Calling math experts: Could you help me describe this card trick? - General

1: Magician shuffles deck of 21 cards, asks member of the audience to take one card, memorize it and place it back into the deck 2: deck is shuffled again, and dealt into three columns of seven cards 3: Audience member tells magician which column the card is in, without exposing which card it is 4: The cards are gathered, and the column containing the card is placed between the other 2 columns 5: repeat steps 2-4 twice again 6: When done, the cards should have been dealt 3 times. if done correctly, their card should be the 11th card in the pile when turned over [URL="http://www.wikihow.com/Perform-a-Card-Trick-Using-Math"]source, with visual aid[/URL] How does this work? How can I mathematically describe this? I'm choosing to include it as part of my "mathematics in games and puzzles" final project for my BC calc class but I can't figure out why it works. Any help is appreciated, i just need a brief explanation. thanks in advance

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Action Replay Max or Gameshark

Let's call P(N) the position of the chosen card in the deck after N deals. P(N) is an integer between 0 and 20 (there are 21 cards after all) and we know its initial value P(0) satisfies the condition [b]0 <= P(0) <= 20[/b]. Now you deal the cards into 3 columns of 7 cards, working row by row. Which means that the position of the chosen card within the column that contains it is floor(P(0) / 3), floor being the "round down" function. When gathering the cards back into a pile, you are making sure that the column that contains the chosen card is placed between the two other columns, meaning that its position is now [b]P(1) = 7 + floor(P(0) / 3)[/b]. Knowing that 0 <= P(0) <= 20, we can deduce that 0 <= floor(P(0) / 3) <= 6. (20 / 3 = 6.666... which rounds down to 6) Therefore, [b]7 <= P(1) <= 13[/b]. (you get that by adding 7 everywhere) Repeat the same operation, [b]P(2) = 7 + floor(P(1) / 3)[/b]. We found out previously that 7 <= P(1) <= 13, so 2 <= floor(P(1) / 3) <= 4. (7 / 3 = 2.333... and 13 / 3 = 4.333...) You can deduce that [b]9 <= P(2) <= 11[/b]. One last time, [b]P(3) = 7 + floor(P(2) / 3)[/b]. 9 <= P(2) <= 11 so 3 <= floor(P(2) / 3) <= 3. (9 / 3 = 3 and 11 / 3 = 3.666...) Turns out floor(P(2) / 3) = 3 no matter what. You can conclude that no matter the initial value P(0), [b]P(3) = 10[/b], meaning that the chosen card will always end up at the 11th position after dealing the card three times.

[QUOTE=_Kilburn;44908168]Let's call P(N) the position of the chosen card in the deck after N deals. P(N) is an integer between 0 and 20 (there are 21 cards after all) and we know its initial value P(0) satisfies the condition [b]0 <= P(0) <= 20[/b]. Now you deal the cards into 3 columns of 7 cards, working row by row. Which means that the position of the chosen card within the column that contains it is floor(P(0) / 3), floor being the "round down" function. When gathering the cards back into a pile, you are making sure that the column that contains the chosen card is placed between the two other columns, meaning that its position is now [b]P(1) = 7 + floor(P(0) / 3)[/b]. Knowing that 0 <= P(0) <= 20, we can deduce that 0 <= floor(P(0) / 3) <= 6. (20 / 3 = 6.666... which rounds down to 6) Therefore, [b]7 <= P(1) <= 13[/b]. (you get that by adding 7 everywhere) Repeat the same operation, [b]P(2) = 7 + floor(P(1) / 3)[/b]. We found out previously that 7 <= P(1) <= 13, so 2 <= floor(P(1) / 3) <= 4. (7 / 3 = 2.333... and 13 / 3 = 4.333...) You can deduce that [b]9 <= P(2) <= 11[/b]. One last time, [b]P(3) = 7 + floor(P(2) / 3)[/b]. 9 <= P(2) <= 11 so 3 <= floor(P(2) / 3) <= 3. (9 / 3 = 3 and 11 / 3 = 3.666...) Turns out floor(P(2) / 3) = 3 no matter what. You can conclude that no matter the initial value P(0), [b]P(3) = 10[/b], meaning that the chosen card will always end up at the 11th position after dealing the card three times.[/QUOTE] Thank you so much! This is exactly what I needed.

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