More than you think!

Of course, there are as many different kinds of spreads, as there are readers – from drawing a single card, to three cards, to the standard ten-card Celtic Cross to huge spreads that use the whole deck. Here I’m asking more specifically once I choose a particular spread how many different results I can get from drawing different cards in a different order.

Start with a simple three card spread: We draw three cards and lay them out in order. How many different three-card arrangements can we get?

There are 78 choices for the first card, 77 choices for the second, and 76 choices for the third. 78 * 77 * 76 = 456,456 different layouts for a three-card spread.

That’s a lot! If you looked at one of these every minute of every hour of every day (no breaks!), it would take you almost 7 years to look at every single possibility.

*Math section:
*

In math, we have a way of expressing this situation called permutations. For a three-card spread, you start with 78 cards and lay down 3 of them, and the order matters. That’s usually expressed as 78 permute 3, sometimes written 78 nPr 3, especially on calculators.

For calculating permutations, we use a math trick called factorials. A factorial is a way of saying “start with the number, then multiply it by all the numbers smaller than it down to 1.” Factorial is written as exclamation point: n! (pronounced “n factorial” or “n bang” if you want to hang with the kewl kids).

So 10! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1. (In more mathy language, n! = n * (n-1) * (n-2) * … * 3 * 2 * 1.) The thing about factorials is that they get big _really_ fast. Really, really fast. 10! = 3.6 million.

And factorials are really useful for calculating permutations. Here’s the trick: notice that 8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1, so 8! includes everything inside 10! except the 10 * 9. So if we take 10! divided by 8!, we can cancel out everything except that 9 and 10, and the answer is 90.

For the calculation above, we can say that the number of permutations possible for a three card spread drawn from 78 cards is 78! divided by 75! = 78 * 77 * 76 = 456,456. In general, to calculate how many ways there are to take n objects and choose permutations of r of them, we calculate n! / r!

**Celtic Cross ten-card spread:**

Now let’s think about the standard Celtic Cross ten-card spread. We can calculate how many different spreads we can get by taking:

78 * 77 * 76 * 75 * 74 * 73 * 72 * 71 * 70 * 69

which is the same as 78 nPr 68 = 78! / 68!

which equals 4.56 times 10 ^ 18. That’s a really, really, really big number. It has nineteen digits. I’m not even going to try to write it all out. Let’s think about that some other ways.

There are four and a half _quintillion_ different possible Celtic Cross spreads. Quintillion, with a Q. (Come on, it’s a fun word!)

Suppose that a person started laying out a Celtic Cross spread, looking at it, shuffling the deck, and laying out a new one. If this person could lay out one spread a *second*, every second, from the moment of the Big Bang until now, that person would have seen only a *tenth* of the possible spreads so far in the lifetime of the known universe.

Or imagine every ten-card Tarot spread was represented by a grain of sand, and all that sand was piled into a giant cone. That cone’s tip would be twice as high as the roof of the Empire State Building, and its base would extend out to cover at least part of Grand Central Terminal and the same distance on the other side. This cone, 781 meters tall, would be almost as tall as the tallest building on earth, and have a base 1159 meters in radius – more than a mile across!

Two more geek comparisons for fun:

Suppose you were going to do a Celtic Cross spread for every cell in your body, plus all the microbes that live in and on your body. If you got together with about a thousand of your closest friends, all your cells together would use up all the available Tarot spreads.

The number of spreads is a few million times larger than the number of stars in our galaxy. Not just the stars you can see in the night sky, but all the stars in our galaxy. You’d need a few million galaxies for every star to have a corresponding Tarot spread. This is smaller than the estimated number of stars in the known universe, but not by a whole lot.

Note that none of this is counting reversed cards!

I got started on this by asking how likely it was to have five cards out of ten be Major Arcana, and calculating the answer was more complicated than I expected. If folks enjoy this, let me know, and I’ll continue with a series of two or three more posts, eventually answering the probabilities of the number of Major Arcana cards.

It took me a while to do these calculations, so if you cite them, please link back.

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NB: All calculations assume that every card has an equal chance of being chosen – if that’s not true, then all bets are off.

Sources:

Size of a grain of sand. Note that sand comprises a range of sizes.

Estimate of number of cells in the body. Then I added in a factor of 100 to account for bacteria.

Estimate of the number of stars in the galaxy.

*Beyond this place there be numbers:*

4.56 x 10 ^ 18 spreads or grains of sand

4.56 x 10 ^ 18 times 2.51 x 10 ^ -10 m ^ 3 equals 1.1 x 10 ^ 9 m ^ 3, or a billion cubic meters

4.56 x 10 ^ 18 times 6 x 10 ^ -5 kg equals 3 x 10 ^12 kilos, or 3 trillion kilos

For the cone, Wikipedia states that dry sand has an angle of repose of 34 degrees.

Thus V = 1/3 pi r ^ 3 tan (34 deg) allows one to solve for the radius.

The rest is left as an exercise to the reader.

as many as you can create!

In one sense, of course that’s true. You’ll see that in the sense I’m using it here, I mean how many different readings are possible within a single example of a spread. Maybe I should have said how many readings are possible, but I’m not sure everyone would use the terms the same way. We seem to use “spread” to mean both the arrangement of possible cards (imagine blank outlines that indicate where you will put cards down) and the actual layout of specific cards in a specific reading (“Her spread had a lot Major Arcana in it this time…”). Anybody want to suggest more specific terminology to differentiate those two?

I’m a little lost outside of the scope of the statistical meaning of the post … the ten-card spread, with both positive & negative definitions of all the cards(78 in total), the possible outcomes for any given reading are 1:1.2 billion!

Gonna spell that, cause (for me) punctuation is hard to see in small print (guess I’m getting old) … that’s one in one-point-two BILLION!

and that’s a ten-card spread! My largest is a fifteen-card spread. Quite a card trick! Ehh?!?

Forgive me if I’m a little unsure of the scope of your post, as I was certain it was mathematical …. correct me if I’m wrong, as I enjoy the conversation!

I mean that I’m using the word “spread” to mean both “the Celtic Cross spread” and “this Celtic Cross spread right here with card 1 this and card 2 that and card 3 the other…”. Of course there are as amny different ways to arrange the cards as the reader can invent, but within a given ten-card arrangement, there are a limited but very large number of possible readings.

The true number of possible readings with ten cards both upright and reversed is actually on the order of 4.6 sextillion – that’s ten to the 21st power. It’s a lot, lot bigger than billions. I’ll get to that – including reversals – in my next post.

I found this extremely fun; please continue. Also, let me know if you want any help with the calculating.

Thank you! It’s odd – the most time-consuming part wasn’t actually the mathematically challenging bit of sorting through the probabilistic thinking but the effort to try different scales to find something to compare the number to that would make it comprehensible. More next week!