Here is a more general mathematical proof because I am annoying and pedantic

Consider our pack of cards, numbered sequentially and starting from partly because it makes the maths simpler but mostly to confuse people.

Let the card that our sucker chose be at position x.

Now, we deal the cards into 3 piles and then recombine them into a single deck. Since we are dealing into three piles and there are seven cards in each pile, it can be seen that the new position of our card in the deck y is given by

y = 7( x mod 3 ) + ( x div 3 )

where the first term gives the base position of the pile that the card is in and the second term gives the position within that pile.

We also perform an extra operation however, which is to take the pile with our card in it and make it the middle pile. In other words if the card is now in the first pile then we want to add 7 to its position, If its in the middle then we want to leave it where it is and if its in the last pile than we want to subtract 7 from its position. i.e. the position is now given by

y = 7( x mod 3) + ( x div 3) - 7( x mod 3 - 1 )

which simplifies nicely to

y = 7 + ( x div 3 )

In the trick we perform this operation 3 times, so after iterating we get

y = 7 + ( 7+ (7+ x div 3) div 3) div 3)
= 7 + 2 + 1 + ((x div 3) div 3) div 3)
= 10 + ((( x div 3 ) div 3) div 3 )

In our pack, x can only range between 0 and 20 so the second term always evaluates to 0