How many major arcana to expect in a Tarot reading?

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.

How reversals create more Tarot arrangements

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…


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/

How many Tarot spreads are there?

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.

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.

How many reversed cards in a Tarot reading?

How many reversed cards should we expect? Many Tarot readers read reversed cards with slightly different meanings; we know that reversals have a lot of nuance in them, and don’t just mean “bad” things, but it can still be disconcerting to see seven or eight cards out of ten upside-down. Actually, people who work with reversed cards should be accustomed to seeing quite a few reversals in any spread, but it’s still hard to quantify whether any given number of reversals is unusual. It turns out that the binomial probability distribution can help us refine our expectations of how many reversed cards are usual in a typical spread.

While we know to expect “about” five reversed cards out of ten, the binomial probability distribution tells us that really anywhere from two to eight cards is not surprising.

The binomial probability distribution applies to situations where we have a number of independent events, each of which has two possible outcomes. In a Tarot spread, each card is an event, and its outcomes can be either upright or reversed. I’m going to assume that each card will be reversed about half the time, for a probability of 0.5. This assumes your method of reversing cards is truly random, but it will be a reasonable approximation for most methods.

The average number of reversals in a given spread will be the number of cards in the spread times the probability of a single card being reversed. For a ten card spread, the expected number of reversed cards is five. This intuitively makes sense; it’s when we get more or fewer that we start to wonder whether it happened by chance.

The number that helps us understand how much things differ from the expected is the standard deviation. For a probability distribution, the standard deviation is a measure of how wide the distribution is. This means it tells us how normal or weird a particular difference (deviation) from the expected value (standard) really is.

For a binomial probability distribution, we find the standard deviation by taking the expected value times the probability of the opposite outcome (which is also one half, or 0.5), and then taking the square root. For a spread of ten Tarot cards, the standard deviation works out to about 1.6.

Almost 70% of the time, when we deal ten cards, the number of reversed cards will be the average (five) plus or minus one standard deviation (1.6). Therefore most of the time, we should expect four to six reversed cards.

For 95% of readings, we’ll be within plus or minus two standard deviations, which means anywhere from two to eight cards reversed. This is important; it’s easy for us to look at six reversed cards and say that that’s close to five, so it’s normal, but when the reversals start creeping up towards eight, a significant majority of the reading, we can get nervous. Don’t!

Understanding that eight reversed cards is normal is an example of how math helps us refine our intuition. 95% of the time is a lot – this is nineteen times out of twenty. In only one out of twenty readings would you expect to get something outside of these bounds. How long does it take you to do twenty full Celtic Cross readings? Only once out of those twenty times would you see a reading with all reversed cards or all upright cards, or even a reading with just a single card upright or a single card reversed.

Of course, this kind of understanding is only a starting place for the real work of a Tarot reading. The specific meanings of the cards which are reversed, the patterns revealed in the layout, and most importantly the meanings that the client reads as applicable to her or his life, are much more important for the interpretation of a given Tarot reading, and they give each reading its unique qualities.

Still, the simple numerical understanding indicates that we should expect to work with what seems like a lot of reversed cards. Hopefully, knowing this means that the presence of these cards won’t necessarily make you disturbed or uncomfortable; they are a way for Tarot to give you a wider range of nuance and information, so engaging with them can lead to even more insight and understanding.

Zombies, tokenism, and greatness

xkcd, a great geeky webcomic, features Zombie Marie Curie in today’s strip. ZMC says she’s tired of being THE token female scientist and spends some time talking about other great women scientists and mathematicians, but her real point is:

You don’t become great by trying to be great. You become great by wanting to do something, and then doing it so hard that you become great in the process.

As someone who did one of the multitudinous Marie Curie projects in grade school, I really appreciate the point about tokenism. I’m glad that I had learned about tokenism by the time, years later, that a guy tried to tell me that the 1950s weren’t that bad about gender roles because: Look! Adm. Grace Hopper had genitals that were an innie and she was a computer scientist! Bugs! Um, ok. Name two other famous female computer scientists – heck, famous female scientific/technical experts from the 1950s. Tokens, yoo haz one. (sic)

But the point about greatness is even more important. We don’t need another Marie Curie. We already had one. The message I’d like to get out to young women, and everybody, is that we need the first one of you doing whatever you’re good at that makes the world a better place.

What do we teach? Where’s the proof?

