34. Five-Card Prediction Alex shuffles an ordinary deck of 52 playing cards (no Jokers) and chooses any five cards. Peter is not watching. Alex hands the five selected cards to You (the Magician). You secretly look at the five cards, pick one of them, and hand that single card back to Alex face-down. You then arrange the remaining four cards in a specific order, place them face-down in a tidy pile, and give the pile to Peter. Peter turns the four cards face-up, studies their order for a moment, and immediately announces the exact suit and rank of the card that Alex is still holding. Assuming Peter and You have agreed on a method beforehand (but have no hidden markings or outside communication), explain in detail how this feat is always possible and how the four face-up cards unequivocally reveal the fifth. Added 26 April 2011 Show Hint Show Solution Hint: Among any five cards there must be at least two of the same suit. Use one of those as a “key” card. The order of the four cards can encode up to 24 different messages. Solution: 1. Guaranteeing a shared suit With four suits and five cards, the pigeon-hole principle ensures that at least two of the five share a suit. Call these two cards S1 and S2. 2. Choosing the key and the hidden card The Magician chooses one of the same-suit pair to keep among the four visible cards (the key) and gives the other to Alex (the hidden card). If necessary the roles of the two cards are swapped so that, when the ranks are considered cyclically in the order A-2-3-…-K-A, the distance from the key to the hidden card is between 1 and 6 inclusive. This is always possible because the two ranks are at most 6 or at least 7 steps apart, and you can pick the shorter of the two directions. 3. Encoding that distance After the key card is fixed, three cards remain. Their order relative to the key card can be permuted in 4! = 24 ways. Six of those permutations will be used to signal the numerical distance 1 – 6 (the other 18 are unused or can be assigned to alternative conventions). A convenient mapping is: distance 1 → key followed by ABC distance 2 → key followed by ACB distance 3 → key followed by BAC distance 4 → key followed by BCA distance 5 → key followed by CAB distance 6 → key followed by CBA (Here A, B, C are the three non-key cards sorted in some agreed order, e.g. alphabetically by suit then by rank.) The Magician simply stacks the four cards so that this ordering appears from top to bottom. 4. Peter decodes Peter turns over the pile. The first card he sees is necessarily the key card, so he knows the suit of the hidden card. He inspects the relative order of the remaining three cards, consults the agreed table above, and learns the distance d (1–6). Finally he counts forward d steps clockwise through the ranks starting from the key card’s rank; the card reached is the hidden card, which he names aloud. 5. Example Suppose the five cards are ♣9, ♥K, ♦4, ♣3, ♠A. Two clubs are present. The Magician chooses ♣9 as key and ♣3 as hidden because the distance 5→3 clockwise is 6. The remaining three cards (♥K, ♦4, ♠A) are ordered as CBA to indicate distance 6. Peter sees ♣9 ♥K ♦4 ♠A, recognises the permutation CBA, deduces distance 6, counts six steps from 9 ♣ (10 J Q K A 2 3), and announces "Three of Clubs"—correct. Thus, by exploiting the forced duplicate suit and the 24 possible permutations of four cards, the trick works every time without any outside communication.
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