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"A F 2 7" puzzle survey results


CraigD

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With poll responses down to 1 in the last 24 hours, I think it’s time to discuss the results of the ”Puzzle survey to confirm a published result”. Per people’s suggestions, the poll remains open, so if you haven’t yet voted, feel free to read it’s first post and vote your answer.

 

Here is the puzzle:

Each card has a single letter on one side, a single number on the other. (As is normal with cards) you can only see one side of each card. You see these 4 cards:

 

A F 2 7

 

You are asked to determine the truth of the statement “every card with a vowel on one side has an even number on the other” by flipping as few cards as possible.

As of the time of this post, the answers given by us hypography members are:

Votes % Answer

6 37.50 #1 and #2 and #3 and #4 (A and F and 2 and 7) – all of the cards

4 25.00 #1 and #3 (A and 2)

3 18.75 #1 and #4 (A and 7)

2 12.50 None of the above

1 6.25 #1 and #2 (A and F)

 

The correct answer is: Flip cards #1 and #4 (A and 7). As described in the literature, less than 20% - 18.75% - of us answered correctly, despite our being, IMO, an unusually logical and precise community of people. The sample size – 15 - is too small to be conclusive, but doesn’t contradict the literature.

 

To see why, consider if card #1, which shows a vowel, has an odd number on its other side. This would make the statement false. If card #4, which shows an odd number, had a vowel, the statement would be false. Since the statement doesn’t say anything about consonants, we don’t care what’s on card #2, and since it isn’t contradicted by a consonant on a card with an even number, we don’t care about card #3 (In formal terms, (V->E)‘->(‘V->’E)).

 

:) So, the 13 people who got the wrong answer, what lead you astray? What insights into our cognitive processes can this provide? :)

 

PS: This puzzle, and the observation that fewer than 20% of most populations get it correct on the first try, was described in 1966 by psychologist Peter Wason.

The book in which I encountered it is William Poundstone’s “How Would You Move Mount Fuji?” Despite it’s subtitle, “Microsoft's Cult of the Puzzle - How the World's Smartest Company Selects the Most Creative Thinkers”, this book actually examines the history of intelligence and aptitude testing in general, and is fairly critical of how smart employers actually are to rely on puzzles in screening potential employees.

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I pointed out to Craig in a PM that I jumped to a couple of conclusions and voted. Then reread that puzzle and immediately recognized that I had presupposed this to be an if and only if type of puzzle. (i.e. that only vowels could have even numbers on the back.) In that case you must flip all. But since that was not the case, I quickly commented on the correct answer.

 

I wonder what causes us to psychologically jump to that conclusion (as 6 of us did.)

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Well, since you are asking...

 

I voted "none of the above" because I thought the correct answer was A, 2, 7. I knew 'A' had to be checked. I also knew 'F' was irrelevant. That just left me with the numbers left. I just couldn't see how both of them would not have to be checked. I also realized that odds were very high (though not impossible) that I was wrong simply because I choose "none of the above", but since I just couldn't see any other answer, I had to choose it.

 

Will go and polish up my brain some. :)

 

Take care!

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well, I was wrong... I just didn't see the logic in the question. I wasn't sure how to go about solving a problem with such general specifications. So, I picked the cop out answer- #1 #2 #3 and #4. I probably could have figured it out if i had taken the time to actually contemplate the situation. Oh well. That was an interesting puzzle.

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I just didn't see the logic in the question.

All that the problem asked was this:

You are asked to determine the truth of the statement “every card with a vowel on one side has an even number on the other” by flipping as few cards as possible.

To do this you only need to see the other side of any cards that have a vowel or an odd number showing. As long as the vowels have an even number on the reverse and the odd numbered cards do not have a vowel on the reverse you can prove the statement true.

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To do this you only need to see the other side of any cards that have a vowel or an odd number showing. As long as the vowels have an even number on the reverse and the odd numbered cards do not have a vowel on the reverse you can prove the statement true.

 

I thought of that... but then I told myself... "naw, it can't be that simple... I must've missed something". It's one of those things that gets you tripped up because you second guess your original thought.

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  • 2 weeks later...

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