Howdy, Inner Circle!
Here’s a fun concept I’ve played with a lot but never really took time to write out.
I first had the idea while reading ‘Bound to Please’ by Simon Aronson.
In this groundbreaking book, he introduced a very clever concept; an entirely new way of categorizing the cards in a deck.
See, playing cards have a number of ‘external’ properties that are visible to everyone—their number, suit, color, whether they’re odd or even, face card vs number card.
Many powerful tricks are based on arranging the deck into a predefined organization based on one of the above properties. For example, Juan Tamariz’ ‘Neither Blind Nor Stupid’ works by alternating the colors of cards in the deck.
Such tricks can be very fooling. But when you work in the memorized deck, they take on a whole new level.
That’s because the memorized deck introduces a number of NEW and ‘internal’ properties that are ONLY known by you.
In a memorized stack, each card also has an ‘internal property’—its stack number.
However, this property is only known to YOU, not the audience.
And since numbers are, well, numbers—there’s an awful lot of cool things you can do with them.
Over the next few days, I’ll share a couple of cool ideas that this generated for me.
For this demonstration, I’ll be using the Tamariz stack, but this effect is stack-independent (i.e it works with any stack.)
If you follow this with cards in hand, it should be easy enough to follow. If not, don’t say I didn’t warn you!
Here’s the first one:
Since the stack numbers are only known to us, we could sort the deck into ‘even stack numbers’ and ‘odd stack numbers’ and the audience wouldn’t see any kind of distinction.
So we could start out with the deck divided into two halves—all of the even stack numbers in one half, and all of the odd stack numbers in the other.
For example, if we were using the Tamariz stack, one half would be made up of the 4C (#1), 7D (#3), 4H (#5) and so on, and the other would be the 2H (#2), 3C (#4), 6D (#6) etc.
Once you’ve got the deck set up like this, you can move into a shuffle procedure that should fool even other magicians.
Hold a break at the separation between the two halves. Give one of the halves to one spectator and one of the halves to the other. Get them to each shuffle that half.
Since the only thing we’re concerned about is that one half contains odd stack numbers and the other contains even numbers, it doesn’t matter how many times they shuffle!
All they’re doing is shuffling odd cards with other odd cards, and even cards with other even cards. (since one pile is only odd cards, it doesn’t matter how much they shuffle, they’re only ever shuffling odd cards! Same goes for the other pile.)
Now tell them you’ll “shuffle both those halves together.”
Now, faro the cards together using the standard handling (or the method we share in Module 4 of Skyscraper).
You’ll be left in the following situation:
Every card of the deck should alternate between even stack number and odd stack number (because of the perfect faro.)
The two halves have been shuffled individually, and it looks like they’ve been shuffled together.
Now you can spread the deck and let someone remove a card freely. Once they take it, square the cards, and tell them to look at their card.
Now have them place it in the deck somewhere. You can show them that you don’t maintain any breaks of holds.
In fact, you can set the deck cleanly down on the table.
Believe it or not, you’re now set up to reveal their card.
If you try this with red and black first, you’ll see exactly what’s going on.
Separate the deck into one half containing all reds, and the other half containing all blacks. You can then have people separately shuffle those halves without actually mixing the red and black together.
Now, when you faro, you should be left with a deck that alternates between red and black. If you remove any card – red or black – and place it somewhere else in the spread, you’ll now be left with the following two situations:
At one point, there’ll be two reds together, and somewhere else there’ll be two blacks together.
Their card is one of those 4.
However, 4 is still quite a few cards.
How do we narrow that down to two?
After you faro, see if the bottom card is red or black.
If the bottom card is red, the top card is black (since it alternates between red and black every card.)
Therefore, every odd-numbered card is black (1st card, 3rd card, etc.), and every even-numbered card is red (2nd, 4th, 6th, etc.)
Then, as we spread the cards, we just count how many cards we’ve spread.
Let’s say they pick the 12th card down. In this case, it’s an even-numbered card, so we know it’s red.
Now when they move it, we know we can ignore the two black cards together – we just need to worry about the two red cards together since we know that’s the color of their card.
All we need to ask is ONE question to know which card it is (i.e “you’re thinking of a red card…hearts? No, you’re right. It’s diamonds. In that case…your card is X” or ‘you’re thinking of a red card…hearts? Yes? Great. In fact, I think it’s the…X”)
When we then transition to using our stack version, it becomes even more deceptive.
NOTE: so things don’t get too confusing, I’ll refer to the odd stack cards (i.e the cards whose stack number is odd) as ‘odd stack number’ rather than just ‘odd.’
