# A 2 Card System (updtd)

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#1 4 January, 2017 - 19:45
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#### A 2 Card System (updtd)

Hi, I am Mark Joseph, 24 and I am from the Philippines. I just want to share my 2 card system. I use the same idea as the two block system where a two pairs of cards could have the same image. The idea of my system is converting two cards into a 3 digit number since some of my images are not phonetically coded like 025, which is for me is a gift, 014 a heart, 901 striker from Mortal Kombat etc.

The following is a 2 card system with a built in 1 card system feature

Suit Values
♠=1 ♥=2 ♣=3 ♦=4

Suit Combination Rules
For the 1st suit in the suit combination:
♣=5
♥ would be the red version of ♠
♦ would be the red version of ♣

For the Face Card (J, Q)
2nd digit = suit combination
3rd digit = index(#) of the 2nd card.

For the Suit Combination
2nd digit = index(#) of the 1st card
3rd digit = index(#) of the 2nd card.

First Digit
Q = 0 (because it is like zero)
J = 1 (because it is like a “t”)
(red ver.)
♠♠ ( ♥♠ ) = 2 (because 1+1=2)
♠♥ ( ♥♥ ) = 3 (because 1+2=3)
♠♣ ( ♥♣ ) = 4
♠♦ ( ♥♦ ) = 5
♣♠ ( ♦♠ ) = 6 (because ♣ is in the 1st position it is 5; 5+1=6)
♣♥ ( ♦♥ ) = 7
♣♣ ( ♦♣ ) = 8
♣♦ ( ♦♦ ) = 9

For the K♠ + 2nd card, the rule is subtracting the suit values to get the first 2 digits and the index (#) of the 2nd card would be the 3rd digit.
(red ver.)
K♠ #♠ (K♥ #♠) = 00# (because 1-1 = 0)
K♠ #♥ (K♥ #♥) = 01# (because 2-1 = 1)
K♠ #♣ (K♥ #♣) = 10# (because 3-1 = 2, converting to binary digits that would be 10)
K♠ #♦ (K♥ #♦) = 11# (because 4-1 = 3, converting to binary digits that would be 11)

Examples:
4♠ 2♠ = 242 = 4♥ 2♠
5♣ 8♥ = 758 = 5♦ 8♥
J♣ 3♦ =193 = J♦ 3♦
K♠ 9♦ = 119 = K♥ 9♦
K♠ 5♥ = 015 = K♥5♥
Q♠ 4♦ = 054 = Q♥ 4♦

The 352 images outside the 1000 digit system would be the J, Q, K in the 2nd card and K♣.
(We are free to use whatever method we like for the 352 images).
# (suit) J (suit) = 100 images
# (suit) Q(suit) = 100 images
# (suit) K(suit) = 100 images
a total of 300 images
K♣ #♠ (K♦ #♠) = 13 images
K♣ #♥ (K♦ #♥) = 13 images
K♣ #♣ (K♦ #♣) = 13 images
K♣ #♦ (K♦ #♦) = 13 images
a total of 52 images

Mine is:

The J,Q and K in the 2nd card:
J = “t,d” sound
Q = “s,c,z” sound
K = “k,g” sound

I use the ben system for my 1000 digit images. So to avoid having images that have the same phonetic code, I use the major system for the face cards.

Example:
811 = “vat” so for me avatar
80J = “f s t” so for me fast and furious; Paul Walker

For the K♣ + 2nd card:
I just use the 2nd card as the clue to create the image or you can use your existing 1 card system.

Example:
K♣ 10♣ (K♦ 10♣) I convert the 2nd card to number so that would be 30, so for me it would be Stephen Curry.

The one disadvantage of 2 block method is sometimes having 8 (or more) images in a locus, which is so very hard when in speed cards discipline , but I have this idea that just came to my mind.

I just actually put a maximum of 4 images in a location. So basically if I have:
1 image in a locus = I know automatically that it is a Red First.
2 images in a locus = that is automatically a 1 Black First + 1 Red First.
3 images in a locus = that is 2 Black First + 1 Red First.
4 images in a locus = that is either a 3 Black First + 1 Red First or 4 Black First.

So in short, if I have 4 images in a locus the last image there is either a black first or red first. To know that if it is a black first or red first, I rely on the 2nd deck in the recall (i.e. speed cards).

Example:
So if I will be going to memorize this:

4♠2♠- 5♣8♥- J♣3♦- K♠9♦(tablet) - Q♠ 5♥- 7♥ 6♠- K♥ 4♣(picture)

It will be :
Locus 1: 4♠2♠- 5♣8♥- J♣3♦- K♠9♦(Tablet)
Locus 2: Q♠ 5♥- 7♥ 6♠
Locus 3: K♥ 4♣(Picture)

When I do the recall :
Locus 1: 4♠2♠- 5♣8♥- J♣3♦- K♥9♦(Tablet)

As you can see, even though I made a mistake there of putting the Red Version of the Tablet, I won't have to worry becase when I get to the 3rd locus and I know that my image there is Picture and my Picture is K♥ 4♣

Locus 2: Q♠ 5♥- 7♥ 6♠
Locus 3: K♥ 4♣ (Picture)
So that's how will I know that the Tablet in Locus 1 should be in the Black Version: K♠9♦ not K♥9♦.

