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glowball
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Purpose: simple algorithmic stack that a few of my magician friends can do acaan with one fellow magician as a shill when I am not present (I normally let them use my Aronson deck with me as a shill and they do the trick for a waiter or waitress). The randomness of the deck is not super important thus Si Stebbins is okay. Don't need any built-in features. I strictly want a simple algorithmic stack (they won't memorize 52 cards). The ease of learning and using the algorithm outweighs just about everything else.

I ruled out about every kind of stack except for the Harding stack, the Quickerstack, the Karma stack(s), the Si Stebbins stack.

I finally decided to create my own variant of Si Stebbins (at the end of this post I give the reasons for rejecting the other three excellent algorithmic stacks mentioned).

Because I'm creating this stack in the reverse direction from the normal Stebbins I'm naming this Snibbets (which is Stebbins spelled backwards). Note that it uses shcd. Have the first four (the key cards) memorized (8S, 5H, 2C, QD).

Snibbets simplified rules for any card:
Any card is named.
1. Determine the number of steps from its key suit card to it (steps within the suit).
2. then multiply the steps by 4.
3. then add the pip points/lobes. The result is the position of the card. Very simple.

Snibbets simplified rules for any number:
Any number from 1 to 52 is named.
1. Divide the number by 4 to get a quotient and a remainder. Note if remainder is zero (means diamonds) steal one from the quotient to make the remainder equal 4.
2. Use the remainder to determine the suit shcd.
3. Add the quotient to the suit's key card value.
4. If the result is greater than 13 then subtract 13.
The result is the value of the card at the
position named by a spectator.

Mental mnemonic memory story for the four key cards: "Eight shovels were used to bury five hearts that were stopped by two clubs while a rich queen watched". If you visually picture this little story in your mind you should never forget these four key cards.

Actually the mental imagery of the order of these four cards is not important, what is important is simply to remember the names (value and suit) of these four important starting point key cards (the sequence of your mental image of these four cards is not important)

Always have the 8 of spades as the starting card (top card) of this stack.

Snibbets Acaan 8S (physical deck):
8S, 5H, 2C, QD, 9S, 6H, 3C, KD, 10S, 7H, 4C, AD, JS, 8H, 5C, 2D, QS, 9H, 6C, 3D, KS, 10H, 7C, 4D, AS, JH, 8C, 5D, 2S, QH, 9C, 6D, 3S, KH, 10C, 7D, 4S, AH, JC, 8D, 5S, 2H, QC, 9D, 6S, 3H, KC, 10D, 7S, 4H, AC, JD
That's pretty much it. A more detailed explanation will be in my next post.

Notes on why I decided against the three excellent algorithm stacks: Harding, Quickerstack, Karma stack for our situation.
First based on my rudimentary research these are the best easy to use algorithmic stacks that I found and they all have a much more random appearance than Si Stebbins, but Stebbins is public domain (free).

I think the Harding stack may be public domain also but I'm not sure. May as well be with all the YouTube videos about it. It's technique of visually thinking of a new deck order (A thru K of each suit) and digit switching is ingenious and fairly easy to do.

The Quickerstack and Karma stack methodology using a tetradistic system and pairs via Harry Riser is pretty easy and ingenious but those stacks are not free and for our limited use situation my friends are probably not going to shell out $60 for the Quickerstack nor $18 for the Karma stack (a more obscure site has Karma for $6.xx).

I was about to go with the Harding stack but decided to see if I could come up with some simple rules to use with the Si Stebbins as the base for my rules.

Note that when doing just a couple of acaan tricks for a waitress the cards are being dealt face up onto a single pile that the spectators are not going to have a good look at the various cards in context with each other thus the randomness objection to Si Stebbins greatly diminishes. Of course we would do some false shuffles and a quick face-up spread before doing the acaan(s).

Another small benefit to the Si Stebbins is that it's pretty easy to tell if a card(s) get out of order and thus the magician can quickly check the accuracy of the stack before performing and then correct the deck.

Thus I believe that my Snibbets stack (and it's rules) is the easiest for my magician friends to learn and use. We will see.
glowball
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Full explanation:
First I want to say that I strongly suspect that someone has done something like this 50 years ago and I just don't know about it, so I apologize If this is the case.

But I haven't seen any explanations on the internet or the magic Café on using a reverse Si Stebbins (this automatically puts the stack in shcd sequence which is a very good thing) to do acaan.

Also making any spade the first card of this stack greatly facilitates the calculations but of course you will have a different four key cards at the beginning.

I have seen several videos that use the bottom four cards and a somewhat awkward backwards count so I wanted something that started from the beginning and was a little more concrete with the instructions and used a set starting point with the stack with a spade as the first card.

