Memorizing the order of black/red in a deck of cards with PAO

#1

Hey everyone, I probably am a bit late to the party but I read this blog post and I am wondering how this could be translated into a PAO system.

Someone commented the following:

How did not I see this before,

we add 6 bits to make two digits (000=0 to 111=7; 64 numbers in total)
that means we have 9 images for 54( or 52) cards, and then we may use the PAO system, which drastically reduces it to 3 images, just 3 images and one per loci, what the hell, we dont even need loci for this, just link or peg the three images, and voila, we have the whole order of black and red order of a whole deck of card in three images encoded.

And in the same way or the other, remembering random binary numbers, 6 digits give an image, and with PAO 18 bits can be encoded in 1 image. Fascinating these are. WOW

Could someone elaborate on this and maybe even take me through this step by step? I’m truly frustrated, since I somehow cannot figure this out by myself, lol. This probably is due to my inexperience in all these techniques and lack of practice - I just started a few days ago and am still busy learning my PAO.

I would like to use this method to get some practice while learning a neat trick at the same time.

Best regards,
Monti

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(Josh Cohen) #2

I think they are suggesting turning each group of three cards into a decimal number and then combining the decimal digits in pairs.

So these binary digits would record the color values for the first 18 cards: 000 010 100 011 100 001

and could be translated into these decimal digits:

  • 000: 0
  • 010: 2
  • 100: 4
  • 011: 3
  • 100: 4
  • 001: 1

which can be memorized with a 2-digit PAO system as: 02 43 41. The rest of the deck could be encoded with another PAO group.

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#3

Hey,

the time this post was pending (because it was my first one) actually allowed me to figure it out myself. For the sake of teaching myself twice (and other people as confused as I initially was), I want to quickly elaborate on how three PAO scenes are already enough to memorize the black/red sequence of a whole deck.

For this, you of course need to have filled in your 00-99 list, depending on which numbers you attribute to which letters.

Using a slighty customized Dominic system (and introducing some small changes into the naming of some of the individual sequences in order for them to work), for me it looks like this:

0O; 1L; 2R; 3E; 4A; 5S; 6G; 7T; 8B; 9P]

All of the sequences:

000 = Sero = S = 5
111 = All = A = 4
100 = Top = T = 7
001 = Bottom = B = 8
110 = Roof = R = 2
011 = Lower = L = 1
101 = Entre (Spanish for between) = E = 3
010 = Outer = O = 0 (I exchanged Middle/Entre and Outer because the 0 stands for the black cards. This way, the 0 means that Outer are actually 0’s, which isn’t the case with the original version. Not necessary but definitely eliminates possible confusion in my case)

In short:

000 = 5
111 = 4
100 = 7
001 = 8
110 = 2
011 = 1
101 = 3
010 = 0

So, how does it look in practice? I would start at the right, with the first cards facing me.
image

In this case:
000 = 5
101 = 3

#53 (Person in your PAO)

011 = 1
010 = 0

#10 (Action in your PAO)

010 = 0
100 = 7

#07 (Object in your PAO)

This results in 53 10 07; the first location in your memory palace where you store the order of the first 18 cards.

The same process now simply has to be repeated and with a lot of practice (which I need but most people here in the forum probably already have), the whole sequence of the deck can be memorized in literally a few seconds. :face_with_monocle: That’s crazy to think about.

Cheers,
Monti

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