Please explain why the following card \"trick\" will always workout mathematical
ID: 2951726 • Letter: P
Question
Please explain why the following card "trick" will always workout mathematically: Take the deck and make piles in the following way. Turn the top card over and leave it face up. Put on topof that card the number of cards it would take to get to 13,placing them all face up. For example if you turned over a 9then you would put four cards ontop of the 9. If you turned aKing, then you're already at 13, no extra cards needed. Anace is one and would require 12 cards, and so on.... Keepmaking piles in this format until you run out of cards orcannot complete a pile (those will form a DISCARDpile). Now, turn over all the piles you've created using the'13' scheme, so that all the piles are facing numbers down. Have someone point to three piles. Leave those threepiles where they are and pick up all the other piles and add themto the DISCARD pile. Ask the person to turn overthe top card of two of the three piles. Whatever the first card's value is, say an '8', then you willplace 8 cards from the DISCARD pile in front of that pileface down. Now, do the same thing for the otherpile with the top card turned over (i.e. place the number ofcards according to the value of that top card) Now,in front of the pile that does not have the top card turned over,place 10 cards. Conclusion: Count the number of cards you have left andthis will ALWAYS be the value of the third pile's topcard. How does this work? I'm looking for mathematical reasonswhy it should never fail. Thanks! Please explain why the following card "trick" will always workout mathematically: Take the deck and make piles in the following way. Turn the top card over and leave it face up. Put on topof that card the number of cards it would take to get to 13,placing them all face up. For example if you turned over a 9then you would put four cards ontop of the 9. If you turned aKing, then you're already at 13, no extra cards needed. Anace is one and would require 12 cards, and so on.... Keepmaking piles in this format until you run out of cards orcannot complete a pile (those will form a DISCARDpile). Now, turn over all the piles you've created using the'13' scheme, so that all the piles are facing numbers down. Have someone point to three piles. Leave those threepiles where they are and pick up all the other piles and add themto the DISCARD pile. Ask the person to turn overthe top card of two of the three piles. Whatever the first card's value is, say an '8', then you willplace 8 cards from the DISCARD pile in front of that pileface down. Now, do the same thing for the otherpile with the top card turned over (i.e. place the number ofcards according to the value of that top card) Now,in front of the pile that does not have the top card turned over,place 10 cards. Conclusion: Count the number of cards you have left andthis will ALWAYS be the value of the third pile's topcard. How does this work? I'm looking for mathematical reasonswhy it should never fail. Thanks!Explanation / Answer
Let the first card turned over have the number 'x' (i.e. if it'sthe 8 of Clubs, x=8, if it's the Jack of Diamonds, x=11 and so on).We will now place 13-x cards on top of this card. Repeat theprocess for a number of other 'counted' piles. When the cards are all turned over, the card with the number 'x'will now be on top of the pile, as it was previously on thebottom. Once all the cards have been placed into either a counted pile orthe DISCARD pile, we eliminate all other piles except for three.Call these piles 'A', 'B', and 'C'. All of these piles will havebeen constructed the same way as 'x' above. Define the 'base'card for Pile A as 'a' and similarly for b and c. As mentionedpreviously, the act of turning the piles upside down has put thesebase cards on top of the pile. Start with Pile A. We will turn over the card 'a', and thus place'a' more cards in front of Pile A. (We will regard the cards placedin front of Pile A as being part of Pile A for convenience.) Thereare now 'a' cards being added to the 13-a cards from before,meaning that there are now (13-a) + a + 1 = 14 cards in Pile A. The'+1' of course comes from the card 'a' itself. We do the same thingfor Pile B. There are now 14+14 = 28 cards in total in Pile A and B. Pile Cstill has (13-c) + 1 = 14-c cards in it (the +1 again is the card'c' itself). Thus, at the moment, we have 28 + (14-c) = 42-c cardsin total in Piles A, B and C. However, we started off with a standard deck of 52 cards. Left inyour hand, in the DISCARD pile, must be enough cards to make up thestandard deck when added to the number of cards in Piles A, B andC. Thus there are currently 52 - (42-c) = c + 10 cards in theDISCARD pile. Dealing out 10 cards in front of Pile C eliminatesthese 10 cards and thus we are left with 'c' cards in our hand,which will be the number on the top of the third pile.