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**April 2006**

Bill Simon's 1964 book *Mathematical Magic* (Dover) contains a wonderful trick called "The Four Queens," based on a clever principle the author devised. It uses just 8 cards, and works for packets of size any power of 2. Simon's '64 principle was the basis for a nice 12 card Karl Fulves effect called "Even Money Proposition" in the April 1969 *Pallbearers Review*. More recently, it has been taken to brilliant heights in Dave Solomon & John Bannon's stunning trick, "The Power of Poker," using 10 cards.

We stick with 8 card incarnations here, starting with a slight twist (modulo a gender change) on the original, before discussing some other variations. In each effect, audience members are offered six completely free choices. You imply--though it isn't true!--that these choices are independent, binary ones, so that any of *2 ^{6} = 64* different things could happen. Yet, without fail, a very predictable separation occurs every time.

Shuffle the deck casually while talking about the number of free choices you will shortly offer a spectator. "We'll need just eight cards," you comment, counting out that many onto the table. Pick them up, and do a little more shuffling while you continue. The packet is held face-down throughout. "Now, I'm going to ask you to make several totally free choices, to determine two piles of four cards. First, you'll decide on two cards for your pile, then two for my pile, then one for your pile, then one for mine. That's the system: *2 + 2 + 1 + 1 = 6 * degrees of freedom for you, as you determine where six of the cards go. So many possibilities, and it's your call every time. Then I'll give you the top card of the two that remain, and I get the last one."

Hold the top card of the packet in front of you, making it clear that nobody can see its face. (The same holds for all of the cards as you proceed.) "Either this card will form the first card of your pile, or we'll tuck it underneath. If you opt not to choose this one, then you must use the next card instead, and if you do choose this one, the second card is tucked underneath. You decide, do you want this card or not?" If the spectator wants the card, deal it to the table to start a pile, and tuck the next card underneath the packet in your hard. If the spectator declines, tuck that first card underneath and deal the second card to the table.

Repeat this procedure with the next two cards; the net result is that two cards are now in a pile in front of the spectator, and two have been tucked underneath. Now do it again for the next two pairs of cards, only this time the two cards selected are dealt into a second pile in front of you. Do it again for the next pair of cards in the packet, dealing the selected card onto the spectator's pile, and finally once more with the next pair of cards, dealing the selected card onto your pile. "That's three cards for each of us," you comment, as you deal out the last two cards--the top one to the spectator's pile, and the bottom one to your own--as claimed at the outset. "And that makes four each." Stand back.

"Let's recap. We had eight cards selected from a shuffled deck, and that packet was then shuffled further. Next, I gave you six completely free choices. As a result, any of *2 ^{6} = 64* things could have happened to the selected packet." Pause to let that sink in. "Yet, I think something rather interesting may await us. Please turn the two piles over." It will be found that one is all Jacks, the other all Kings. "Congratulations!" you proclaim. "You've just separated the men from the boys!"

The shuffling referred to above is all carefully controlled. The 8 cards used are of course the 4 Jacks and the 4 Kings, which can be maintained at the top while shuffling other parts of the deck. The rest depends on an ingenious yet surprising mathematical principle due to Bill Simon: * If the chosing and dealing is done as above, then regardless of what choices are made, the pile the spectator ends up with contains the top half of the original packet. You, by default, end up with the bottom half. * (Simon's trick started with 4 Queens on top followed by 4 indifferent cards, and his climax was the revelation that the spectator's pile contained the Queens. He also used a magician's force when dealing out the last two cards, to give the illusion of another free choice.)

To pull off the trick as described, all you have to do is ensure that the packet starts off with the 4 Jacks followed by the 4 Kings (or vice versa). You can even do additional shuffling to this 8 card packet and preserve that top/bottom separation. In addition to shuffles which only mix up the top of bottom halves of the packet, following option is worth considering. *Run off k cards (where k at least 4), thus reversing their order, and drop the rest of the packet on top. This can be repeated any number of times--perhaps with different values of k each time--and the desired separation is maintained.* The easiest way to convince yourself of the above is to work with a face-up packet consisting of four red cards followed by four black ones. (An odd number of this kind of shuffle will put the black cards of top.)

Furthermore, working through Bill Simon's free choices and dealing concept with such a packet, initially ordered R R R R B B B B, reveals what is really going on. After the first free choice, there is a red card in front of the spectator, and the packet is ordered R R B B B B R. After the second free choice, there are two red cards in front of the spectator, and the packet is ordered B B B B R R. Continuing to track the possibilities after each of the six free choices, and your final deal, leads to the following table.

Stages \ Cards |
Packet |
Spectator's Pile |
Your Pile |

Start | R R R R B B B B | ||

Result of 1st free choice | R R B B B B R | R | |

Result of 2nd free choice | B B B B R R | R R | |

Result of 3rd free choice | B B R R B | R R | B |

Result of 4th free choice | R R B B | R R | B B |

Result of 5th free choice | B B R | R R R | B B |

Result of 6th free choice | R B | R R R | B B B |

End | R R R R | B B B B |

Note that after the fourth free choice, we arrive at a scaled down version of the original, so one could appeal to induction. The final deal results in the two piles being totally monochromatic; so much for free choice!

