For the last topic in my series on mathematical investigations of Tarot, I want to discuss how many major arcana cards should be expected in a typical spread of ten cards. The answer to this is a little difficult to calculate, and it draws on the work we’ve done so far. We established that the number of distinct ten-card arrangements (spreads) without reversals is 78! / 68!, which is approximately 4.56 * 10 ^ 18.

In order to find out the probability of having, say, 4 major arcana cards in a ten-card spread, we can figure out the number of different arrangements of four majors and six minors are possible. That gives us the number of different spreads with exactly four major arcana cards in it. If we divide that number by the total number of ten-card spreads, we get the probability of having four major arcana out of ten.

To find out how many ten-card spreads have exactly four majors, imagine that we split the deck into two groups: majors and minors. First we draw four majors at random, then draw six minors at random. How many different ways can we do that? We’re going to use a concept called *combinations *to figure out. Combinations are like permutations, except in combinations order does not matter. Combinations are like a salad – all the ingredients are mixed up together. Permutations are like a sandwich – which thing is on top matters.

To figure out how many combinations of four major arcana cards are possible, say we draw four. There are 22 possibilities for the first card, 21 possibilities for the second, 20 for the third, and 19 for the fourth, so there are 22*21*20*19 = 22! / (22-4)! = 22! / 18!. But this way of counting would assume that the order of drawing matters, when in fact we haven’t arranged the cards into the final spread yet at all. So this number is counting it as different if we draw the Fool, the Star, the World, and the Hanged Man as opposed to the Hanged Man, the Star, the World, and the Fool. These are different permutations (different orderings) but the same combination. Each combination can occur in 4*3*2*1 different permutations, so each one is being counted 4! = 24 times. If we take the result above and divide by 4!, we’ll get the correct number of combinations of four major arcana cards: 22! / (4! * 18!)

For the combination of six minor arcana cards, we can do the same thing and get 56! / (6! * 50!)

So now we have a combination of four majors and six minors and we have to make a Tarot spread out of them. For any set of ten cards there are 10! (= 10*9*8*…*2*1) ways to arrange these into a ten-card spread because the positions of the cards matters.

Then we can multiply these results together to figure out how many Tarot spreads there are with four majors and six minors: (22! / (4! * 18!)) * 56! / (6! * 50!) * 10! = 8.62 * 10 ^ 17.

Now that number isn’t really useful to us. All it tells me off the top of my head is that the number of spreads with four majors is smaller than the total number of spreads (and if we had gotten the opposite result, we’d know something was wrong with our math!). How much smaller is it? Well, we can divide: the number of spreads with four majors divided by the total number of spreads is 0.1887.

That number is a probability! Specifically, it’s the probability that a randomly-dealt spread of ten cards has exactly four major arcana in it. If you remember that percentages are probabilities times 100, then that number tells you that about 18% of your ten-card Tarot spreads will have four major arcana cards.

I used the same process to calculate the probability of having zero, one, two, and so on majors in a ten-card spread. Here are my results:

- For a spread with no major arcana cards the probability is 0.0282
- 1 major has probability 0.1324
- 2 majors has probability 0.2607
- 3 majors has probability 0.2838
- 4 majors has probability 0.1887
- 5 majors has probability 0.07994
- 6 majors has probability 0.0217
- 7 majors has probability 0.0037
- 8 majors has probability 0.00039
- 9 majors has probability 0.00002
- 10 majors has probability 5.138 * 10 ^ -7 (that’s a number starting with six zeroes after the decimal place)

Interestingly, you’re most likely to get three majors in a ten-card spread, and almost as likely to get just two majors. Anytime you see more than four majors it is going to be very unusual, and anything with more than half majors is extremely unusual.

The most surprising result to me was that you have less than a 3% chance of getting a spread that is all minors. What surprises you about these results?

*Note these results are rounded. If you add them up you will get almost (but not quite) 1.000, which is what we should get (there’s 100% probability of one of these conditions occurring when we deal a ten-card spread).*

*Again, these results took some effort to calculate, so if you use them, please link back.*