Counting Ill The value of the secret card is between 1 and 6 hops clockwise from the value of the first card revealed All that remains is to communicate a number between 1 and 6. The Magician and Assistant agree beforehand on an ordering of all the cards in the deck from smallest to largest such A2..K品A◇29..Q◇A29..QA·2·..Q The order in which the last three cards are revealed communicates the number ac- cording to the following sche eme sma meaium, arge medium ((((( medium, small, sma medium)=5 large, medium, small )=6 In the example, the Assistant wants to send 6 and so reveals the remaining three cards in large, medium, small order. Here is the complete sequence that the Magi clan sees 10Q◆J9 The Magician starts with the first card, 109, and hops 6 values clockwise to reach 39, which is the secret card! 4.3 Same Trick with Four Cards? Suppose that the audience selects only four cards and the assistant reviews a sequence of three to the magician. Can the magician determine the fourth card? Let X be all the sets of four cards that the audience might select and let y be all the equences of three cards that the assistant might reveal. Now, one on hand, we have 52)=270,725 y the Subset rule. On the other hand, we have Y|=52.51.50=132,600 by the Generalized Product Rule. Thus, by the Pigeonhole Principle, the Assistant must reveal the same sequence of three cards for some two different sets of four. This is bad news for the Magician: if he hears that sequence of three, then there are at least two possibilitie for the fourth card which he cannot distinguish� � 12 Counting III – The value of the secret card is between 1 and 6 hops clockwise from the value of the first card revealed. • All that remains is to communicate a number between 1 and 6. The Magician and Assistant agree beforehand on an ordering of all the cards in the deck from smallest to largest such as: A♣ 2♣ . . . K♣ A♦ 2♦ . . . Q♦ A♥ 2♥ . . . Q♥ A♠ 2♠ . . . Q♠ The order in which the last three cards are revealed communicates the number ac￾cording to the following scheme: ( small, medium, large ) = 1 ( small, large, medium ) = 2 ( medium, small, large ) = 3 ( medium, large, small ) = 4 ( large, small, medium ) = 5 ( large, medium, small ) = 6 In the example, the Assistant wants to send 6 and so reveals the remaining three cards in large, medium, small order. Here is the complete sequence that the Magi￾cian sees: 10♥ Q♠ J♦ 9♦ • The Magician starts with the first card, 10♥, and hops 6 values clockwise to reach 3♥, which is the secret card! 4.3 Same Trick with Four Cards? Suppose that the audience selects only four cards and the Assistant reviews a sequence of three to the Magician. Can the Magician determine the fourth card? Let X be all the sets of four cards that the audience might select, and let Y be all the sequences of three cards that the Assistant might reveal. Now, one on hand, we have 52 |X| = = 270, 725 4 by the Subset Rule. On the other hand, we have | | Y = 52 · 51 · 50 = 132, 600 by the Generalized Product Rule. Thus, by the Pigeonhole Principle, the Assistant must reveal the same sequence of three cards for some two different sets of four. This is bad news for the Magician: if he hears that sequence of three, then there are at least two possibilities for the fourth card which he cannot distinguish!