In my first piece on the number of Tarot arrangements I only looked at how many different arrangements are possible considering cards in positions of a given spread, and I didn’t take into account reversals. Now, not everyone reads with reversals, (I typically don’t) but a lot of people do, and there’s an interesting bit of math related to powers of two when we throw in reversals.

For a three card spread, the first card could be reversed, which would double the number of possible spreads, right? Imagine that you were “counting” the number of possible arrangements by writing them all down on a giant (REALLY giant!) piece of paper. You’d have to write down all the possible arrangements, then write them all down again with the first card reversed. Then when you counted how many you had, it would be the number of things you originally wrote down times two.

If we call the original number of arrangements N, then when we think about the possibility of the first card being reversed, the new number is N*2.

When you consider the possibility of the second card being reversed, we have to double the number again: N*2*2.

And the third card could also be reversed, which makes N*2*2*2. Do you see the pattern?

Since N (the number without reversals) is 456,456, the number of possible arrangements for a three card spread *including* reversals is N*2^3 = N*8 = 3.65 million. Just by considering reversals we’ve gone from 456,456 arrangements – fewer than half a million – to over three and a half million.

A couple of alternative approaches:

Notice that every time we allow a single position to be reversed, we double the number of possible arrangements. The number of positions in the spread affects how many more possibilities are allowed when each individual card can be upright or reversed. One way to approach the original question that’s tempting but incorrect is to say that we should be able to just double the number of arrangements without reversals. But that would only work if all the cards had to be upright or reversed together. Since each individual card can be upright or reversed, the bigger the spread, the more possibilities reversals create.

It’s also tempting but incorrect to imagine that we are drawing from a deck that has twice as many cards. It seems like it should work: if we allow for reversals, we have twice as many possible entries in each position of the spread, right? But this doesn’t work because it is imagining that we are drawing from a deck where the Fool upright and the Fool reversed are two completely separate cards. If we did that, we could draw the Fool upright in one position *and* the Fool reversed in a different position – and that’s obviously not possible with regular Tarot cards. This highlights the fact that we’re drawing cards *without replacement*, which will be important for future calculations.

The Celtic Cross, and what exponents are good for:

Once we figure out how many for every position in the reading we have to multiply that times two for every position in the spread. We have to consider each position separately, so for a three card spread, it’s the original number N times 2 three times because there are three positions, each of which can be either upright or reversed.

Exponents are a shorthand for “multiply repeatedly.” So instead of writing 2*2*2 we can just write N*2^3, and it means the same thing.

When we start looking at the Celtic Cross, with ten positions, there are ten individual opportunities for each position to be reversed. So we have to take the original number of arrangements and multiply by 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2. That kind of notation makes me cross-eyed! This is when it is easier to write 2^10, which means “multiply by 2 ten times.” Or we can do the multiplication out and discover that 2^10 = 1024. That means that including reversals in a ten card spread gives you a thousand _times_ as many possible arrangements. Not just a thousand more – not added – but a thousand *times*!

Since the original number of arrangements (without reversals) for the Celtic Cross spread was 4.56 x 10 ^ 18, when we include reversals there are 4.56 x 10 ^ 18 * 1024, which is about 4.67 x 10 ^ 21, or 4.6 sextillion, with a “sex.”

According to at least one back-of-the-envelope estimate, that would be about as many grains of sand as there are on all of Earth’s beaches; it’s also in the range of estimates of number of stars in the known universe.

That’s a lot of Tarot!

What does this have to do with the probability of getting several Major Arcana cards in a given spread? Stay tuned for part 3…

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PS: If you have ever wondered why numbers that have to do with computers tend to come in these unusual sizes – 1024 instead of 1000, 256 instead of 250, and so on, the reason is that computers work in binary, which means that the number of numbers they can deal with is expressed in powers of two, just like the powers of two that we’re working with here.

PPS: This estimate used a slightly different size of grains of sand than my original calculations did. It’s within an order of magnitude, though, and really depends on your definition of sand. http://www.universetoday.com/106725/are-there-more-grains-of-sand-than-stars/