It is now clear that if you start with a Jack and King separated packet, face-down, you end up with a pile of Jacks and a pile of Kings. You could, of course, openly start with these 8 cards, rather than maintaining such a packet at the top of the deck through some less than effective shuffles, and spin a tale of four monarchies whose enlightened Queens who gave their husbands and sons lots of free choices, yet always managed to maintain a certain hierarchy.

Shuffle the deck, and hand it out for additional shuffling. Ask that 8 cards be handed to you, sight unseen. Hold them so that only you can see the faces, and make some comments as you move a few around. Then turn the packet face-down, and shuffle a bit while asking for a pen and piece of paper. "I'm going to ask you to make some decisions shortly, but first I'm going to write down a prediction about the outcome." After you have writen your prediction and folded the paper, proceed as earlier, offering the spectator six free choices as two piles of three cards are determined. Deal the last two cards yourself, without comment.

Review the scenario which has just unfolded. "I was handed eight cards from a shuffled deck. I made a prediction. You determined the two piles, using a series of totally random choices. I did not influence your decisions in any way." Pause for effect. "Please turn over your pile." Pause. "Now open and read my prediction." The cards revealed, needless to say, match the prediction exactly.

The secret here, in addition to using Bill Simon's wonderful principle, is that you have to tailor the prediction to the cards you are given. You *do* shuffle the cards thoroughly at the outset, so that regardless of what the spectator does, you end up with 8 random cards. If all seems so fair. But before the chosing and dealing starts, you arrange the packet split into top and bottom halves according to some observation you make very quickly, on the fly (see below). You can do some more shuffling as described earlier, and still preserve that essential upstairs/downstairs separation.

With probability roughly *8!/(4!4!)* over *2 ^{8}*, or

Since the above possibilities are essentially independent, then with probability roughly * (1-0.2734) ^{3} = 0.3835* none of those cases holds for your 8 cards. But there are so many other possibilities: exactly 4 of some suit, exactly 4 of some two suits (e.g., Hearts and Clubs), etc., that it's hard not to find something special about some 4 of the 8 cards to enable you to make a prediction. For instance, if you get 3 prime values and a single Ace, you can focus on the 4 composite values. If you notice that the cards naturally split into two groups of 4 with the same sums, go with that. Or predict a certain sum. You can always predict that the four smallest value cards will separate from the four largest values. Use your imagination, and note that if you repeat this trick for an audience, the fact that your prediction varies with each performance actually adds mystery to the effect, because nobody will suspect the underlying principle! (If you don't trust your imagination to bail you out, ask for nine cards, and find eight you can work with, retaining the ninth as your lucky card, under which you can hide your prediction.)

"No doubt you have heard of Six Degrees of Separation?" you ask, as you shuffle the deck and hand it out for more shuffling. "It's a theory that given any two people in the world, there is a chain of acquaintances of length at most six connecting them. Does that six include them? A very good question, I've always wondered about that too. Let's find out. Please hand me any eight cards. I'll need two volunteers to assist me. You'll each select a card, with a lot of freedom, and then have even more freedom to mix up all of the cards."

The same principle as before is employed. Look at the cards briefly, decide on some upstairs downstairs arrangement, and split the packet accordingly. Hand four of the cards to one spectator, and the other four to a second spectator. Ask for more shuffling, making sure that the cards are kept face-down. Then have each spectator select a card from their pile. Take the three card piles back and have the selected cards noted and memorized while you discretely switch the packets in your hands. Distract the audience by claiming that these two cards represent two randomly chosen people in the world. "We're going to do some card mixing in a moment and see if these two people end up within six of each other."

Have the cards replaced in the piles--actually the first spectator replaces a card in the second pile and vice versa. Have the piles shuffled again. Take them back, placing one of top of the other, and go through the Bill Simon routine as above. At the end of this, pick up one of the piles and glance at the card faces. Repeat with the other pile. Spot both selected cards--they will be the ones that do not match their companions according to whatever split you are using--and note their positions. Don't let anyone else see any card faces. Earnestly discuss the difficulty of the task ahead of you.

"I'm going to try to verify that your two cards are indeed at most six apart." Continue, "It won't be easy, as I have no way of knowing what two cards were selected." This, of course, is not true at all, but nobody has any inkling that you already know the identities of the selected cards. Reassemble the packet by dropping one pile on top of the other. *The secret here is to do this in such a way that from the top down, there are at most four cards between the selected cards.* (If this is not true when you place the spectator's pile on top of yours, then it will be if you place yours on top of the spectator's.) You must remember the positions of both selected cards. The second one, from the top down, is of course in position 5, 6, 7, or 8. You are ready for the kill.

"Let's spell the work MAGIC." Do this, counting out 5 cards into a pile. "I have a feeling that this last card is one of the selected ones." Have the card turned over and confirmed as one of the selected ones. This works if the second card is fifth; if it's the sixth, say, "The *next* card is one of yours." If the second selected card is seventh or eight, use the word `MAGICAL.' In any case, the audience should be surprised that you have located one of the selected cards, in view of all that has gone before. Now comes the grand finale. "The other card must be close by, please count back three cards" (here you specify whatever is necessary to get back to the first card). Have it turned over, and its identity confirmed. Congratulate yourself out loud on verifying that the two selected cards, representing two randomly selected people in the world, turned out to be at most six cards apart. The audience, hopefully, is more impressed that you knew exactly where to find the selected cards, and that there were no more than four cards between them after all.