E-Book Market & Sales – Analysis Pool

On this page you can find a collection of all my statistical analysis and research regarding the Kindle ebook market and sales. I’ll keep the page updated.

Mathematical Model For E-Book Sales

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Mathematical Model For (E-) Book Sales

It seems to be a no-brainer that with more books on the market, an author will see higher revenues. I wanted to know more about how the sales rate varies with the number of books. So I did what I always do when faced with an economic problem: construct a mathematical model. Even though it took me several tries to find the right approach, I’m fairly confident that the following model is able to explain why revenues grow overproportionally with the number of books an author has published. I also stumbled across a way to correct the marketing R/C for number of books.

The basic quantities used are:

• n = number of books
• i = impressions per day
• q = conversion probability (which is the probability that an impression results in a sale)
• s = sales per buyer
• r = daily sales rate

Obviously the basic relationship is:

r = i(n) * q(n) * s(n)

with the brackets indicating a dependence of the quantities on the number of books.

1) Let’s start with s(n) = sales per buyer. Suppose there’s a probability p that a buyer, who has purchased an author’s book, will go on to buy yet another book of said author. To visualize this, think of the books as some kind of mirrors: each ray (sale) will either go through the book (no further sales from this buyer) or be reflected on another book of the author. In the latter case, the process repeats. Using this “reflective model”, the number of sales per buyer is:

s(n) = 1 + p + p² + … + pn = (1 – pn) / (1 – p)

For example, if the probability of a reader buying another book from the same author is p = 15 % = 0.15 and the author has n = 3 books available, we get:

s(3) = (1 – 0.153) / (1 – 0.15) = 1.17 sales per buyer

So the number of sales per buyer increases with the number of books. However, it quickly reaches a limiting value. Letting n go to infinity results in:

s(∞) = 1 / (1 – p)

Hence, this effect is a source for overproportional growth only for the first few books. After that it turns into a constant factor.

2) Let’s turn to q(n) = conversion probability. Why should there be a dependence on number of books at all for this quantity? Studies show that the probability of making a sale grows with the choice offered. That’s why ridiculously large malls work. When an author offers a large number of books, he is able to provide list impression (featuring all his / her books) additionally to the common single impressions (featuring only one book). With more choice, the conversion probability on list impressions will be higher than that on single impressions.

• qs = single impression conversion probability
• ps = percentage of impressions that are single impressions
• ql = list impression conversion probability
• pl = percentage of impressions that are list impressions

with ps + pl = 1. The overall conversion probability will be:

q(n) = qs(n) * ps(n) + ql(n)* pl(n)

With ql(n) and pl(n) obviously growing with the number of books and ps(n) decreasing accordingly, we get an increase in the overall conversion probability.

3) Finally let’s look at i(n) = impressions per day. Denoting with i1, i2, … the number of daily impressions by book number 1, book number 2, … , the average number of impressions per day and book are:

ib = 1/n * ∑[k] ik

with ∑[k] meaning the sum over all k. The overall impressions per day are:

i(n) = ib(n) * n

Assuming all books generate the same number of daily impressions, this is a linear growth. However, there might be an overproportional factor at work here. As an author keeps publishing, his experience in writing, editing and marketing will grow. Especially for initially inexperienced authors the quality of the books and the marketing approach will improve with each book. Translated in numbers, this means that later books will generate more impressions per day:

ik+1 > ik

which leads to an overproportional (instead of just linear) growth in overall impressions per day with the number of books. Note that more experience should also translate into a higher single impression conversion probability:

qs(n+1) > qs(n)

4) As a final treat, let’s look at how these effects impact the marketing R/C. The marketing R/C is the ratio of revenues that result from an ad divided by the costs of the ad:

R/C = Revenues / Costs

For an ad to be of worth to an author, this value should be greater than 1. Assume an ad generates the number of iad single impressions in total. For one book we get the revenues:

If more than one book is available, this number changes to:

R = iad * qs(n) * (1 – pn) / (1 – p)

So if the R/C in the case of one book is (R/C)1, the corrected R/C for a larger number of books is:

R/C = (R/C)1 * qs(n) / qs(1) * (1 – pn) / (1 – p)

In short: ads, that aren’t profitable, can become profitable as the author offers more books.

For more mathematical modeling check out: Mathematics of Blog Traffic: Model and Tips for High Traffic.

Inflation: How long does it take for prices to double?

A question that often comes up is how long it would take for prices to double if the rate of inflation remained constant. It also helps to turn an abstract percentage number into a value that is easier to grasp and interpret.

If we start at a certain value for the consumer price index CPI0 and apply a constant annual inflation factor f (which is just the annual inflation rate expressed in decimals plus one), the CPI would grow exponentially according to this formula:

CPIn = CPI0 · f n

where CPIn symbolizes the Consumer Price Index for year n. The prices have doubled when CPIn equals 2 · CPI0. So we get:

2 · CPI0 = CPI0 · f n

Or, after solving this equation for n:

n = ln(2) / ln(f)

with ln being the natural logarithm. Using this formula, we can calculate how many years it would take for prices to double given a constant inflation rate (and thus inflation factor). Let’s look at some examples.

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In 1918, the end of World War I and the beginning of the Spanish Flu, the inflation rate in the US rose to a frightening r = 0.204 = 20.4 %. The corresponding inflation factor is f = 1.204. How long would it take for prices to double if it remained constant?

Applying the formula, we get:

n = ln(2) / ln(1.204) = ca. 4 years

More typical values for the annual inflation rate are in the region several percent. Let’s see how long it takes for prices to double under normal circumstances. We will use r = 0.025 = 2.5 % for the constant inflation rate.

n = ln(2) / ln(1.025) = ca. 28 years

Which is approximately one generation.

One of the highest inflation rates ever measured occurred during the Hyperinflation in the Weimar Republic, a democratic ancestor of the Federal Republic of Germany. The monthly (!) inflation rate reached a fantastical value of r = 295 = 29500 %. To grasp this, it is certainly helpful to express it in form of the doubling time.

n = ln(2) / ln(296) = ca. 0.12 months = ca. 4 days

Note that since we used the monthly inflation rate as the input, we got the result in months as well. Even worse was the inflation at the beginning of the nineties in Yugoslavia, with a daily (!) inflation rate of r = 0.65 = 65 %, meaning prices doubled every 33 hours.

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This was an excerpt from “Business Math Basics – Practical and Simple”. I hope you enjoyed it. For more on inflation check out my post about the Time Value of Money.