Two cards system

Hi. In two cards system there are around 1200 combinations to memorize or twice that amount?

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Depends on how you use your loci… if you keep adding to the same location as long as the first card is a black suit and then move to the next location only when the first card is a red suit you can get away with just half (i.e., 1,326); otherwise, you’ll need all 52*51=2,652 card pairs.

The reason here is that two card values can share the same image and the location then tells you which of the two possible suit pairs it is. Say your image for ace of spades and two of hearts is banana and your image for ace of hearts and two of spades is the very same banana, you know which of the two it is by looking at where you’ve stored it.

If the banana is the first thing at that location is has to start with a red suit and will thus be the ace of hearts followed by the two of spades. If it’s not the first item at that location, it’ll have to start with a black suit, so it’d be the ace of spades followed by the two of hearts.

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Thank you for the answer. I get what the issues with pairs of the same cards is, we have one pair and the inversed variant. However, I don’t get how location can tell me which variant it is out of the two. After all, when I memorize a deck of cards, I use only one selected location, or are you refering to that I can tell somehow by the location in which I try to memorize that I can recall which variant of the pair it is?

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I’m not sure what you mean by that… there isn’t a fixed numbers of items (card pairs) in a location if you use this approach. Let’s go from location A to location B:

Scenario 1:

A) tiger, butterfly, banana
B) car, laptop

Scenario 2:

A) tiger, butterfly
B) banana, car, laptop

In scenario 1, we didn’t change locations, so the card pairs starts “black” and is thus the ace of spades followed by the two of hearts. In scenario 2, we changed locations because the card pair starts “red” and is thus the acce of hearts followed by the two of spades.

Also, the car in scenario 1 starts “red” and the “car” in scenario 2 starts “black,” but that just as a side note. If that car was supposed to remain “red” you’d have to move to location C right after placing the banana in location B.

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Aha. So that means I may have to change a location several times and use many locations, or when a change happens again after going from scenario 1 to scenario 2 I go from scenario 2 to scenario 1, or do I go from scenario 2 to scenario 3, and then scenario 3 to scenario 4, etc.?

Also, there are pairs of cards of the same colour for example spades and clubs, do I also change the location for the inversed pair of them?

Ps. scenario = location

Edit:
So, after a moment of thinking I think it would be like that: I look at each pair first whether it’s inversed or normal (according to numbering of cards from my one card system, lower value card, higher value card = normal, higher value card, lower value card = inversed). If it’s inversed I set it in one location, if it’s normal I set in the other. I set breaks in the path in both locations according to switching between two locations.
I am considering whether
a) try to make switches in more locations, but then I can get unlucky and end up with 26 locations each having one locus, if I have to switch every time
b) twist the image of the inverted pair, like imbibe with darkness, negative energy, etc. so that I can use only one location.

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Scenario = scenario
Location = location

I was trying to show you how the banana has two different values depending on where you place it; hence, two scenarios.

As far as double black pairs, just use a double red as what you call an “inverse,” so that two of spades followed by three of spades is the same image as two of hearts and three of hearts. If it’s the red pair you move to the next location and if it’s the black pair, stay in your current location.

Yes, if the first, third, fifth, etc. is red you’ll need 26 loci. Similarly, if every second card is black, all images would be in one loci. Luckily, the law of large numbers makes that highly unlikely; however, there are always people that win the lottery too, so…

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What if it’s two of spades and three of clubs vs three of clubs and two of spades? I suppose in this case I have separate images for both of two sets of them?

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Well yeah, I mean 23 and 32 is not the same thing. I mean I’m not sure if I fully get what you mean by “inverse” but we’re not talking “reverse”

The “red”-version of two of spades and three of clubs would be two of hearts and three of diamonds.

Obviously, that’s only true if you associate spades with hearts and clubs with diamonds. You’re better off cutting the suit pairs in half, seeing how there’s 16 of them and half is then eight. Plus you get the red indicator to know when to move on.

You can also treat it like a red traffic light and stay in the same place and move to the next location when you get black. Either way, there are 13 values per suit so it’d be really hard to cut that in half.

That said, how you let two card pairs share the same image is entirely up to you. The convention I’m using above is a pretty commonly accepted one.

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Okay. So I think I’ve gotten the grasp of the method. Just one question about this quote. You mean 26 memory palaces where each will have one locus used?

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26 loci in 1 memory palace…

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Actually, no. You switch to the next palace to remember that suddenly a pair is reversed. A logical approach (for me) is to hm make two memory palaces, one begins with red cards, second with black cards, I switch between them making appropriate jumps (like when I switch back to the palace with red cards after beginning with red cards and then having a beginning black card in the second pair of cards). Appropriate jumps, meaning I would omit some loci like 2-5 in the memory palace with red cards to know that there was a jump. Is it about right?
Edit. Yeah. wonder is right. I’ve misread what bjoern wrote. Bjoern wrote about a situation where I have to use one memory palace because every first card of every pair is red or black. I am typing about a situation where every second pair has a beginning black or red card. That is the case that troubles me.
Edit 2: It’s just, do I use for this method two memory palaces, one for pairs beginning with black cards, one for pairs beginning with red cards? I dont use any more memory palaces than that? Just two?

for the second card of a pair: red or black is already specified in the encoding if you are using 1326 images. there would only be one memory palace.

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No. Im talking about every second pair, not about second card of the pair, so first card of a pair changes —> black —-> red —-> black —-> red ——> black etc. Two memory palaces in that case and I simply jump between them?

oh no, you just keep placing the images in the same loci until the red card comes.

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Okay. Now I am just confused. You mean I can place 48 cards in a row in one locus? I am afraid my memory won’t take it too well.

it’s the method that is commonly used.

not 48 in a row though, the odds of that are astronomically small.

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Okay. Could you give me a link or sth because I still lack some basic understanding?

You can look up Lance Tschirhart’s Shadow System, it’s described in there although Shadow System may confuse you more lol.

There’s other systems that can be used with this method like Major or a modified Ben system which is what I use.

What’s this modified Ben system that you use?

I don’t know how other people do it, but I switch location after placing the image for the current card pair. So before I see the card pair, I know which location it’s going into. The suit of that card pair determines whether I move to the next location after that card pair or not.

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