Order amidst Chaos

By on 01/02/2016
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Order amidst chaos. That could be the motto of the mathematical magic trick that I would like to present in this installment. You will need a deck of playing cards with equal numbers of red and black cards. A standard 52-card deck should be fine for our purposes. As a preparatory move, you should arrange the cards so that colors alternate, as shown in Figure 1.p4120012

 

Figure 1: This is how the cards should be arranged.
 

And now we let chance intervene three times in this pack of cards. In step one, someone cuts the deck somewhere approximately in the middle. In the second step, someone else shuffles the two halves together. Finally, a third person cuts the pack at a point such that two cards of the same color are separated. See Figure 2:

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Figure 2: Cut, shuffle, and cut again…
The two halves are placed one on top of the other and given to you. A naive person would think that these three random processes have resulted in a chaotic mix of the cards, the cards arranged totally at random. And so it appears at first glance. However, a remarkable phenomenon is at work. It turns out that cards 1 and 2 have different colors, as do cards 3 and 4, 5 and 6, and so on. As the magician, you may now place the pack of cards under a cloth and murmur mystical incantations and then pull out pairs of oppositely colored cards as if by magic, even though you are actually just drawing the cards from top to bottom. See Figure 3:

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Figure 3: … and now present the cards a pair at a time.
The mathematics behind this is interesting. The fact that the cards are arranged pairwise with different colors after the three random events can be proved with combinatorial methods. In this connection, mathematicians speak of an invariant. The magician Gilbreath, who invented the trick at the beginning of the previous century, seems to have discovered it by trial and error.

A Variant of the Trick

For those who would like to add this trick to their repertoire, here is a variant. Recall that the original goes according to the following outline:

  • Prepare the pack of cards (even number of cards of alternating colors).
  • Cut the deck, and then shuffle the cards.
  • Cut the deck at a point where there are two cards of the same color and reassemble the deck.

Then each pair (cards 1 and 2, cards 3 and 4, etc.) contains two cards of opposite colors.

And now the variant: The deck of cards is prepared just as in the original version, and it is again cut. Warning: This time you must somehow inform yourself as to whether the cards on the bottoms of the two halves are of the same color or different colors. This could be done, for example, as the cards are handed to the shuffler.

The next step is again as previously: the two halves are shuffled together. And that is it: you do not need to cut the pack again.

The advantage over the first variant is that you don’t need to have someone look at the cut deck to see where two cards of the same color reside. Thus no one will get the idea that the colors red and black are more regularly distributed than would be the case in a well-mixed pack of cards.

There are now two cases: Case 1 is that the two bottom cards had different colors. Then no adjustment is necessary. It is guaranteed that every pair-cards 1 and 2, cards 3 and 4, and so on-are of opposite colors. The second case is that the two cards on the bottom were both red or both black. Now things are a bit more complicated. While you are muttering your magic formulas, move the top card to the bottom of the deck. Then again, all pairs will contain one red card and one black. Of course, you don’t have to move the top card to the bottom. Instead, your first pair should consist of the top and bottom cards. Thereafter, things proceed as before. Good luck with your magic!

And where is the mathematics in all of this? It guarantees that the trick will always work. It can be proved that the cards end up as described here. However, the rather complex theory necessary for the proof is beyond the scope of this column.

This is an article from the book “Five-minute mathematics” by Ehrhard Behrends which was published in 2008 by the American Mathematical Society (AMS). It is reproduced here with the kind permission of the AMS.

 

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