Can someone please explain the math behind this card trick:
https://www.youtube.com/watch?v=Kdaoc_jEKUM
I am not sure why the 2 face up cards need to be in the 1oth place for the trick to work.
I know that it has to do modulo 26 but not sure about the remainder of the math behind the trick.
Thanks.
The initial part of the trick is about having the top half of the deck and the bottom half of the deck matching, and staying that way through various cuts and when removing the helper cards. You then remove the chosen card and manipulate the target card to the middle before replacing the the two face up helper cards. So ...
Once you have set up the deck, you have a chosen card set aside and $51$ cards in your hand, with the target card at position $26$ and the face up cards at positions $10$ and $42$ ($10$ from the bottom)
Deal the $51$ cards into two piles, and one of the piles will have $25$ cards with the target card at position $13$ and the face up cards at positions $5$ and $21$ ($5$ from the bottom)
Deal the $25$ cards into two piles, and one of the piles will have $13$ cards with the target card at position $7$ and the face up cards at positions $3$ and $11$ ($3$ from the bottom)
Deal the $13$ cards into two piles, and one of the piles will have $7$ cards with the target card at position $4$ and the face up cards at positions $2$ and $6$ ($2$ from the bottom)
Deal the $7$ cards into two piles, and one of the piles will have $3$ cards with the target card at position $2$ and the face up cards at positions $1$ and $3$ ($1$ from the bottom)
So you end up with the two face up cards and the target card between them
This works because you have an odd number of cards to deal each time ($51,25,13,7,3$) and so the middle card goes to the middle of the pile with an odd number. $10$ happens to be the position in $51$ card case which also survives to the $3$ card case, while $42=51+1-10$ is its symmetric equivalent