If there’s an even stack number on the bottom, we know that the top card is an odd stack number. In this case, every odd-numbered card is an odd stack number. (1st card of the deck is odd stack number, so is the 3rd, so it the 5th, etc.)
If there’s an odd stack number on the bottom, we know the top card is an even stack number. In this case, we just remember that everything is the OPPOSITE. Every even-numbered card is an odd stack number (2nd card, 4th card, 6th card, etc.) and every odd-numbered card is an even stack number (1st card, 3rd card, 5th card, etc.)
For example. We might have the 3D on the bottom, 9S on top, QD second, 8C third, etc.
The 3D is even, which tells me that the top card is an odd stack number.
The top card is the 9S, which is the 9th card — so I was right.
Since the deck now perfectly alternates between odd and even stack numbers, I know that the second card is an even stack number. It’s the QD, which is 46th, so I was right.
This means the third card must be odd. It’s the 8C, which is 33rd in stack, so I was right.
And so on.
If, however, the 9S was on the bottom, I’d know that the top card is even stack number (the QD is 46, so we were right.)
And so on.
Once we know this, we can determine if their card is odd or even based on how many cards we spread before they select one.
If they are a particularly difficult spectator and reach deep into the deck beyond your count, hold a break in the place they removed the card from and count to that break while they look at the card.
If you fail this, don’t worry. In the worst-case scenario, you’ll have to ‘fish’ their card from 4 possible cards.
But if you do, you can narrow it down to 2 – allowing you to guess their card after just one question.
Let’s look at an example:
They shuffle the halves. You faro them together. You glimpse the bottom card and see that it’s an odd stack number (for example, the 3 of Spades – #21).
We know that the top card is an even stack number. Therefore, every card in an odd position is an even stack number (for example, 1st card is the 2H, 3rd card is the JD, 5th card is the 10H, etc.) and every card in an even position is an odd stack number (for example, 2nd card is JS, 4th card is AS, 6th card is AC.)
Now, let’s say they took the card that we counted as #16 when we spread through the deck.
Since the position (16) is even, their card is an odd stack number.
Now, we get them to place the card back in the deck.
It’s important we get them to place the card back somewhere DIFFERENT to where they initially removed it. Putting it back in the same position is the only thing that would make this not work.
They put it back.
Now we can cut the deck a couple times (don’t worry, the order will stay the same) and spread through the cards with the faces toward us.
Let’s imagine at some point in the stack we see these two anomalies:
‘QH, 7H, KC’ and ‘2D, 4S, 9D’
In the first example, we have QH (stack number 11) followed by 7H (stack number 41) and then KC (stack number 18.)
Notice anything?
Yep. Two odd cards next to each other.
Notice anything?
Correct again, buckaroo.
Two even cards together.
Now, because we were able to initially determine that their card was an odd stack number (see above) we know that their card is either the QH or the 7H.
At this point, we might say:
“I think your card is a red card. Probably a hearts. Is it a picture card?”
If the answer is yes, you know that their card is the QH. If no, you know their card is the 7H.
Either way, you can divine their card with just one key question – to help you decide between suit, color, picture or number, high number or low number, etc.
And 50% of the time you’ll get that question right.
Either way, it’s a stunning display of mindreading and the shuffle procedure is very deceptive.
Other comments:
If you start in the stack and do 7 out faros, you’ll arrive in a position naturally where all of the even cards are in one half and all of the odd cards are in the other half.
Alternatively, you could prepare the deck by giving your stack 5 or 6 faros before you perform. Then when it comes to performance, you just need to give the deck 1 or 2 faros to bring yourself to faro 7.
So to the audience, it looks like you shuffle, give them half the deck each to shuffle, then shuffle those shuffled halves together.
Quite the deception!
You could, if you wish, have the two spectators swap halves so they each feel they’ve shuffled the whole deck. Again, as long as the two halves are isolated it really doesn’t matter.
Alright, that’s enough for today.
Next week, I’ll show you an effect that looks like this…
You hand out the deck, letting various spectators shuffle it. You then retrieve the shuffled packets and shuffle them together further.
You then invite two audience members up to help you.
The first spectator cuts the shuffled deck in half, giving the other half to the second spectator. They shuffle their packets and are instructed to place them in their pockets and keep an eye on them, ensuring you can’t do anything sneaky!
Next, you remove another deck from your pocket. This deck again is shuffled by the audience.
You act as if presenting ‘Out Of This World’ and get them to try to sort the cards by red and black by ‘feel.’
When you turn over the cards at the end, it appears they messed up.
However, you reveal that in fact, each spectator dealt themselves the exact same cards as the ones previously placed in their pocket (from a previously shuffled deck!)
It seems to be an impossible coincidence!
And best yet, you end STILL IN STACK!
Speak soon,
Benji