In a recent tutoring session, I told my student the following riddle:

A father is driving his son to a baseball game when they are involved in a car accident. The man is killed instantly; the boy is badly injured and taken to a nearby hospital for emergency surgery. While reviewing the case before surgery, the surgeon suddenly cries, “I can’t operate on him! That’s my son!” Who is the surgeon?

I used this simple riddle as an example of unexamined assumptions. One of the great contributions that math makes to a liberal arts education is teaching people how to examine assumptions and construct sound logical arguments. Learning how to write proofs can be frustrating at first, because it means learning to break down every hidden assumption, even the ones that seem pretty basic.

We use the commutative and associative properties of addition and multiplication every day; learning that they have names, and that they’re not necessarily a given in some systems, is so counter-intuitive that it’s hard to absorb. They don’t seem like things that need to be stated or counted among our assumptions. It’s hard to imagine that in some systems a times b doesn’t give you the same answer as b times a. It gets easier after you’ve seen a few examples; not necessarily simple, but easier.

Learning to find the hidden assumptions and then imagine alternatives in math prepares people to do the same thing in the world of human interactions. In human interactions, when cherished assumptions are challenged, and alternatives imagined, the assumptions often get defended as “facts of nature,” or “God’s law,” or “just the way it is.” But once people learn that hidden assumptions aren’t necessarily true, and once they see a few counterexamples, those arguments sound weaker and weaker.

That’s why a proposal to ban teachers from talking about homosexuality will ultimately backfire. The legislator sponsoring the bill claims that he’s trying to give teachers more time to teach core subjects, like math. People trying to defend their unexamined and unsupportable assumptions about human relationships would be better off trying to stop teachers from teaching math, because math is what teaches people to examine their assumptions, imagine alternatives, and recognize counterexamples.

This is why teaching keeps my hope alive, especially when my student was able to come up with three different possible answers to the riddle. It makes me hopeful that it will get easier to imagine that a boy has two fathers or that women, even mothers, can be surgeons.

The Foolish Student’s Journey through Calculus

For fun, I decided to do a quick comparison of the Major Arcana of the Tarot with the chapters of the first book I came across with approximately 22 chapters. Since I am an eclectic soul, the first such book that came to hand was Calculus Made Easy by Silvanus P. Thompson. This book was originally written in 1910, revised and enlarged in 1914, and the third edition was edited posthumously in 1946. It’s still in print because it is a concise summary of the geometric approach to calculus. I was astounded at how well the Major Arcana related to the various chapters, so I present a quick summary for your amusement and possibly edification:

