How to cheat at cards

Kevin Houston teaches us how to deal ourselves the best hand

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As a child I didn’t want to be a mathematician—I didn’t even know that was a job. Instead, for a short time, my dream was to be a crime-fighting card cheat. My interest arose from the TV series Sword of Justice where the main character obtained some help from card cheats in his quest to take down the despicable crime syndicate that had framed him and sent him to prison. (The opening credits on YouTube will attest to the programme’s brilliance.)

Despite the incompatibility of combining crime-fighting and card cheating I told my family that that was to be my chosen career. Unfortunately, I wasn’t very good at it, usually announcing my intentions to cheat before offering to play a quick game with someone. Nonetheless, to this day my family refuse to play cards with me—even Uno.

Along the way, I did learn a thing or two though. So allow me to explain to you an impressive trick that all the best card cheats know and that has some interesting mathematics hidden behind it.

The perfect shuffle

Every card cheat needs to be able to give the appearance of shuffling a deck of cards so that it remains in order. However, there is an entirely legitimate shuffle that allows us to cheat: the perfect shuffle. It mixes the cards but in a perfectly known and determined way.

A perfect (out) shuffle.

A standard deck has 52 cards. Split them into two equal piles and then interweave the piles so that the new order of cards alternates between the two piles.

The effect on the deck is shown on the right, where the cards are labelled 0 to 51 for reasons which will become clear later.

Instructions on how to do the shuffle are given below. Admittedly, it is not easy and plenty of practice is required but is satisfying once learned and has an impressive ‘wow!’ reaction when performed.

Eight perfect shuffles

What can we do with the perfect shuffle when we have learned it? First we have an interesting—and hopefully surprising—fact:

Perfect shuffle a deck eight times. It is now back in the order you started with!

Modular arithmetic is also used by clocks: they count the time mod 12.

With this you can perform a surprising trick. Put the deck in new deck order. If you perfect shuffle it five times, then the deck looks shuffled. There are patterns if one looks closely but you can show it to someone and with a straight face claim it is mixed up. Do the perfect shuffle three times for them—so now you have done it eight times in total—and to your spectator’s bafflement the deck is suddenly in order!

How can we use mathematics to prove this surprising fact? We start by numbering our cards. As above, from top to bottom, we shall number them 0 to 51, rather than the usual 1 to 52.

Next we use modular arithmetic. Arithmetic modulo $n$ is where we consider all numbers that have the same remainder after division by $n$ to be equivalent numbers. So $5=17 \mod 4$ means that 5 and 17 are equivalent modulo 4 as they both have remainder 1 after division by 4. Sometimes people write $5\equiv 17 \mod 4$ rather than using the equals sign.

With our 0 to 51 numbering the formula for finding where a card goes to after a perfect shuffle is relatively easy:

(Except for card 51) the card at position $x$ goes to position $2x \mod 51$.

For example, card 7 goes to $ 2 \times 7 \mod 51 = 14 \mod 51$, ie position 14. Note that this is the 15th physical card in the deck as we started counting at 0.

Similarly card 35 goes to \begin{align*} 2 \times 35 \mod 51 &= 70 \mod 51 \\ &= 51 + 19 \mod 51\\ &= 19. \end{align*} That is, card 35 goes to position 19.

Note that the card 7 gets lower in the deck and card 35 gets higher. So immediately we see that a perfect shuffle is beginning to mix the cards.

The formula has the exception of card 51. It is easy to see in the picture on the previous page that this card does not move and so we don’t have to worry about it in the formula—we will ignore it from now on.

But why does this $x\mapsto 2x \mod 51 $ result hold? Consider first the top half of the deck, ie all the cards in positions 0 to 25. During the shuffle, the two cards in a pair of consecutive cards get a card between them. So the new positions of the top half cards will be multiplied by 2. Since we can have at most $2\times 25=50$, which is less than 51, we get $x\mapsto 2x\mod 51$ since $2x$ is in this case actually equal to $2x\mod 51$ (rather than just congruent).

For the bottom half cards we see that their positions should also be multiplied by 2 since they get cards between them. However, we note that card 26 (the top card of the bottom half of the deck) needs to go to position 1 (the second physical card position!). One way to do this is to take mod 51 (since $2\times 26 \mod 51 = 1$). It is easy to check that we can use this for all the other cards.

