Quote:
Originally Posted by pdurrant
another from Richard Wiseman.
I blindfold you, and give you a pack of cards. I tell you that ten of the cards in the pack are face-up, and the rest are face-down. I ask you to use the whole deck to make two piles of cards, and each pile must contain the same number of face-up cards.
How do you do it?
(No, the cards aren't marked, and you can't feel the difference between a face-up and face-down card by touch.)
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Split the deck into a pile of ten cards and a pile of all the rest of the cards. Turn the pile of ten cards over. Done.
I never said the piles had to be the same size, just that they must contain the same number of face-up cards.
How does it work? Like this.
Some of the face-up cards are in the pile of 10, and some are in the rest. Let's call the number of face-up cards in the pile of 10 'U'. Since we know there are 10 face-up cards altogether, there are 10-U face-up cards in the bigger pile.
Since there are 10 cards in the pile contain U face-up cards, it contains 10-U face-down cards.
By turning the pile of 10 cards over, you change the face-up cards to face-down cards and vice-versa. So now the pile of 10 cards has 10-U face-up cards in it.
And we already know that the bigger pile also has 10-U face-up cards in it! Problem solved!
I think HIMS must give us the next question.