There are numerous card tricks which use mathematics to work instead of manipulations. I try to collect some here. The page will grow as I find more interesting ones. Let us start.
Trick 1
You take the 4 aces from deck plus 12 cards, and put the aces face-up between the face-down twelve cards. You can shuffle as much as you like. The result is a stack of 16 cards like this:
X A X X A A X X X X X X X A X X
X means that the card is face down.
The next step is to create chaos by turning every second card. You can do that by drawing the cards from the stack and putting them on another stack alternating turned and not turned. The result is a stack like this:
? A ? X X A ? X ? X ? X ? A ? X
Here, ? is an arbitrary face up card and X is a face down card. If you follow my instruction to put the card from the top to the other stack, this sequence will be reversed. But that does not matter.
Now you create even more chaos by arbitrarily turning pairs of cards, again putting the pairs from your stack to another stack. You take the pair, and either lay it down on the new stack, or flip it over and lay it down after that. Note, that the pair changes order when you flip it.
Finally you lay out the cards in 4×4 grid from the top in the following order:
1 2 3 4
8 7 6 5
9 10 11 12
16 15 14 13
Now, you fold either left or right column, or the top or bottom row inward, turning the cards. E.g., if you select the left column, you take card 1, turn it over, and put it on card 2. Then you take card 8, turn it over and put it on card 7. Same with cards 9 and 16. If you selected the top row, you’d turn card 1 onto card 8, card 2 onto card 7 and so on.
You do this kind of folding, always with the left or right column, or the top or bottom row, until you are left with one stack of cards. Magically, this stack will have the aces faced up and all other cards faced down.
How does that work?
You will see how, if you just take a stack of 16 turned-down cards and follow the instructions. You will end with a single stack, all faced in the same direction. Thus, the tricks simply works, because the flipping in the first two steps is just reversed by the collection at the end.
After the first step, you have a stack with each second card flipped like this:
X ? X ? X ? X ? X ? X ? X ? X ?
That does not change by flipping over pairs of cards, because the pairs change their order during the flip.
Consequently, your final layout will look like this:
X ? X ?
? X ? X
X ? X ?
? X ? X
During the turning, each card processes through a path towards the final remaining stack. E.g., assume that the final stack is on the card in the second row and column at position (2,2). Then the top left card at position (1,1) is turned exactly twice in the process, because the length of the path of the card is 2. The card at position (3,4) in the third row and last column is turned 3 times.
It not hard to see visually, why all cards end in the same direction now. The X card turn an odd times, and the ? cards turn an even times, or the other way around. But let us prove that mathematically.
The number of turns is actually the sum of the absolute differences in column and row indices.
\text{Turns between $(i_1,i_2)$ and $(j_1,j_2)$} = |i_1-i_2| + |j_1 - j_2|The signature of the card (0 or 1), meaning turned or not turned, is
\sigma(i,j) = (i-j) \mod 2
Consequently the X all end up in the same direction as the ?. The number of turns modulo 2 is determined by
(|i_1-i_2|+|j_1-j_2|) \,\text{mod}\, 2 = (\sigma(i_1,j_1) + \sigma(i_2,j_2)) \,\text{mod}\, 2The equality is based on some facts of mdulo-arithmetic modulo 2, such as:
|a| \,\text{mod}\, 2 = a \,\text{mod}\, 2, \\ (a+b) \,\text{mod}\, 2 = (a \,\text{mod}\, 2 + b \,\text{mod}\, 2) \,\text{mod}\, 2 \\ (-a) \,\text{mod}\, 2 = a \,\text{mod}\, 2In the end facing of the card at (i,j) moved to (k,l) is:
(\sigma(i,j) + (|i-j|+|k-l|)) \,\text{mod}\, 2 = \sigma(k,l)Trick 2
You take a number of red cards (hearts or diamonds) and the same number of black cards (clubs or spades).
R R R R R R R
B B B B B B B
Then you sort the two piles into one taking one alternatively, face down. If you can do that with riffle shuffle, that is impressive.
R B R B R B R B R B R B R B R B
Now, you do one riffle shuffle of this pile. This is done by parting the pile into two, bending the two piles, and let the piles fall into one another in an irregular manner. The result might look like this:
R B R B R B R B R + B R B R B R B
-> R B R B R B R B B R B R R B R B
I have marked the cards from the second pile with 4 cards in bold italic. This looks like a good shuffle. You can even cut the cards as many times as you want.
You now show the cards face up and convince the spectator that the cards are really shuffled concerning red and black cards.
The next step is important. Secretly, you make sure that the first and the last card are in the same order by cutting the pile if necessary. Our example might look like this:
R B R B | R B R B R B R B B R B R
Now we separate the new shuffled pile into two piles putting one card alternatively in each of the two new piles. With the example above, the result looks like this:
R R R R R R B B
B B B B B B R R
You notice that the second pile is a mirror of the first. This is always the case. You can use this effect in many ways. E.g., hand one pile to the spectator and let him uncover the cards putting the red cards in one pile and the black cards in the other. Do the same movements on your side with the other pile face-down. Surprise the spectator with the fact that you sorted your pile into red and black blindly while he was looking at the cards.
Why does this work? Why doesn’t the riffle shuffle matter? And why does it matter that the same color is on both ends?
To repeat, we start with a well ordered pile P with alternate colors. For this discussion we assume the top card is red. We then divide that pile into to piles P1 and P2, P1 being the top one, and riffle shuffle P1 and P2 into a pile PS. The final step is to cut PS such that the first and the last card have the same color, if that step is necessary, and get a final pile PF. We need to show that this pile consists of pairs with mixed color only. In our example, the pairs are
R B R B R B R B R B R B B R B R
We treat the simple case first. Assume P1 and P2 have both an even number of cards. In that case, we will always have to cut PS, because the top card will be red and the bottom card will be blue. We also note that the cards in PF from P1 and the cards from P2 are still in perfect alternating order. The pairs in PF which contain only cards from P1 or only cards from P2, are thus mixed, and we can remove them. We are left with a pile of pairs, where each pair has one card from P1 and another one from P2, like one of the following:
R B B R R B
R R B B R R
Still, the sequence of cards from each pile is in alternate order. Thus, we have only to show that remaining pile does not start with two red or two blue cards as in the second case above. In that case, we could not have removed any pairs from one of P1 and P2 only because the alternation of the cards from P1 and P2 is still intact. Thus the whole pile PF must look like this, and that contradicts the different colors at both ends.
The other case of an even and an odd pile is quite similar.
Trick 3
Take any number of cards and have the audience select an integer C larger than half of the number of cards. Then count C numbers, one after the other, from the top to a new pile, and put the rest of the cards simply on top. Do that three times. Now you can predict the top card.
For the trick, simply remember take a glimpse to the bottom card. It will be on top after this procedure, no matter what C is.
This looks really mysterious, but it is trivial. Three times counting down C cards, one after another, with a number C randomly chosen, is certainly confusing. But what happens, is simple.
Assume that L cards are left after the counting and dropped on top. Then the bottom card will become card number L on top. Counting down from the top makes this card card number L from the bottom. In the third count, the bottom L cards will be dropped on top, bringing the card to the top.