Now let’s see what happens when we do eight perfect shuffles on a particular card, say card 12. We just repeat the process that the card in position $x$ will go to position $2x \mod 51$: \[ \begin{array}{lccrcr} {\text{once}} & 12 & \to & 24 \mod 51 & & \\ {\text{twice}} & 24 & \to & 48 \mod 51 & & \\ {\text{three times}} & 48 & \to & 96 \mod 51 & =& 45\mod 51 \\ {\text{four times}} & 45 & \to & 90 \mod 51 & =& 39\mod 51 \\ {\text{five times}} & 39 & \to & 78 \mod 51 & =& 27\mod 51 \\ {\text{six times}} & 27 & \to & 54 \mod 51 &=& 3 \mod 51 \\ {\text{seven times}} & 3 & \to & 6 \mod 51 & & \\ {\text{eight times}} & 6 & \to & 12 \mod 51 &=& 12 \mod 51. \end{array} \] Of course, that doesn’t prove that this happens for the whole deck. I may have been cheating by picking one that worked—and you should never trust a card cheat! So let’s do the calculation for a general $x$. If we repeat the $2x\mod 51$ operation we get $2(2x)\mod 51$. Continuing in this way for the full 8 times we see card $x$ goes to position $2^8x\mod 51$. Simplifying mod 51 we get \begin{align*} 2^8 x \mod 51 &= 256 x \mod 51 \\ &=\left( (5\times 51) x + x \right) \mod 51\\ &=\left( 0 + x \right) \mod 51\\ &=x. \end{align*}

Hence, we have proved that eight shuffles move the card in position $x$ back to position $x$. If we have a different number of cards in our deck, then the number of perfect shuffles needed to return the deck to its original order varies. For example, a deck with 50 cards needs 21 perfect shuffles. The number of shuffles required for the deck is called the order. The following table shows the order of the perfect shuffle for a selection of decks where $N$ denotes the number of cards:

$N$ order
4 2
6 4
8 3
10 6
12 10
14 12
16 4
18 8
50 21
52 8
54 52

The word order arises due to a connection with group theory and the order of elements of a certain group. The details of this and many more hardcore mathematical facts can be found in Magic Tricks, Card Shuffling and Dynamic Computer Memories by S Brent Morris: one of its appendices is a list of all orders for $N$ up to 200.

Binary fun with in and out perfect shuffles

Up to this point I have talked of a perfect shuffle, ie singular. In fact, there are two types:

  • A perfect out shuffle: the top card remains the top card.
  • A perfect in shuffle: the top card becomes the second card.

A perfect in shuffle.

The perfect out shuffle is just what have been calling the perfect shuffle. The in shuffle is the new one. Its effect on deck order can be seen below—the top card ends up ‘in’ the deck.

By combining out and in shuffles we can move the top card to anywhere in the deck. And, as we shall see, from a mathematical perspective we get a surprising appearance of binary.

Alex Elmsley (1929–2006) was a mathematician and magician. He studied physics and mathematics at the University of Cambridge before becoming interested in computers. Among magicians he is known for inventing a convincing false count of cards which is now known as the Elmsley Count.

Elmsley was very interested in the perfect shuffle (which among magicians is also known as the Faro shuffle) and investigated the mathematics behind it. In particular, he found a method for moving the top card to any position in the deck through a sequence of in and out shuffles.

Let us suppose we wish to move the top card to position $k$ in the deck. (We keep the numbering of the cards from 0 to 51 so the top card is in position 0.) To move the card via a sequence of in and out shuffles we proceed as follows:

Translate $k$ to binary. Then do out and in shuffles where $\text{out} = 0$ and $\text{in} = 1$.

Note how brilliant it is that ‘o’ looks like 0 and ‘i’ looks like 1.

For example, suppose we want to move the top card to position 13. Well, 13 is written as 1101 in binary. So we do ‘i i o i’: in in out in.

Moving the top card to position 13.

So why does this work? Imagine we have a card in position $x$ in the top half of the deck. (The case of the bottom half is similar and is, of course, left as an exercise.) What does an out shuffle do to the binary representation of the number? Since an out shuffle is $x\mapsto 2x \mod 51$ it just adds a zero to the representation. So, a card at position 101 moves to 1010. This is no different to when we tell children that to multiply by 10 they can just place a 0 at the end.

