# Vowels and Even Numbers

“If there is a vowel written on one side of a card, then there is an even number written on the other side.”

How many of these four cards do you need to flip in order to check the validity of this sentence?

What would the answer be if you know that each card contains one letter and one number?

**SOLUTION**

You need to flip all cards except for the second one. If each card contains one letter and one number, then you need to flip only A and 7.

Fliping B is Irrelevant because it can’t dissaprove the statemant. Same with 6, because there is nothing that says that even number must have vowel on other side. A has to be fliped, of course, but 7 too, because, if it’s vowel on other side, the sentence is not valid.

Hi Moronski. If there is a vowel on the other side of B, then the statement would be wrong. If we stipulate that each card has one number and one letter written on it, then you are correct.

wow

you need to flip all the card except B , we are talking about vowels , so obviously we need to check card with A , and see what is written on its back , then we are talking about even numbers , so again card with number 6 is of prime importance , we need to check what is written on its back, now we need to test the hypothesis by checking the card with number 7 with the assumption to see a consonant on its back to validate our conclusion

Following on the previous comment from Corinne, if the only condition to prove is that “IF there is a vowel on one side of the card, there is an even number on the other side” then the ONLY card you have to flip to check the validity of this statement is the third card with “A” on it. There are no stipulations saying there must be odd numbers on the back of consonants, or numbers at all, therefore the first card showing “B” is irrelevant to the puzzle. There are also no stipulations there can ONLY be vowels on the flip side of even numbers, so the second card showing “6” is also irrelevant. And given there are no stipulations at all about odd numbers, the fourth card showing “7” is also irrelevant.

This is the same conclusion i got to, even assuming there has to be a number on one side and letter on the other

if you flip the 7 and there is a vowel on the other side then it disproves the statement.

if you flip the b and there is a vowel on the other side then it disproves the statement~

neither b, nor 7 are even numbers.

This answer is incorrect. There is no need to flip the B because the puzzle makes no mention of which numbers must be on the back of a consonant. It was simply that Vowel->even NOT ~vowel->~even. A common mistake in mathematical logic is equating the inverse to the original statement.

Hello Corinne, thank you for your feedback. You are right that the solution to the original four-card trick by Peter Wason was simply to flip the cards with A and 7. However, in his experiment from the sixties, he specified that each card has a letter on one side and a number on the other side. Without this assumption, you need to flip the card with letter B as well.

We thought that dropping the assumption makes the puzzle trickier, but maybe adding both variations would be even better.

Hello!

I think you also need to flip the 6 because if there is a vowel at the back of it, then the assumption is false.

Hello Eloik. “6” is an even number, so even if there is a vowel on the other side, the statement will still hold true:)

Ah I see now, that is tricky