WAIT! I wasn’t done with Effect #1 yet, take me back…
Take Me to Effect Three
The Memorized Deck Starter Pack – Effect #2
“Yo Doc, this is heavy…”
I’ve always found ‘weighing the cards’ to be among the most boring effects you can do with a deck of cards.
When they’ve seen us make cards appear at freely chosen numbers, inside sealed envelopes, and after spelling THEIR names…do audiences REALLY care that we know how many cards are in a given packet?
I just never found it very impressive in its most basic form.
Magician: “Cut the deck and place it in my hand.”
The spectator does as the magician asks.
Magician: “15 cards.”
Magician counts the cards to show it’s 15.
…polite applause.
If anything…it feels like more of a ‘puzzle’ than a miracle.
That’s due in part to the way the cards are handled. Since most versions of this routine rely on a stacked deck, the cards can’t be shuffled by the audience.
In fact, that’s probably one of the biggest things going ‘against’ this routine.
My goal this week is to figure out two things:
- How to perform ‘weighing the cards’ with a deck that the audience feels they shuffled
- How to perform ‘weighing the cards’ with a deck that the audience really DID shuffle
(as you’ll see, I do this by going progressively ‘deeper’ each time, even when it feels we’ve reached the logical conclusion of where we can take this.)
Once we have those factors, we’re well on the way to making this routine something more powerful. Of course, the final piece of the puzzle is if YOU can make it compelling by the way you present it.
That’s the one bit I can’t figure out for you, but with the methods I’m going to show you, you should be set for a pretty compelling presentation.
This week, I’m going to attempt to get progressively ‘deeper’ into what I hope will be a more fooling version of this routine.
You can read my first version below…
Back to Basics:
First, we need some common ground.
How does this routine usually work?
As I mentioned, using a memorized deck makes this task VERY easy.
Since each card can be thought of as its corresponding stack number, all we need to do is glimpse the bottom card of the cut-off portion and we’ll instantly know the card.
For example, let’s say we’re in Mnemonica stack.
Someone cuts the deck. We glimpse the bottom card and see the 3S. We know that 3S is 21, so they must have 21 cards in their packet.
That’s really all there is to the basic method.
But let’s go deeper…
How do we make it FEEL like they shuffled the deck?
Here’s the first idea I had:
Start in stack. Break the deck exactly in half (by feel, or just hold a break at KD) and hand one half to your spectator.
Let them shuffle that half while you ‘do the same’.
In fact, you false shuffle.
Then swap halves and tell them to ‘cut the deck’ while you do the same.
Then swap back. It’s easy enough to undo their cut with a couple of estimation cuts and bring the stack back to starting order.
Now you can reiterate that ‘you shuffled and cut the deck’ – which is somewhat true, but not the full picture.
Now you’re going to make the whole thing feel even MORE shuffled by doing an in-faro shuffle.
THIS IS THE HARD MOVE WE TOLD YOU ABOUT ON THE SALES PAGE!
If you don’t know how to do the faro shuffle, there’s plenty of tutorials on YouTube,
Here are a couple of links to detailed tutorials on the faro shuffle:
https://www.youtube.com/watch?v=2TyA0CInQH4
https://www.youtube.com/watch?v=134LJpwU5lE
(I will also say that I’ve found the most reliable cards for hitting perfect faros are the Richard Turner Gold Seal decks.)
Remember, the important part is that the in-faro will means that the initial top card is now the card second from the top, and the card on the bottom portion of the top half is now the bottom card.
In fact, what this faro is going to do is place every card of your stack in a position DOUBLE the initial stack number.
Don’t worry, that’s not as complicated as it sounds.
It just means that the 4C (card 1) is now going to be second from top (aka card 2), the 2H (card 2) is now going to be the 4th card (card 4), all the way down to the KD (card 26) which will now be the 52nd card.
Now, if you cut the deck, 50% of the time the card on the bottom of the bottom is one of your stack (26/52 = 50%).
If this is the case, you just take that stack number and 2X it to know how many cards are in that packet. For example, if the card on the bottom was the 6D (6) we know there are 12 cards in the packet.
What if the card on the bottom is not one from your stack?
In this case, all we need to do is glimpse the top card of the remainder of the deck, and subtract 1 from that result.
(that’s because, if the bottom card of the cut-off packet isn’t one of your stack, the top card of the remainder MUST be since the deck alternates between stack/non stack.)
For example, if the top card of the remainder of the deck is the 5S, we take the stack number (16), 2X it (32) and subtract 1 (31.) That final number is the number of cards in the packet.
And that’s all there is to it!
Ok…one other thing.
Remember to note how the spectator counts the cards. If they reverse the order as they count, you’ll need to reverse their reversal to bring your self back to initial order.
(that’s a lot of reversals…makes me feel like a driving instructor).
To return to stack, just deal the cards into two piles – this will leave you with a half stack.
So there we are: a simple way to make it FEEL like the cards were shuffled, but still retain the initial effect.
NOTE: I realized after typing this out that Juan Tamariz also discusses this idea on page 199 of Mnemonica.
But can we go deeper?
I think so.
Let’s do it!
“We need to go deeper…”
Here’s an idea that lets you let the audience genuinely shuffle all of the cards in the deck.
It uses a quarter-stack.
(Or in other words, 13 cards.)
My initial idea was inspired by a very clever shuffle procedure in Pit Hartling’s ‘Unforgettable’ (in his book Card Fictions).
