Math card trick 2

It is the explanation of the trick shown below in the embedded video.

Figure 1: showing  how the cards are arranged to start with. It shows the card (red) which is the randomly chosen card by the observer from the set of randomly chosen 9 cards. It is always kept in the 44th position from the top of the deck.

1. Now there are 4 piles created starting from the top of the deck. ( counting 10 downwards and stopping when I get a card match).
2. Let x be the number for the card match in the first pile, such that 0<= x <=10.
3. It means that the number of cards making a pile can range from 1 to 11 .
4. So if you get a card match at x, you stop there, so the number of cards in Pile 1 = 11 - x.
5. Similarly let y be the card match for Pile 2, number of cards in Pile 2 = 11 - y
6. Let z be the card match for Pile 3 , number of cards in Pile 3 = 11 - z
7. Let a be the card match for Pile 4, number of cards in Pile 4 = 11 - a
8. Now lets link x,y,z and a.
9. We know that in the worst case each pile should have gotten 11 cards each from the deck.
10. So now if Pile 1 card match is x then, it leaves out x  cards, that should have  otherwise  belonged to Pile 1 , in the deck. That leaves a total of (41 +x) cards in the deck.
11. Now again we shell out (11 - y) cards out (since in the second pile, card match was y). That will leave a total of (30 + x + y) cards in the deck.
12. Similarly we give away (11 -z) and (11 - a) number of cards for Piles 3 and 4 respectively. Finally that will leave us with (8 + x+ y+ z+ a) number of cards in the deck.
13. Now lets find out what will the position from the top of the deck of our random card.
14. Our random card is in the 9 th position from the bottom of this new deck too. ( since we will have disturbed it all along).
15. In a pack of  (8 + x+ y+ z+ a) cards , the number of cards above the 9th card is (  x+ y+ z+ a - 1)
16. So (x + y + z + a) th  card from the top will our random card, which was purposefully kept in the top of the 9 card pile previously.

Math card trick 1.

This is an explanation of the card trick shown in video  embedded below.

Lets go step by step.

Figure 1: The initial split up of the cards

1. So to start with the pack is split up as shown above.
2. You are asked to place one of your random cards over Pile 1 and take choose a random number less than  15 ( let it be x) and then chose that many number of cards from Pile 2 and place it over the newly created Pile 1 ( I call it newly created since, we have altered the pile, hence forth that will be the convention).That means you have placed your 1st random card in the 11th position from the bottom in pile 1.
3. Now you place your 2 nd random card over the top of the remains of the Pile 2.That means your second random card is in (16-x) th position in Pile 2 from the bottom. Now again chose a random number less than 15 (lets call it y) and chose that many number of cards from Pile 3 and place it over the new Pile 2 to create yet another new Pile 2.
4. Now place your 3 rd random card over the new Pile 3 and then place all of Pile 4 over new Pile 3 to create another new Pile 3. Now the 3 rd random card will be in (16 - y) th position in new Pile 3.
5. Now all the new Piles are placed in the following order bottom to top. Pile 1, Pile 2 and Pile 3.
6. Now lets see where our random cards are from the bottom. 
1. 1 st  random card - 11 th postion
2. 2 nd random card - (11 + x + (16 - x)) = 27 th position  [ number of cards in Pile 1 + random number of cards kept over Pile 1 from Pile 2 + position of second random card with from bottom in Pile 2 from bottom].
3. Similaryly 3 rd random card - (11 + x + (16 - x) + y + (16- y)) = 43 rd  position.
7. It is seen that the position of the cards are a constant number and it is not dependent on any variable we chose before.
8. Now I shift 4 cards from above and put it in the bottom of the deck.
9. The new position would be 15, 31, 47 for the 1 st, 2 nd, 3 rd cards respectively from the bottom of the deck.
10. They can also be numbered from the top of the deck and they are 6, 22, and 38  for 3 rd, 2 nd and 1 st card respectively. 
11. Now i start putting the cards down from the top of the deck as follows. I alternatively put the cards facing up and facing down. I start with first card on the top of the deck, i put it face down, second card face down, i continue to do this till the end.
12. Now i end up putting all the cards in the odd position (numbering from top of the deck) facing upwards. Its obvious that our random cards, all of them were in the even position. So they all face down.
13. Now i collect all the cards that went face down. It is to be noted that the order of the even cards will have reversed now. So card in that was previously in position 52 will be in the top of the new deck and card that was in position 2 will be in the bottom of the new deck.
14. Now lets see where our random cards are in the new deck from the top.
1. 1 st random card - 24 th position.
2. 2 nd random card - 16 th position.
3. 3 rd random card - 8 th position.
15. Now I again repeat what i did in step 11. I eliminate all odd positioned cards ( numbering from top of the deck). 
16. Now new deck we have 13 cards with the position of our random cards (from top of the deck) as follows
1. 1 st random card - 2 nd position
2. 2 nd random card - 6 th position
3. 3 rd random card - 10th position
17. On repeating step 11 we eliminate all odd positioned cards from the deck.
18. We have now 6 cards facing down with our random cards positioned as follows (numbering from the top of the deck)
1. 1 st random card at 6th position
2. 2 nd random card at 4 th position
3. 3 rd random card at 2 nd position.
19. Now if we repeat step 11, we end up eliminating all odd positioned cards leaving our chosen random cards facing down.
20. So that shows it works for any three cards chosen at random.