# The Automatic Gauge Method

This works best on a hard surface so don't expect good results on an overly soft working surface.

1. The spectator having shuffled the pack places it on the table. Arrange matters so that the side of the deck is towards you. Have him cut the cards but not complete the cut.

2. As the cut is seldom equal, one packet will contain more cards than the other. The handling is such that exactly which packet is the greater makes no difference. For the example let's assume the packet A nearest you is larger as in Figure 2.

Figure 2

3. The right hand is always placed on the smaller packet, in this case packet B, the one furthest away from you. The left hand is always placed on the larger packet, in this case, packet A.

4. Both packets are now moved towards each other till the sides meet; however, both are jogged as seen in Figure 3, a side view, with the hands omitted.

5. The right 1st finger presses down firmly onto its smaller packet. The left 1st finger also presses down but lightly on its packet. The left thumb gently pushes its top cards forward. What happens is that only those cards which do not jam against the smaller packet, will move forward as in Figure 4 which shows the action from the end view. The left fingers pick up just this sliding portion of the left hand cards and at the same time the right hand moves its packet away. The action is seen in Figure 5, another end view.

Figure 4

Figure 4

Figure 5

6. The right hand action in Figure 5 is accompanied by the request of "Take the packet you cut off and shuffle it." Meantime the packet, held by the left hand, is shuffled by yourself.

7. Naturally you Run Shuffle your packet counting the cards at the same time. This Run Shuffle, in the event of a large number of cards, can be broken up by say, running 10 cards and throwing the rest of the packet onto these in an injogged condition. Get a break at the injogged packet and Run-Shuffle the rest, up to the break, as you finish the count to give you the total of cards in this packet.

8. Suppose the total is 19. To this, add one to make 20 and then divide this by two to give you 10. Next subtract 10 from 26 to give you 16 which will be the estimated total of not only the packet the spectator cut off but also of the packet still on the table.

At most you will be off by one card on one packet due to the fact that your Run-Shuffled packet may have had an odd number of cards; therefore, one of the tabled packets has to contain that extra card.

9. Drop your Run Shuffled packet; 19 cards in this case to the table alongside the other packet. At this point you know the exact number of cards in your packet, 19 cards, and the estimated number, 16 in this case, of the other two packets. It does not matter how the spectator decides to assemble the packets you can't be off more than one card.

For example - He notes the bottom card of his packet and places it on the other estimated 16 card packet then all onto your packet. Suppose you then shuffle off the top 15 cards leaving the estimated 16th on top. If he had 16 cards in his packet the card is then on top but if he had 17 then it is the second card from the top or only one card off. Later, in the effects section, I will show you how a three and even four card error can be allowed and yet terminate the effect successfully.

10. I have given the handling when the larger packet is nearest you but sup-

¥ose it is the one furthest from you. he handling is still the same with right hand grasping the smaller packet and left hand the larger; however, in this case the left 2nd finger pulls the top cards back as the two halves are jammed together. This action is shown in Figure 6.

Figure 6

11. The left fingers, of course, lift up the Sliding Packet as the right hand moves its packet forward for the spectator to shuffle. From here everything is as already explained in order to estimate the number of cards in the tabled packets.