With this trick, it doesn't worry me if there are 10 consecutive Black First in a deck.
For the Multi Deck Discipline I continue the story from deck 1 to 2. Any thoughts?

Thanks, I hope this will help.

10 January, 2017 - 23:47
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Joined: 6 years 1 week ago

Hi Mark Joseph,
So are you saying you haven't used this system to memorise a deck of cards yet?

Jimbo

11 January, 2017 - 20:04
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Joined: 10 months 2 weeks ago

Hello. I have used it. But i am not yet fast.

24 February, 2017 - 00:54
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Joined: 9 months 2 weeks ago

hmmm. sounds like the 52 images from the Kclubs+2nd card can be used also as an image for a single card and is useful for those surprise event in xmt like the card image event.
nice work.

4 March, 2017 - 18:00
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Joined: 10 months 2 weeks ago

QUOTE:
hmmm. sounds like the 52 images from the Kclubs+2nd card can be used also as an image for a single card and is useful for those surprise event in xmt like the card image event.
nice work.

I guess so.

20 June, 2017 - 20:49
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Joined: 5 months 2 days ago

Thanks for sharing. Definitely helps since I use Category System for my Numbers ;)

20 June, 2017 - 22:19
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Joined: 5 months 2 days ago

Hi Mark Joseph, I'm Asiah (18 years old), I'm still the beginner of this memory technique. I just want to ask, how much time did you take to train your brain since you start ? What's the first thing you trying to memorize ?

21 June, 2017 - 16:54
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Joined: 10 months 2 weeks ago

Hi Asiah.
I just want to ask, how much time did you take to train your brain since you start ?
I just started practicing in Sep. 2016 so, I am just still a beginner like you.

What's the first thing you trying to memorize ?
It was 40 words.

28 September, 2017 - 19:25
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Joined: 5 months 2 days ago

Hi, I have experimented your system and I have found some problems.

Case 1:
Let's say for example for this string of cards in a deck.
4♠2♠- 5♣8♥- J♣3♦- K♠9♦ ............................
Q♠ 5♥- 7♥ 6♠ - K♥ 4♣(last pair).

When we do the memorization using your idea of putting a maximum of 4 images, it would be something like this:
Location 1 : 4♠2♠- 5♣8♥- J♣3♦- K♠ 9♦
..........
Location # : Q♠ 5♥- 7♥ 6♠
Location # : K♥ 4♣(last pair).

But when in the recall:
Location 1 : 4♠2♠- 5♣8♥- J♣3♦- K♥ 9♦
..........
Location # : Q♠ 5♥- 7♥ 6♠
Location # : K♠ 4♣(last pair).

The problem is that in recall the last pair there can be either a black first or red first thus it can use either K♠ or K♥. So there would be no way distinguish which is which.

Case 2:
For this string of cards in a deck.
4♠2♠- 5♣8♥- J♣3♦- K♠ 9♦-
3♣ 4♥ - 1♠3♠- 1♣1♦- K♥ 4♣....... .

When we do the memorization using your idea of putting a maximum of 4 images, it would be something like this:
Location 1: 4♠2♠- 5♣8♥- J♣3♦- K♠ 9♦
Location 2: 3♣4♥ - 1♠3♠- 1♣1♦- K♥ 4♣

But when in the recall:
Location 1: 4♠2♠- 5♣8♥- J♣3♦- K♥ 9♦
Location 2: 3♣4♥ - 1♠3♠- 1♣1♦- K♠ 4♣

This is also the same as in case 1 the last pair in Locus 1 and Locus 2 can also be either in black version or red version. Thus, the K♠ and K♥ can be interchanged and you will not be able to tell which is the correct one.

So what do you actually do for these cases?
Spot.

4 October, 2017 - 16:13
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Joined: 10 months 2 weeks ago

Hi Spots in the moon.

For Case 1 :

Case 1 is not so much a problem. When I do speed cards I always remember the color of the last pair .
For the multi deck discipline, I continue the story from deck 1 to deck 2 so that you won't have to remember the last pair in every deck.

For Case 2:

This is the risk I take but the good news is that it doesn't always happen. What I did before was I put maximum of 3 images per location and Case 2 always happen but when I switched to 4 images per location the probabiltiy that Case 2 would occur lessens.

For the multi deck discipline, I put maximum of 5 images per location so that Case 2 would never happen I guess. Creating 5 image story won't hurt so much ,Besides you will still review it.

That's all.

22 October, 2017 - 18:05
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Joined: 5 months 2 days ago

Cool.

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