Snibbets ACAAN (8S is first card):
Any Card:
Spectator names any card then the magician
mentally calculates the number of steps (within the suit) from that first suit key card (from the appropriate of these key cards 8S, 5H, 2C, QD) to the target card then multiply that number of steps by 4 and then add the pip/lobes points that are on the target card suit symbol. Simple.

Example 1: spectator names seven of clubs: magician remembers that 2C is the first club. then calculates 7 minus 2 (use 2 because 2C is clubs key card) equals 5 then multiply the 5 by 4 giving 20 then add the pip points of three (because clubs has three lobes) thus 20 plus 3 equals 23 as the final answer. Easy peasy.

Example 2: spectator names 6 of diamonds:
magician remembers that QD is the first diamond. Then one step to get to the king then six steps to get to the six thus one plus six equals seven steps then multiply the seven by four equals 28 then add the four points on the diamond giving 32 as the final answer.

Example 3: spectator names 9 of diamonds:
magician remembers that QD is the first diamond. Then one step to get from the queen to the king then 9 steps to get to the 9 thus one plus 9 equals 10 steps then multiply the 10 steps by four equals 40 then add the four points on the diamond giving 44 as the final answer.

Note about the number of steps within the suit. If the target card is higher in value than the starting top card then the calculation is super simple as in example 1.
If the target card value is lower than the starting top card then you must go across the king to ace barrier thus calculate the number of steps as in example 2.

Any number:
Spectator names any number from 1 to 52 then the magician calculates what card is at that number (here is how):
divide the number by 4 giving a quotient and a remainder. The remainder indicates the suit (If remainder is zero (diamonds) steal one away from the quotient to make the remainder 4 (reduces the quotient by one). If remainder is 1 then suit is spades, if remainder is 2 then suit is hearts, if remainder is 3 suit is clubs, if remainder is 4 suit is diamonds (the remainder tells us the pip points/lobes thus the suit ie: SHCD method).

Now that you know the suit then think of which one of the top four key cards has that same suit and then use the quotient to add forward from that starting key card to tell you the value of the target card. The number of steps forward is the value of the target card.

Example A: the number named is 27:
we always divide the number by the constant 4 and we get a quotient of 6 with a remainder of 3. The three tells us the card at position 27 is a club. We then remember the first club in the deck is the key card 2C. Therefore we then add the quotient 6 to the 2 to get 8. Thus the card is 8C. That was easy because we did not have to go across the king to the ace boundary.

Example B: the number named is 49:
we have to determine what card is at that position.
First we divide 49 by the constant 4 and we get the quotient 12 with a remainder of 1. The remainder 1 tells us the suit at the 49th position is a spade so we next remember the first spade in the deck which is the 8S key card so we then add the quotient 12 to the 8 and we get 20 (If this value is over 13 we must subtract 13 to get the card value down in range of 1 to 13). So 20 minus 13 is 7. Therefore the 49th card is the 7 of spades.

Example C: the number named is 44:
First we divide 44 by the constant 4 and we get the quotient 11 with a remainder of 0 (If remainder is zero (means diamonds) steal one from the quotient to make the remainder 4). The quotient is now 10. The remainder 4 tells us the suit at the 44th position is a diamond so we next remember the first diamond in the deck which is QD so we then add the quotient 10 to the Q (12) and we get 22 (If this value is over 13 we must subtract 13 to get the card value down in range of 1 to 13). So 22 minus 13 is 9. Therefore the 44th card is the 9 of diamonds.
glowball
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What if one of the four key cards is named:
well, the rules still work, you just have to be alert when you're using zero. Example suppose the spectator says five of hearts and you do the mental process and realize the number of steps from the key card to the target card is zero then you mentally multiply 0 by 4 and you get 0 and then you add the pip number of lobes (the 5H is hearts and hearts have two lobes) so you add the 0 plus 2 and you get the position 2 as the final location answer. Of course the simpler way in this situation is that when you immediately recognize that a key card is named you just immediately use the pip points/lobes to get the number.

What if one of the key card numbers is named IE 1 or 2 or 3 or 4? Again you could use the rules and it will work. Example let's say the spectator says the number 3. Then the calculation is: divide by the constant 4 and remember the quotient and the remainder. In this case the quotient will be zero and the remainder of course will be three therefore the three specifies the suit that has three points/lobes which is clubs. Then you remember the key card for clubs is 2C then you advance through the clubs a quotient number of times which in this case is zero which means you don't move at all thus the card is 2C. A much simpler way if 1 or 2 or 3 or 4 is the number named by the spectator is to just use that number to determine the suit and remember that suit's key card and name it.
hcs
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The seemingly simple idea to reverse the Si Stebbins System has already been described by Ernest Hammond [1] in 1919. He used – as I do – the suit order shocked from bottom to top. (Remember, the suit values are ♠ – 1; ♥ – 2; ♣ – 3; ♦ – 4 or 0.)

Hammond already called his reverse stack SNIBBETS!