Now, an in shuffle is given by \[ x\mapsto 2x +1 \mod 52 . \] That is, multiply by 2, add 1 and take mod 52 (note, not 51 this time). This can be checked in the same way as for the out shuffle. For a binary representation this will just add a 1 to the end. Thus, the card in position 101 moves to position 1011.

To put this all together, consider what happens in the above example where we move the top card to position 13. That is, we apply in in out in and track the progress of the initial top card. The first in shuffle moves the top card to position 1 in the deck. The second in shuffle moves it to 11. The out moves it to 110 and the last in moves it to position 1101.

OK, so this is an unexpected and charming appearance of binary but it only moves one card. That’s useful, but let’s see how we can move more cards and cheat in a real game.

Dealing four aces

An impressive bit of cheating is to deal yourself four aces in a game of poker—that’s a hand that is hard to beat. For our version we need a game with three other players. Take out the four aces and place them on top. A card cheat can do this secretly with ease. Now, do two perfect out shuffles. Deal the cards starting with you, ie go round in a circle giving each person a card until everyone has five cards each. You now have a hand with the four aces!

The reason that this works is simple. At the start, the pack (starting at the top) looks like

AAAAxxxxxxx…

where A denotes an ace and x denotes an indifferent card (whose value is unimportant).

Two perfect out shuffles give you all the aces (green).

Do the first shuffle and indifferent cards get placed between the aces. So we get

AxAxAxAxxxxxxx…

Do the second shuffle and we get a set up ready for a good hand when there are three other players:

AxxxAxxxAxxxAxxxxxxx…

Well, of course if you are going to cheat and win all the time, people are going to notice. To dodge this professional card cheaters would often work in pairs and the winning would alternate between the two cheaters, reducing the chances of discovery. So in this poker game of four players you need an accomplice. Suppose you are sitting in a circle with you dealing first to the person on your left and then going clockwise.

AAAA!

Wherever your accomplice sits you will be able to deal them the four aces. As before arrange the four aces on top. Suppose your partner is sitting at position 3. You need to deal them the third card—the card at position 2 in our numbering. Let’s move that top ace to position 2. In binary the number 2 is 10 so we do ‘io’ perfect shuffles. Hey presto, the top ace is now in the right place and furthermore so are the other aces as the two shuffles put three cards between each of them. A normal deal now will give your partner-in-crime all four aces. Let’s see that in a diagram:

start: AAAAxxxxxxx…
in shuffle: xAxAxAxAxxxxx…
out shuffle: xxAxxxAxxxAxxxAxxxxx…

Wherever your partner sits you can do the right sequence of in and out shuffles to ensure that they receive the aces!

If you wish to know more about card shuffling and mathematics, then the best book to read is (as mentioned already) Magic Tricks, Card Shuffling and Dynamic Computer Memories by S Brent Morris. And remember, card cheating is a crime—so don’t get caught!

How to do the perfect shuffle

Splitting the deck. Image: C Houston

The perfect shuffle requires cards in good condition so buy a new deck on which to learn and don’t use it for anything else—cards buckled in card games will not weave very well. There are various methods to achieve a perfect shuffle. This version is done ‘in the hands’ but some can be done on the table and don’t look in any way suspicious.

Squaring the cards. Image: C Houston

Hold the lower half of the deck in your left hand with fingers arranged around as above. The right hand has a similar grip but does not go all the way round and the index finger sometimes contacts the top of the deck, sometimes not.

Butt the cards together to square them up. If the cards are not square, then perfect shuffling is impossible. Press the halves together, with the right index finger touching both top cards so that the halves can be put in position. For an out shuffle the right hand part should be slightly higher so that the left hand top card will go under the right hand top card at the end of the shuffle. For an in shuffle the right hand part needs to be lower.

A perfect shuffle. Image: C Houston

Remove the index finger and begin applying pressure at the bottom corner where the two packs meet. Once interweaving has started let the deck ‘do the work’. It took me many weeks to be able to do this consistently so don’t be discouraged if you can’t do it straight away. Like many skills it needs practice!

Kevin is a senior lecturer in the School of Mathematics at the University of Leeds. He is the author of the bestselling textbook How To Think Like A Mathematician and is surprisingly lucky at card games.
@k_houston_math    kevinhouston.net    + More articles by Kevin

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