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This trick relies on three nice ideas from discrete mathematics: pigeonhole principle 5 pigeons (cards) in 4 holes (suits) means at least two pigeons are in the same hole - i.e., at least two of same suit. mod arithmetic Arrange cards on 13-position clock. One of the same-suit pair can be used as a base card occurring at most 6 cards before the other card, which is chosen to be the hidden card. permutations There are 6 distinct arrangements of the 3 remaining cards - the one chosen indicates whether to add 1, 2, 3,4, 5, or 6 to the base card. Math Card Coding Mindreading Trick 1. Accomplice leaves room. 2. Audience picks out 5 arbitrary playing cards from deck. 3. Presenter arranges 4 (carefully chosen) face-up, and one face down. 4. Accomplice returns to the room, and announces the face-down card. +1 +2 +3 +4 +5 +6 Permutation is fifth possible (123,132,213,231,312,321) +5 Base card gives suit of hidden card, and starting rank on card clock

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Idea #1 pigeonhole principle + = at least two of same corollaries: socks others among five cards, at least two have same suit

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Using Idea # 1 = Base card gives suit of hidden card. Idea #2: Card Clock (mod arithmetic) +5 Base card also gives starting point on card clock. Now only need to signal ADD 5

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Idea #3: permutations There are exactly 6 permutations of 3 objects There are 3 objects to be ordered (3 middle cards) I can order the permutations systematically perm-1, perm-2, perm-3, perm-4, perm-5, perm-6 Depending on which permutation, I can indicate to add 1, 2, 3, 4, 5, or 6, to the base card. 123 132 213 231 312 321 = +1 = +2 = +3 = +4 = +5 = +6 if first card is the lowest, then add 1 or 2, depending on whether the remaining two cards are in or out of order if first card is the mikddle, then add 3 or 4, depending on whether the remaining two cards are in or out of order if first card is the highest, then add 5 or 6, depending on whether the remaining two cards are in or out of order

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Additional Considerations (a)What if two of the three middle cards are the same rank? Answer: Use a natural ordering on suits, with clubs < diamonds < hearts < spades, so that 7D 7H 7C would indicate add 4, for example. (b)What if the base card is J and hidden card is 5? You cant reach 5 from J in 6 or fewer steps. But then this means you can reach J from 5 in 6 or fewer steps. Choose 5 as the base card, J as hidden. This is the job of the main presenter - from the two cards of the same suit chosen, figure out which card should be the base, from which the hidden card can be reached in at most 6 steps. Notes The first page is a summary page that can be printed out for handy reference. The next four pages are useful for a presentation of the trick to an audience. This trick has been around in the recreational mathematics and teacher communities for a while, and I do not know the origin. This presentation prepared by Lenny Pitt, pitt@uiuc.edu

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