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**December 2004**

"How often have I said to you that when you have eliminated the impossible, whatever remains, however improbable, must be the truth?" Those famous words uttered by Sherlock Holmes to Doctor Watson long ago ring very true for the tricks considered below. (This isn't the first time they have been exploited in this context either!)

We begin by reviewing a well-known elimination deal first published in 1936 as "The Spectator's Choice," by Stewart Judah.

A volunteer is invited to shuffle a deck of 52 cards. You take the deck back and fan the cards face down from hand to hand, requesting that one be chosen and remembered; this card is then replaced in the deck. Have the volunteer deal the cards into two piles on a table, A to your left, B to your right, starting with pile A. Have pile B picked up, and again dealt into two piles, by first adding to the existing pile A, while creating a new pile B beside it. Have this repeated, always dealing to pile A first, until one card remains; in the hand or preferably in a new and much deminished pile B.

Before this card is revealed, stress that the deck has been in the volunteer's hands throughout the dealing process. One pile grew steadily thereby increasing the chance that the chosen card was in that pile. The chances were 26 in 52, or 50%, after the first deal, then 39 in 52, or 75%, and now that the elimimation is complete the chances are 51/52, which is over 98%, that the card is in this big pile here. "The *very last* place you'd expect it to be is over here alone," you say teasingly, pointing to the lone card beside the larger pile.

Ask the volunteer to look through the larger pile of 51 cards seeking the card selected. It will not be found. Say, "As Sherlock Holmes once said, `When you have eliminated the impossible, whatever remains, however improbable, must be the truth'" and have the card named. The last card is revealed to be precisely the named card. "Sheer luck!" you say, feigning surprise.

The secret is simple:

Claim 1: The card eliminated by this deal will always be the one that started in the 22nd position from the top of the deck.

The easiest way to verify this is just to note the 22nd card of any full deck and do the elimiation deals. We can trace that card's journey as follows: the 22nd card from the top in the original deck becomes the 11th from the bottom of the initial 26-card B pile, and is thus 16 from the top at the start of the second deal. After that deal, which adds to pile A, this card ends up 8 from the bottom of the new 13-card B pile, and thus 6 from the top. After the third deal it's 3 from the bottom of the next 6-card B pile, and thus 4 from the top. After the fourth deal it's 2 from the bottom of a 3-card pile, and hence is the sole remaining card in pile B after a fifth and final deal.

To perform the trick indicated, the chosen card is in position 22 at the outset. One way to achieve that is to fan the cards from hand to hand in clumps of 3 or 4, silently keeping count, until 21 cards are in one hand. With the hands still together, and maintaining a clear (to you and you only!) break under the 21st card, invite the volunteer to take a card. It's not too hard to force the card to be chosen from the remaining cards past this break, and while it is being inspected and remembered, simply break the deck at the 21st card and have the card returned on top of the lower portion, before reassembling the deck.

If the volunteer insists on taking a card from the first 21 cards, just add one card from the lower portion before having the card returned. If the card is placed in any position *near* the 22nd, that's fine too, as long as you know exactly where it is. You can casually shuffle some cards from one end of the deck to the other to adjust for this, if necessary, and following that you can shuffle large clumps of cards on either side of the replaced card (which is near the middle of the deck) with impunity before directing the elimination deal.

Let's look this more analytically, and see what else we can learn. Given a packet of *k* cards, there exists a unique position *i* from the top such that the card that starts in that position ends up being the last card dealt. We can determine *i* as follows: Order the cards from top to bottom in some specific order, such as: A♣, ..., K♣, A♥, ..., K♥, A♠, ..., K♠, A♦, ..., K♦---so that, e.g., the 15th card is 2♥, omitting the cards past the *k*th one if fewer than 52 cards are used. Next, perform the elimination deal and observe what card is dealt last.

For instance, with a full deck of 52 cards, ordered as above, the last card is 9♥, which was originally the 22nd card. This establishes Claim 1.

Imagine now placing this eliminated card on top of pile B, so that we have a full deck again, and performing a second round of elimination dealing.

