Writings of Charles S. Peirce: A Chronological Edition, Volume 8 - Charles S. Peirce

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      I will now explain to you the object of counting the rounds. But first let me remark that the last round, which consists in placing the ultima upon the final pile, should always be considered as a round of the odd kind. When you dealt into 12 piles and gathered them up, with the first 48 cards you performed 4 rounds of 12 cards each, and had 5 cards left over. These five you dealt out, making the first round of the second set; and then you transferred these five piles over to the tops of the second five, making another round of the second set. Then from these five piles you dealt to the other two piles twice, making two rounds of the third set. Next the ultima was placed upon the next pile, making a round of the fourth set. Finally the ultima was placed on the last pile which, being a round of an odd set, belonged to the fifth set. So the numbers of rounds were

      4, 2, 2, 1, 1.

      From this row of numbers, which we will call the M’s, we make a second row, which we will call the N’s. The first two N’s are 0, 1, the rest are formed by multiplying the last by the first M not already used and adding to the product the last N but one. Thus the N’s are

      0, 1, 4, 9, 22, 31, 53.

      The last N is 53. It will always be the number of cards in the pack. Reversing the order of the M’s

      1, 1, 2, 2, 4

      will make no difference in the last N. Thus, the N’s will be

      0, 1, 1, 2, 5, 12, 53.

      Leave off the first M, and the last N will be the number of piles. Thus from

      2, 2, 1, 1

      we get

      0, 1, 2, 5, 7, 12.

      Leaving off the last, will give the number of piles into which you must deal to restore the order. Thus from

      4, 2, 2, 1

      we get

      0, 1, 4, 9, 22, 31.

      If you deal 53 cards into 37 piles, the numbers of rounds will be

      1, 2, 3, 4, 1.

      If you deal into 34 piles the numbers will be

      1, 1, 1, 3, 1, 2, 1.

      If you deal into 33 piles, the numbers will be

      1, 1, 1, 1, 1, 5, 1.

      If you deal into 32 piles, the numbers will be

      1, 1, 1, 1, 10, 1.

      If you deal into 30 piles, the numbers will be

      1, 1, 3, 3, 2.

      You perceive that the object of counting the rounds is to find out how many piles you must deal into to restore the proper order, and consequently by multiplication how many piles you must deal into to make any given card the first.

      Going back to 10 cards, if we were to deal them into 5 piles or 2 piles, the piles could not be taken up so as to conform to the rule. The reason is that 5 and 2 exactly divide 10; so that the last card falls on the last pile, and there is no pile to the right of the last card upon which to pile the others. To avoid that inconvenience, we had best deal only with packs having a prime number of cards, or one less than a prime number; for, in the last case, we can imagine an additional last card which remains in the zero place, as long as there is only multiplication, no addition; that is, as long as the pack is not cut.

      If we deal a pack of 10 cards into 3 piles twice or into 7 piles twice, we multiply by − 1; for 3 × 3 = 9 and 7 × 7 = 9, and 9 is one less than 0 or 10. Suppose, then, starting with 10 cards in their proper order we deal them into 3 piles (or 7 piles) and, taking them up according to the rule, next lay them down backs up in a circle, thus:—

      Then, my dear Barbara, you can say to your little friend Celarent, who is so fond of denying everything, “Celarent, what number do you want to find?” Suppose she says 6. Then, you count 6 places from the 0, say in the right-handed direction. You turn up the 6th card, which is the 8; and you say: “If the 8 is in the 6th place clockwise, then the 6 is in the 8th place counter-clockwise.” Thereupon, you count 8 places from the 0 to the left and turn up the 8th card, and lo, it is the 6. Or you might have counted, at first, 6 places to the left and turning up the 6th card, have found the 2. Then you would say “If the 2 is in the 6th place counterclockwise, then the 6 is in the 2nd place clockwise.” And counting 2 places from the 0 to the right, you would again find the 6. The same would hold good if Celarent were to call for any other number.

      If you want to do this little trick with 13 cards, you must deal them into 5 or 8 piles. You might begin by asking Celarent how many piles she would like the cards dealt into. If she says 2 (or 11), deal them as she commands, and having done so, ask her whether she would now like them dealt into 4 or 9 piles. If she makes you first deal them into 3 or 10 piles, give her her choice afterward between 6 and 7 piles. If she makes you first deal into 4 or 9 piles, give her then a choice between 2 and 11. If she makes you first deal into 6 or 7 piles, give her her choice afterwards between 3 and 10 piles. If she makes you deal them into 5 or 8 piles, lay them down in a circle at once. In doing so let all be face down except the king, which you place face up. The order will be

      You ask: “What spade would you like to find?” If she says “The knave,” reply “Then we count to the knave place.” You count and turn up the 3. Then you say “If the 3 is in the place of the Jack counting clockwise, then the Jack is in the place of the 3 counting counter-clockwise.” You can then count round to this place clockwise, and find it is the 10th. So you continue: “And if the Jack is in the place of the 10 counting clockwise, then the 10 is in the place of the Jack counting counter-clockwise.” You count, turn up and find it so. Then you count up to this card clockwise, and go on “And if the 10 is in the place of the 2 counting clockwise, then the 2 is in the place of the 10, counting counter-clockwise.”

      The same thing can be done with a full pack of 52 or 53 cards.

      We have thus far considered addition and multiplication separately. Now let us study them combined. Take a pack of 11 cards in their proper order. Cut it so as to carry 3 cards from back to face of the pack. That adds 3 to the face-value of the card in any given place.

      Now deal them into 5 piles and gather up the piles according to rule. This by itself would multiply the face-value of the card in any given place by 5. But acting after the other operation, if x be the place and y the face-value (or original place) we have

      y = 5x + 3.

      On the other hand, starting again with the cards in their proper order, if we first deal into 5 piles and then carry 3 cards from back to face, we have

      y = 5(x + 3).

      In short, the order in which the operations are to be taken in the calculation of the face-values is just the reverse of that of their actual performance. The reason is too obvious to require explanation.

      It is easy to see that before dealing the cards out in the little trick I proposed your showing Celarent you can perfectly well allow her to cut the pack first, provided that after the dealing, or at any time, you recut so as to bring the zero card to the face of the pack. This will annul the effect of the cutting.

      I want to call your attention, Barbara, to the fact that there is another way of effecting multiplication besides dealing out into piles and gathering in. Suppose for instance you hold in your hand the first 11 spades in their proper order while the first 11 diamonds in their proper order are lying in a pack face down upon the table. We will now effect upon the spades the operation

      5x + 3

      and simultaneously upon the diamonds the inverse operation

      For

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