trouble before on tests.
This is incorrect, only due to lack of information in the question.
At no point in the explanation does it say that on both sides of the card, there is 1 even number, 1 odd number, 1 vowel, 1 constanent. This means that the numbers and letters on the opposite side could be 2 constanents and 2 odd numbers.
The only thing that turning the 1st and last card can show, is that the theorum is false (so far)(if there is an even or vowel) but even if you use logic stating that because it is not false (yet), it must be true... it could still infact be false.
At no point will you know the opposite of the two cards in the middle, so in theory, you could state that the theorum is not false, when infact, it is false.
You could come to the conclusion that the theorum is false after either 1, 2, 3 or 4 cards have been turned, depending on if a card in the order that you've turned them, breaks the statement.
You can only come to the conclusion that the theorum is true (not false) after all 4 cards have been turned as any one of the 4 cards could still break the theorum, reguardless of the others, as it is still an independent event (the other cards have no relevence to each other).
Unless all the cards are linked to each other (e.g. both sides of all the cards contain 1 odd, 1 even, 1 vowel, 1 constanent) then the answer will always be that all 4 cards must be turned.
You can't have any form of assumptions being made in a question like this.
Nope, that's wrong.
at no point is there any information about constanants NOT having vowels as well.
The hypothesis ONLY states that if it is a vowel then it will ahve an even number. This does not explicitly rule out the possibility of a constanant also having an even number, this would be a false assumption.
Therefore to prove the hypothesis as stated, the answer is correct.
Logic 101
acid- with regard to your statement about flipping the first and fourth card over can only prove this false, not true. Well, that is the basis of this kind of logic, you can only prove things false, you cannot prove them to be true. The answer is correct, though, cause those 2 cards are the only ones that can be flipped to prove it false.
My mistake, the question is phrased asking you to prove the statement true, which is not possible through this kind of reasoning.
As a response to Guanxiong, your reasoning is sound. However, the test is usually presented with an axiomatic statement which establishes that each card has a letter on one side and a number on the other side. In this case that statement was missing.
My first try was too long, I'll keep it shorter:
A, because Modus Ponens, and 7, because Modus Tollens
The theorem only states that if it is a vowel then the other side will be an even number.
This means only that:
a) if one side is a vowel then the other is an even number and
b) if one side is not an even number then the other side is not a vowel
To say that if one side is an even number means that the other must be a vowel would treat the theorem as a biconditional (a common logico-"intuitive" error)
Good to meet you.
I'm really enjoying your posting.
I am univ. student from Korea.
I'm majoring Computer Science too.
Wason's card problem is really interesting.
Thanks :D
You have rephrased the question to this problem incorrectly so the logic is wrong. In the original question on the Skeptics Dictionary site it says:
"Which card(s) must you turn over to determine whether the following statement is false?"
In your question you say:
"Which cards do you have to flip over to verify that the theorem is not false?"
There is a big difference here. One question says "is false" and the other (yours) says "is not false".
I think you need to modify the article :)
I decided to post a comment, however I rarely due. It seems a lot of people are saying that you need to flip over the 4 or B card, which is wrong. That is they never say whether a consonant has an even or an odd. They could have either one. So by this, you would have to just check two cards. A and 7. because if the A card has an odd number it would void the rule, and if the 7 card has a vowel, it too would void the rule
One thing that have to be stated is that "each card has a letter on one side and a number on the other". If you don´t state this then you could turn over the B-card to falsify the rule. If you turn over the B-card and find a vowel on the other side, what happens then?
One theorem about the deck states: If a card has a vowel on one side, then it has an even number on the other side. Theorem does not prohibit cards with even number to have either consonant or vowel.
Therefor you have to check card with a vowel and card with ODD number (to see if a vowel is on other side).
Card A MUST have an even number
Card 7 MUST NOT have a vowel
Card B - do nothing - we have no concern with consonants
Card 4 - do nothing - We do not know if there are even numbers with consonants
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