Claim 2: The original 36th card is the one remaining if the elimination deal is repeated with the full deck.

To see why this is true one need only observe what the 22nd card is after the first elimination deal: it's 10♠, which was the 36th card originally. This will be the last card if the elimination deal is repeated.

This suggests another "out" if the volunteer resists placing the card at the desired position 22: have the card inserted in position 36 instead (note this is position 17 counting up from the bottom), and then have the elimination deal done as before. It will fail, and must be seen to fail, so simply flip over the last card, and have it confirmed that it is not the selected card. Now place it on top of the other 51 cards, and repeat the elimination deal yourself, commenting, "Let me show you how a real magician does it." Or simply have the volunteer do it again, having first incanted some magic words, such as *Abracadabra* or *Banach-Tarski Paradox*. At the end, proceed with the "Sheer luck" finale as suggested before.

Of course you may wish to leave the failure card to one side and not include it in the second round of elimination dealing, and if so, you will be working with a packet of 51 cards, which is quite different. That provides you with another option (just why it works we will discover in due course):

Claim 3: The card in position 48 (i.e., 5 from the bottom) in a 52 card deck is the one remaining when a double round of elimination dealing is done, and the first "failure" card is placed to one side before the second round of dealing.

(You may wish to wrest triumph from the jaws of defeat, by using the failure card to tap - as if it were a magic wand - the last remaining card after the second elimation dealing, "thereby" turning it into the selected card!)

All of this also suggests a trick in which two volunteers pick one card each, and then find each other's selections.

Proceed as before, having two card selected and remembered, and returned to the deck in positions 36 and 22, respectively, and direct the elimation dealing for each volunteer in turn. When only one card remains for the first volunteer, say, "Wouldn't it be amazing if that were your card?" It turns out not to be that card of course, but the second volunteer should recognize it!

Congratulate first volunteer, saying "*You* clearly have some magic talent, let's see if you both do! There was 1 in 52 chance of finding one card the way you did, and so there is only a 1 in 52 squared, i.e., a 1 in 2704 chance, of both cards turning up. That's pretty unlikely!"

Now have the second volunteer do the elimination deal; needlesstosay, this turns up the first card selected. Congratulate both volunteers for a job well done.

One way to gain some insight into the Judah elimination deal applied to the whole deck is obtained by considering it as a permutation of cards numbered 1--52, assuming the last card is placed on top of the others at the end of the deal. After a single deal, the cards are in order 22, 38, 6, 14, 30, 46, 50, 42, 34, 26, 18, 10, 2, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 51, 49, 47, ..., 5, 3, 1.

This permutation turns out to be the product of a 49-cycle, a 2-cycle and a 1-cycle; specifically, using standard cycle notation, it's:

(1 52 26 10 12 16 17 44 24 19 43 31 37 34 9 48 25 40 23 41 32 21 42 8 15 45 30 5 50 7 49 28 20 18 11 47 29 38 2 13 46 6 3 51 27 39 33 36 22) (4 14) (35).

From this we can see that the original 22nd card becomes the last one dealt (i.e., the 1st one in the reassembled deck) after one deal, and also that the original 36th card is the last one dealt after two deals.

Any of the above methods — a simple verification deal is hard to beat — confirms that when the elimination deal is applied to a packet of just 51 cards, the last card remaining is the one that started 24th from the top. This observation, in conjunction with an examination the cycle decomposition above for the elimination deal on a full deck, establishes Claim 3.

Note also, that for a full deck the cards in positions 4 and 14 switch positions with each deal, and hence return to their original positions after two deals. Furthermore, one card is fixed by the deal: the 35th, which is 9♠ in the order suggested earlier. These observations can be used to come up with numerous other card effects.

Note that this type of elimination deal partitions the deck into clumps of 26 even and 26 odd numbered cards, and hence tranforms a deck in which the cards alternate black/red into a black and red separated deck.

Another twist on the basic plot when going for a double revelation is to include the original 22nd card the second time around, but to add it to the bottom (not the top) of the other 51 cards. That would, of course, require some advance planning!