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## The card trap

A classic trap was used by Wason and Johnson-Laird (1972).Four cards are laid out as below:

E | K | 4 | 7 |

The conditional statement is now given: 'If a card has one vowel on one side, then it has an even number on the other side.'

The question is to decide which are the minimum cards that need to be turned over to prove that the conditional statement is true.

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Without going in search of the

*answer*, what do you think?

1 2

Comments

Two. (E and 4).

Clive

Clive"To affirm the antecedent, E is correct. E is a vowel and thus should have an even number on the other side. If there was an odd number on the other side, the statement would be false, so E must be turned over to check for this.

But choosing 4 is affirming the consequent. Even though 4 is even, it can have a vowel or consonant on the other side and the statement is not falsified."

milkyMrP

MrPedanticmilkyBut where is the rest of the deck? The card with A or I? Don't you need to look on the other side of all the ones with vowels to be sure?

So maybe the answer is 5 - for the 5 vowels.

Or maybe I didn't understand the question.

Anyway, interesting, but where's the linguistics question?

CJ

CalifJimmilky## So what?

Be careful about if-then statements, both in your own use and in those that others use. It does, of course also mean that you can make statements that are logically false and few people will challenge you."milky<The question is to decide which are the minimum cards that need to be turned over to prove that the conditional statement is true.>

Is that supposed to be the "minimum number of cards" or "the minimum card numbers and letters"? I feel it's the latter.

milky