Last time we talked about Weighing the Cards, we discussed an interesting way to perform ‘weighing the cards’ after the audience shuffles the cards.
After we do the shuffle procedure discussed in that routine, we end up with one of our 13 memorized cards in every 4th position (4th, 8th, 12th, etc).
In essence, we’ve engineered the whole thing to place a certain number of ‘known cards’ in known positions.
But what if we could skip all that procedure and get right to that same end result?
After all, the actual identity of the cards doesn’t matter. All that matters is that we know the overall group of cards and can spot them in the deck.
So, in theory, we could do this with a shuffled, borrowed deck if we had a way of knowing which cards were in positions 4, 8, 12, 16 and so on…
Some of you might see where I’m going with this…and you’re shaking your head.
Yep…I never said it was easy. But it’s definitely possible.
Alright, enough teasing.
How do we do this?
Simples:
We memorize every 4th card as we deal through the deck.
You might think that sounds crazy. It’s not. In fact, for those of you who own the Skyscraper Method, you can watch me do this on camera in module 5. I think it took me about 40 seconds to commit the cards to memory.
I do that by using the method we show you in Module 1 of Skyscraper ‘on the fly’ as I deal through the cards.
I note every 4th card (cards 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52) as I deal through the deck under the cover of a different effect.
As I note the cards, I create images based on the ‘PAO’ formula and place them in a memory palace.
Since you’re only memorizing 13 cards, you’ll be using 4 locations (3 cards per location) and just relying on your short term memory to recall the 13th card.
Again, all of this is covered very much in-depth in Module 1 of our Skyscraper course (which you should have access to, as an Inner Circle member.)
How do I justify the dealing?
I would cover it up as part of a different effect. For example, I might say:
“I want you to pick a card, but I don’t want you to name one. I know that there are certain cards you’re more likely to name, so I don’t want you to pick them.
Instead, I’ll deal through the deck like this, and I want you to just LOOK at any one card, but don’t tell me when you do. Really concentrate on the cards and THINK about this. I’ll give you some silence. You’re ready? Great. Let’s go…”
As you deal, build the images and place them in your memory palace.
Of course, after this you can reveal the card in any way you like (Mnemonicosis, Invisible Deck, card to impossible location, etc.).
But that really doesn’t matter all that much.
All that matters is you now know the 4th card, the 8th card, the 12th card, and so on. I’ll call these cards your ‘KNOWN’ cards.
Now, you could get someone to cut the deck anywhere.
Using the same cover we talked about last time (briefly spreading the deck as you say “it’s nearly impossible for me to count this by just glancing at them, so instead I’ll try to figure out how many cards there are by how much they weigh”) just find your closest KNOWN card.
Wherever they cut, one of your KNOWN cards will be either the bottom card, second from bottom, third from bottom, or fourth from bottom. You only need to spread fractionally to spot the index.
The minute you see that card, you’re set. You actually have an EASIER time than the method we previously discussed (which relied on an estimate.)
Since you know the exact number of the KNOWN card, you can just add on how many cards are on top of it, and the total gives you the total number of cards in the cut-off packet.
For example, let’s imagine when you were memorizing the cards the 3rd location in your memory palace contained the image of Dani DaOrtiz (AS) jumping (6C) into a cave (8D.)
That means that the 7th card of your 13 is the AS, the 8th card is the 6C, and the 9th card is the 8D.
However, since each card is actually positioned at multiples of 4 throughout the deck, the actual arrangement is:
AS is 28th (7 x 4 = 28), the 6C is 32nd (8 x 4 = 32) and the 8D is 36 (9 x 4 = 36.)
If someone was to cut the deck and the bottom card was the AS, you’d know that that packet has 28 cards in it. If the AS was second from the bottom, you’d know it had 29 cards in it. If the AS was three cards from the bottom, you’d know it had 30 cards in it. If the AS was four cards from the bottom, you’d know it had 31 cards in it.
If the AS was five cards from the bottom…you wouldn’t need to worry, since the new bottom card would be the 6C, and you’d instantly know there were 32 cards in the packet!
Alternatively, you could perform the glimpse on the top cards of the remaining packet, and then work backward from there. For example, if the 2nd card down in the remaining packet is the 8D (36), you know that the top card of that packet is the 35th card. Therefore, the packet that was cut off is 34 cards.
Anyhow, it’s a fun concept that is most definitely challenging but WILL enable you to perform this with an entirely shuffled, borrowed deck.
Which, I think, is deep enough for this week.
Let me know if I’m out of my mind on this one…
Benji
NOTE – you can make this easier on yourself by performing the dealing procedure early, and then using the remaining time between that effect and this one to run over your memory palace and make you have the cards memorized. There’s also nothing stopping you quickly spreading through the deck to ‘plug’ any holes in your memory.
Trust me. This is very attainable as long as you follow the ‘PAO’ formula we showed you.