However, the fact that the reversal of the system represents a significant contribution to the Si Stebbins System has never been properly appreciated. E.g., Hammond has neither been mentioned in the historical depictions from Joval [2,3] nor Tamariz [4].

Snibbets being hardly known despite its age is an oddity since its application offers many advantages...

References
[1] Ernest Hammond: „The magic of to-morrow” (with H.C. Mole and E.C.P. Medrington), Liverpool, 1919
[2] Martin Joyal: „The Six-Hour Memorized Deck, “1997
[3] Martin Joyal: www.joyalstack.com
[4] Juan Tamariz: „Sinfonie in Mnemo-Dur,” 2005 Smile
glowball
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Hcs, I had never heard this. Thank you, thank you, thank you! This is great to know! So I'm going to refer to this stack as the Hammond Snibbets stack.

Do you know if he had written up the rules to do the acaan calculations?
glowball
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Does his book "The Magic of Tomorrow" talk about Snibbets?

I just want to be sure before I buy the book.
Thanks for your input.
glowball
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A normal Si Stebbins is CHaSeD from top to bottom. Just a slight correction to hcs wording. When you reverse the order of a normal Si Stebbins deck it becomes SHoCeD from top to bottom.

This really surprised me when I was experimenting (surprised that by dealing the cards face down onto a pile changes the deck from CHaSeD to SHoCeD and vice versa).
hcs
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Quote:
On Apr 7, 2022, glowball wrote:
Hcs, I had never heard this. Thank you, thank you, thank you! This is great to know! So I'm going to refer to this stack as the Hammond Snibbets stack.

Do you know if he had written up the rules to do the acaan calculations?
You got an email.
... and, of course he did!
ddyment
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To answer the question posed in this topic's title: this is not the easiest calculation. Considerably easier are the calculations for my SnapStack and Michael Weber's very similar Card Kindergarten. Of course, both of these are even less random in appearance than the Snibbets approach (but they are easier to calculate than new deck order).

That's always the tradeoff, of course: the less random the stack appearance, the more simple the value/position calculation. Snibbets is just one more point on the spectrum.

My Q Stack, incidentally, is both more easily calculated and more random in appearance than the Harding stack (or any other algorithmic stack, to my knowledge).
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Claudio
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Displaying the stack this way makes its tetradistic nature clearer:

8S, 5H, 2C, QD, 9S, 6H, 3C, KD, 10S, 7H, 4C, AD, JS
8H, 5C, 2D, QS, 9H, 6C, 3D, KS, 10H, 7C, 4D, AS, JH
8C, 5D, 2S, QH, 9C, 6D, 3S, KH, 10C, 7D, 4S, AH, JC
8D, 5S, 2H, QC, 9D, 6S, 3H, KC, 10D, 7S, 4H, AC, JD

Being aware of it is important as one efficient strategy, applicable to regular SS and to Snibbets, is to memorize the first 13 cards to easily calculate any card from that basis. I know of a couple of magi who have taken that route to learn Si Stebbins and they are very efficient a it. Furthermore, as they can easily set-up the first 13 cards from a shuffled deck, they use that “innocent” sequence to great effect.

But I suspect most people will be happy with the simple calculation method described by Glowball. With a bit of thinking you can simplify some of the maths. For instance, if you take these 4 cards 10S, 7H, 4C, AD as the start of the stack (from top to bottom), which only involves moving the top eight cards to the bottom of the deck, you end up with:

10S, 7H, 4C, AD, JS, 8H, 5C, 2D, QS, 9H, 6C, 3D, KS
10H, 7C, 4D, AS, JH, 8C, 5D, 2S, QH, 9C, 6D, 3S, KH
10C, 7D, 4S, AH, JC, 8D, 5S, 2H, QC, 9D, 6S, 3H, KC
10D, 7S, 4H, AC, JD, 8S, 5H, 2C, QD, 9S, 6H, 3C, KD

Or again, ordered by quadruplets (where 'X' means '10' for formatting purpose). The numeric value in front of each quadruplet is the position of the first card of the quadruplet. It’s only here for illustration purpose.

01. XS 7H 4C AD
05. JS 8H 5C 2D
09. QS 9H 6C 3D
13. KS XH 7C 4D
17. AS JH 8C 5D
21. 2S QH 9C 6D
25. 3S KH XC 7D
29. 4S AH JC 8D
33. 5S 2H QC 9D
37. 6S 3H KC XD
41. 7S 4H AC JD
45. 8S 5H 2C QD
49. 9S 6H 3C KD

With this, when a Diamond is called out, you only have to multiply its value by 4 to obtain its position. Ex. 8D -> 8 x 4 = 32.

You could leave the other calculation unchanged, but here’s something that works well for me for Clubs:
1. Subtract 3 from the value called out
2. Multiply by 4
3. subtract 1 from the result.

