A very old mathematical divination

By on 01/02/2016

The performer gives to three spectators three (different) items to distribute and some instructions about removing counters while he is absent, and when he returns he instanteneously knows who took what … This is the essence of this fun trick based on basic combinatorics, more precisely on permutations.

by Franka Miriam Brueckler

Before we explain to you the exact routine, first some maths… If you know what permutations are, you may skip this paragraph. Otherwise, start thinking about people in a queue. If there is only 1 person, he can only stand alone in one way. If there are two, Ann and Bob, then Ann can be in front of Bob or vice versa: there are two queues possible with two people. With three, Ann, Bob and Chris, there are more ways: 6. Can you find them all? Four people can arrange themselves in a queue in 24 ways, five in 120, … Generally, there are … .nqueues, or arrangements, of n persons, or items. We call each such arrangement an permutation.

We illustrate the 6 permutations of 3 items as permutations of 3 colors:
In our trick, the three items are permuted – distributed to three positions (i.e. spectators). Thus there are 6 possible ways in which the spectators could have distributed the items and the rules have to be devised correspondingly, so that the instructions about removing the counters result in different situations for different distributions. We perform this trick using 18 pebbles as counters – these are left with the spectators in the beginning, as are the three items to be distributed. They are instructed to distribute and hide the three items and then each obtains his instructions about taking some counters. In order to avoid misunderstandings, we use preprepared instruction cards like the ones below (on each, the first row corresponds to the first of the items, the second to the second one, and the third to the last one).


If each spectator follows his instructions correctly, when the performer returns, he will find 1, 2, 3, 5, 6 or 7 pebbles left. Each number corresponds to a specific distribution of the items. To demonstrate this, we shall call the items S (small), M (medium) and L (large). Then the six permutations and the instructions performed by the first spectator (yellow card), the second (red) and the third (blue one) lead to the following results:

Permutation Pebbles removed Pebbles left

SML 1 + 4 + 12 1
MSL 2 + 2 + 12 2
SLM 1 + 8 + 6 3
MLS 2 + 8 + 3 5
LSM 4 + 2 + 6 6
LMS 4 + 4 + 3 7

As you see, the maths here is very simple. Still, there is a complicated side of the trick for the performer: he must remember which result corresponds to which permutation. For this, people have devised mnemonic sentences, and our personal favourite is the one of those suggested in M. Gardner’s “Mathematics, magic and mystery”: “SaM MoovS SLowly since MuLe LoSt LiMb”. How does this help? If you assign the names S(mall), M(edium) and L(arge) to your items, as we did above, you can find two of them in each of the words in the sentence, except for the fourth one. For example, in the 1st word you have S and then M. This means: if you see 1 pebble left (1st word) the S(mall) item is held by the 1st spectator, then the M(edium) item is held by the 2nd spectator and the remaining, L(arge) item is held by the 3rd spectator. If you see 5 pebbles left, you check the 5th word (MuLe) and find 1st M, i.e. the 1st spectator has the medium item, then L, i.e. the second spectator has the large item, and finally the third spectator has the remaining, small, item.

BTW, this trick is now about 500 years old. You can read more on it, including some more on mathematics related to it, in “Magical Mathematics: The Mathematical Ideas that Animate Great Magic Tricks” by P. Diaconis and R. Graham

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