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Hand your partner an ordinary deck of 52 playing cards. Ask
him then to remove, secretly, three cards from the deck,
except that no two of the three cards may sum to 10. (So, he
may not remove a three and a seven, say.) Tell him that you
will now determine the values of these three cards simply by
removing cards from the deck quickly, emphasizing that you
do so without counting or memorizing any cards.
For purposes of this trick, we consider each card solely according to the face value, with each ace as I and each picture card as 10. Thus, for example, six of clubs is just 6, ten of clubs is just 10, and queen of hearts is just 10. Have your partner layout the cards, in no particular order, in any convenient fashion, such as a 7x7 array. (The random order is so the removal of cards won't make it obvious what was taken.) Now, remove cards in pairs that sum to 10 or 20, such as four of spades and six of diamonds, or ace of clubs and nine of hearts, or ten of hearts and queen of spades.When you have done this as much as possible. you will be left with either l or 3 cards. Try it now if you don't believe me! But, you will be able to tell the values of the three cards that were initially removed. Here is how you do it: If there is one card left, with value x, then x < 10 and two of the picked cards have value 10 and the third has value 10 - x. If there are three cards left with values x, y, and z each under 10, then the cards chosen have values 10 - x, 10 - y, 10 - z. However, it is possible that one of x, y, and z has value 10, say x. In this case, the three card values are 10, 10 - y, 10 - z. (For instance, if there are three cards left, a ten, a two, and a five, then the three cards removed have values 10, 8, and 5.) |
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