The advantage of this formula is that you don’t have to first refer to the base number to subtract.
Also, if you don’t mind dealing with negative numbers, the method works this way when the card value is less than 3.
Add the negative value to 13. Multiply the result by 4 and subtract 1.

Example, 2C:

a. 2 – 3 = -1
b. 13 – 1 = 12
c. 12 * 4 = 48
d. 48 -1 = 47.

In real life, the mental process is slightly different: as soon as you hit a negative number you only deal with its absolute value that you subtract from 13.

Another strategy that I’ve used successfully with the regular SS is to use key cards which don’t necessarily match the physical deck. For example, instead of using 10S, 7H, 4C, AD, you could use 10S, 7H, 3C, AD. Having the 3C as base for Clubs, would simplify some calculations. The general rule to obtain the position of a Clubs consists in subtracting the base value (4, because of 4C) from the card value and multiplying the result by 4 and finally adding 3.

In using 3C as base, you could simplify this way: Subtract 3 (because of 3C) from the card value, multiply by 4 and subtract 1.

Example, 7C is called out:

With 4C (physical key card) as base value:
7 – 4 = 3;
3 x 4 = 12;
12 + 3 = 15;

Witch 3C (decoupled key card) as base value:
7 – 3 = 4;
4 x 4 = 16;
16 – 1 = 15;

With a bit of digging, it’s possible to simplify the maths. The only con I see with the methods I described (maybe some of them original) is that you’ll use different rules, albeit very simple, instead of some uniform rule for every suit. Whether it’s worth it will depends on individual tastes.
glowball
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To all, Thanks for your contributions. This has been more fun than Sudoku!
landmark
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For the original purpose of the thread--to teach those who want a minimum amount of work to do an ACAAN--I would suggest Barrie Richardson's approach in Act Two. It is a stack of the red suits only with a very simple formula; the black cards are distributed at random to the remaining positions. This results in a very random looking stack, yet easily memorized card to position and position to card. The only compromise is an initial dead easy equivoque of the red cards.
ddyment
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As also mentioned in my online introduction to full-deck stacks, Lewis Jones has published a half-deck stack as well.

Unlike Barrie's (with its focus on color), Lewis' uses the odd/even differentiator.
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glowball
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Act Two sounds very clever.

"This results in a very random looking stack, yet easily memorized card to position and position to card."

To equivoque card to position I understand, but how can you equivoque position to card?

In other words how could you get a spectator to say a number that would for sure be on a red card?
Claudio
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With this specific Richardson's stack, you don't memorize the stack. You use a very simple algorithm to calculate the position of the red cards. And, as already mentioned, the performer starts with a classic magician's choice about which card colour to use.
glowball
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When using Aronson we have the spectator name any card and then the shill (me) names the position.

Then we do the trick a second time where the spectator names the position and the shill names the card.

The same two effects can be done with a Snibbets deck or a Harding stack deck but I don't understand how the second trick can be done with the Richardson stack?

Is the spectator aware of which position numbers are valid for the red cards?
glowball
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OZ MD (Hans-Christian) has ebook Snibbets info available at:

https://www.lybrary.com/snibbets-p-925182.html
landmark
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Quote:
On Apr 12, 2022, glowball wrote:
Act Two sounds very clever.

"This results in a very random looking stack, yet easily memorized card to position and position to card."

To equivoque card to position I understand, but how can you equivoque position to card?

In other words how could you get a spectator to say a number that would for sure be on a red card?


One basic way would be to begin: "Let's play a game of imagination-I've got 52 cards in this card case--which should I take out, the red or the black?"

If they say Red: "Okay imagine I'm taking out the red cards. Name one.
If they say Black: Okay imagine I take out the Black cards. Now name one of the remaining cards in the case.

At this point they can name a number (there 's a restricted range, but it's always reasonable) and the magician can count to the card--in fact the spec can count the last few cards. Richardson's idea here is very clever and essentially sleight free.
Claudio
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My opinion is the Richardson Stack, Lazy Magician's Memorized Deck, is ill-suited for the effect Glowball has in mind, especially when going from number to card. There are ranges of numbers which won't work and it won't be easy to fix: sleight of hand will probably be required.

You're much better off sticking with Snibbets or another full stack for this effect as it works every time without complication.

I have the whole series of Rusduck's The Cardiste and and I've read again #4 which deals with Snibbets. Very interesting.
glowball
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To landmark: Thanks for putting me on to the Barrie Richardson's Lazy Man's stack and to Claudio thanks for pointing out that it only goes from card to number but not number to card.

This led me to develop the LAZY-B1 stack which goes both ways:

https://themagiccafe.com/forums/viewtopi......orum=205

It is simply for the red cards to multiply by 4 and then if diamonds add one, if hearts subtract one.

This is so rudimentary and simple surely this has been developed and published somewhere before.
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