  1. Card: The Magician. Book chapter: To deliver you from the preliminary terrors.
    Thompson’s whole point in this book is that once a student’s intuition is properly engaged, calculus becomes much more easy and natural. Thus he starts out by explaining that dx means “a little bit of x” and the integral symbol means “add up all the little bits.” This is in fact a neat way of introducing students to the basics of calculus, worthy of a Magician, and an example of the magical powers that calculus promises to the student.
  2. The High Priestess/Different degrees of smallness.
    Thompson enlarges (sorry) on the idea of dx by using real-world examples to convince readers of some algebraic properties, especially that dx of dx is negligibly small. Since the High Priestess (or Papesse) is usually about internal knowledge or hidden information, as dx is about functions, the comparison is apt.
  3. The Empress/On relative growings.
    The Empress, a card about fertility, presides over Thompson’s chapter on comparing one rate of growth to another.
  4. The Emperor/Simplest cases.
    Basic differentiation is the first example of students gaining the power or authority of calculus, when they begin to become Emperors over relationships between functions and rates of change.
  5. The Hierophant/What to do with constants.
    The chapter about basic rules of differentiation shows up at the same point as the card that alludes to hierarchical power structures and established authority – the constants in our lives.
  6. The Lovers/Sums, differences, products, and quotients.
    When two (or more) functions are involved in a differentiation, how we handle them in calculus becomes more complicated; thankfully it’s still less complex than the kind of human relationships the Lovers card usually refers to.
  7. The Chariot/Successive differentiation.
    Learning about second (and more) derivatives means that students start making lists of successive derivations and charging right through them – although sometimes one of the horses pulling the Chariot goes off in a different direction and they get confused.
  8. Strength/When time varies.
    With physical examples, Thompson explains the idea of a rate and introduces equations describing acceleration, force, and work, while the Tarot shows a woman closing a lion’s mouth – but gently.
  9. The Hermit/Introducing a useful dodge.
    In this chapter, the chain rule lets students ignore one part of an equation while working on another, then return to the ignored part to get the whole solution – rather as the Hermit takes time away to focus on some things first.
  10. Wheel of Fortune/The geometrical meaning of differentiation.
    Just as the wheel shows people at different points in its rotation, this is the chapter when we finally start looking at curves and how the curves change between different points. This is also a level when students generally start to internalize (or not) the intuitive aspects of calculus; if they do, they stand a good chance of completing the course successfully, but they’re only halfway through.
  11. Justice/Maxima and minima.
    The lady with the sword and scales takes a hard look at what’s going on as students learn to find the turning points – where a function is at its highest and its lowest.
  12. The Hanged Man/The curvature of curves.
    In this chapter, students have to take a different perspective on differentiation and its geometrical meaning, rather as the Hanged Man is all about a changed perspective.
  13. Death/Other useful dodges – partial fractions and inverses.
    This correlation is appropriate on two levels: partial fractions are one of the more technically difficult algebraic techniques taught in basic calculus, and can be the “death” of many students’ patience, but the process of using partial fractions or inverses is also all about transforming from one form into another.
  14. Temperance/On true compound interest and the law of organic growth.
    Temperance is usually depicted pouring liquid from one vessel into another. In my favorite deck, he’s juggling. These two very practical applications of calculus are all about pouring and juggling – money and populations and how they are constantly in flux even when they seem to stay still.
  15. The Devil/How to deal with sines and cosines.
    Enough said.
  16. The Tower/Partial differentiation.
    Here I actually dislike Thompson’s technique, because he starts in to a topic usually reserved for Calculus III in today’s teaching style, and it runs the risk of shattering students’ still-precarious understanding and self-confidence without ever getting to the second half of basic calculus, integration, so I find the card appropriate.
  17. The Star/Integration.
    Some readers see the Star as a kind of healing experience after the tumultuous change of the Tower; she is usually pictured as pouring out two vessels of water. The idea of reuniting what has been broken is not a bad metaphor for integration (adding up all the “little bits” that were broken up in differentiation). It’s also comparable to the way tiny individual drops of water can make an entire sea when added together.
  18. The Moon/Integrating as the reverse of differentiating.
    In some ways of teaching calculus, this is “the big secret” – that integration and differentiation are not just seeming opposites but literally opposite processes that reverse each other. Most students start to pick up on this a little earlier, and have subconscious suspicions of the relationship even before it’s presented as a theorem. Since the Moon is about hidden information and the subconscious, it’s appropriate.
  19. The Sun/On finding areas by integrating.
    Finally, students put it all together and start doing problems that can have real-world equivalents. The ability to find areas under curves is another of the major accomplishments for students of calculus, so it certainly can feel like a triumph, or coming back out into the sun after a long night.
  20. Judgment/Dodges, pitfalls, and triumphs.
    Thompson pulls together a useful assortment of ways to transform an insuperable problem into a solvable one, “resurrecting” it or giving the student ways to emerge triumphant.
  21. The World/Finding solutions.
    Here Thompson pulls together the techniques of the entire subject so far to launch the student on the next major topic of mathematical studies: solving differential equations. As in the Tarot, this is both an ending, the completion of the journey that began at #1, and the beginning of a new exploration. I feel justified in combining chapters 22 and 23 under this card as well, because they too are side-notes or brief introductions of other directions the study of calculus can be expanded.

0/22: The Fool. Prologue/Epilogue and apologue. I cannot put it better than Thompson himself did in his Prologue:

Considering how many fools can calculate, it is surprising that it should be thought either a difficult or a tedious task for any other fool to learn how to master the same tricks.

Some calculus-tricks are quite easy. Some are enormously difficult. The fools who write the text-books of advanced mathematics – and they are mostly clever fools – seldom take the trouble to show you how easy the easy calculations are. On the contrary, they seem to desire to impress you with their tremendous cleverness by going about it in the most difficult way.

Being myself a remarkably stupid fellow, I have had to unteach myself the difficulties, and now beg to present to my fellow fools the parts that are not hard. Master these thoroughly, and the rest will follow. What one fool can do, another can.

In his Epilogue, he explains why mathematicians may condemn the book and admits the charge that he has simplified the subject beyond what professional but impractical mathematicians regard as necessary. He defends his deliberate choices as being the best way to  introduce “fools” like himself to the subject – a journey that any of us can find ourselves taking part in.