The idea is to hand out packets of cards to each member of the audience but keep the first 13 for yourself. They shuffle, you cut (but eventually cut card #1 of the stack back to the top.)
I was then going to in-faro those cards into the first 26, then in-faro those 26 into the remaining half…then perform mathematical calculations (i.e each stack number is now in a position 4X that original number) based on the position of whichever of those cards ended up closest to the bottom of the cut-off packet to arrive at the number of cards.
However, I have a suspicion I’m not the first person to have thought of this.
(Woody Aragon has a routine called ‘The Human Scale’ that I imagine may be similar to this.)
So I decided to throw caution to the wind and go even ‘deeper.’
The idea I landed on is an interesting one.
Some will like it.
Some won’t.
And when it’s all said and done, it takes a certain ‘familiarity’ with the cards that only time can give you.
But done right, it’s very impressive.
Here’s what goes on:
You start in stack.
(Or, if you’d rather, you start with the first 13 cards of the stack in place and the rest of the deck shuffled.)
You remove the first 13 cards of the deck (easily done if you hold a break under them beforehand – if not just thumb count them off) and hand them to your first spectator to shuffle.
You then split up the remainder of the deck into various packets and hand them out to be shuffled further. It doesn’t matter how many cards are in each of these packets.
This way, the entire deck really has been shuffled by the audience. However, there’s still a factor you have control over—the 13 cards in the first packet you handed out are still the first 13 cards of your stack.
However, the ‘image’ is one of complete chaos.
It’s important to note: those 13 cards won’t be in stack ORDER. But they WILL be the same cards. It doesn’t matter how much they shuffle the cards, those cards will always be the first 13 cards of your stack.
Here’s what comes next:
You take the first packet back from the first spectator, then the second packet from the second spectator.
Now you’re going to in-faro the first packet into the second, perfectly weaving the cards together.
Either way, make sure it’s an IN-faro.
In-faro means that the bottom card of the first packet becomes the bottom card of the combined packets, and the top card of the second packet becomes the top card of the combined packets.
Here’s some more info on the difference between an IN-faro and an OUT-faro:
https://www.magicalapparatus.com/card-techniques/in-and-out-faro-shuffles.html
You’ll be left with what I’ll call ‘Packet 1’.
Now take back the third and fourth packets, combining them into what I’ll call ‘Packet 2.’
You’re now going to in-faro Packet 1 with Packet 2. Again, make sure the bottom card of Packet 1 becomes the new bottom card and the top card of Packet 2 is the new top card.
Once you’ve done this (and done at speed just looks like you’re further shuffling the cards as you retrieve them) you’ve placed the first 13 cards of your stack at positions 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48 and 52.
(again, the magician I first saw make excellent use of this technique is Pit Hartling in Card Fictions – although my addition is the idea that the cards themselves don’t need to be in stack order for this particular effect.)
However, these cards aren’t in stack order. You could have the 4C at 24, the 5H at 4 and the QH at 36.
But that’s not the important part. The important part is that every 4th card is one that you’ll be able to instantly recognize as belonging to the first 13 cards of your stack.
Why is that useful?
Here’s why:
When the spectator cuts the deck, I’m going to estimate how many cards he cut.
I’m then going to take the cards, spread the top 4, and ‘check’ my estimate.
How do you justify this?
You could briefly spread the cards as you say:
“One way of figuring out how many cards you cut off could be to try to count them REALLY fast just by looking at them. But I’m not going to do that…”
I’m just looking to see one of my 13 cards. It doesn’t matter which one – I just want to see one.
When I see one (and there’s always going to be one within the top 4) I just look at how many cards deep it is.
Now, I’m going to adjust my initial estimate based on this new information.
I know this sounds ‘vague’, so let’s look at an example:
For example, let’s say I reckon they cut about half the deck (26 cards or so) off.
I already know that I have cards from my initial stack in positions 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, and 52.
In this case, I think they’ve cut about half the deck (26.) If that’s the case, one of my stack cards should be the third card down (the card in position 24).
Let’s say I spread the cards briefly and spot the 3D fourth from the top.
In this case, I’m going to guess the number ‘27’. That’s because we saw the 3D, which we believe is more likely to be in the 24th position than 20 or 28. We then take 24 and add 3 (there are 3 cards on top of the 3D) to get 27.
On the other hand, if the packet looked to be slightly under half, I’d guess that the 3D was position 20, which would mean you were holding 23 cards (20 + the three cards on top of 3D.)
If it looked to be over half, I’d guess the 3D to be in position 28. When we add the three cards on top, we get 31.
If you REALLY wanted to be sure, you could have run through all of those estimates in your mind before making a call. Since you were left with the numbers 23, 27 and 31, we should be able to fairly easily see which one of those numbers looks the most likely to be the right one and call out that one.
You can apply the same process wherever they cut – estimate how many, and then estimate which position your stack card should be, then adjust your number based on where it actually is.
As you may be starting to realize, for this method to work optimally, you need to have a pretty good feel for ‘estimating’ how many cards are in a cut off packet. If you’ve been working with the mem deck for a while, you probably will have this. If not, don’t worry—it comes over time.
So, there’s a method for performing this that really makes it FEEL like the whole deck was utterly mixed beforehand.
It’s one you’ll need to practice and hone your ‘feel’ for, but once you do, it’s very clean.
That said, I think I prefer the method we’ll talk about next (just for its sheer ‘cojones’)
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.
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.