Literature

Quantitative Analysis of Top 60 Kindle Romance Novels

I did a quantitative analysis of the current Top 60 Kindle Romance ebooks. Here are the results. First I’ll take a look at all price related data and conclusions.

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  • Price over rank:

pricerank

There seems to be no relation between price and rank. A linear fit confirmed this. The average price was 3.70 $ with a standard deviation of 2.70 $.

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  • Price frequency count:

pricescount

(Note that prices have been rounded up) About one third of all romance novels in the top 60 are offered for 1 $. Roughly another third for 3 $ or 4 $.

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  • Price per 100 pages over rank:

pricerank

Again, no relation here. The average price per 100 pages was 1.24 $ with a standard deviation of 0.86 $.

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  • Price per 100 pages frequency count:

PPP1

About half of all novels in the top 60 have a price per 100 pages lower than 1.20 $. Another third lies between 1.20 $ and 1.60 $.

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  • Price per 100 pages over number of pages:

PPP2

As I expected, the bigger the novel, the less you pay per page. Romance novels of about 200 pages cost 1.50 $ per 100 pages, while at 400 pages the price drops to about 1 $ per 100 pages. The decline is statistically significant, however there’s a lot of variation.

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  • Review count:

reviewscount

A little less than one half of the top novels have less than 50 reviews. About 40 % have between 50 and 150 reviews. Note that some of the remaining 10 % more than 600 reviews (not included in the graph).

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  • Rating over rank:

rankreviews

There’s practically no dependence of rank on rating among the top 60 novels. However, all have a rating of 3.5 stars or higher, most of them (95 %) 4 stars or higher.

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  • Pages over ranking:

pagesrank

There’s no relation between number of pages and rank. A linear fit confirmed this. The average number of pages was 316 with a standard deviation of 107.

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  • Pages count:

pagescount

About 70 % of the analyzed novels have between 200 and 400 pages. 12 % are below and 18 % above this range.

Statistics and Monkeys on Typewriters

Here are the first two sentences of the prologue to Shakespeare’s Romeo and Juliet:

Two households, both alike in dignity,
In fair Verona, where we lay our scene

This excerpt has 77 characters. Now we let a monkey start typing random letters on a typewriter. Once he typed 77 characters, we change the sheet and let him start over. How many tries does he need to randomly reproduce the above paragraph?

There are 26 letters in the English alphabet and since he’ll be needing the comma and space, we’ll include those as well. So there’s a 1/28 chance of getting the first character right. Same goes for the second character, third character, etc … Because he’s typing randomly, the chance of getting a character right is independent of what preceded it. So we can just start multiplying:

p(reproduce) = 1/28 · 1/28 · … · 1/28 = (1/28)^77

The result is about 4 times ten to the power of -112. This is a ridiculously small chance! Even if he was able to complete one quadrillion tries per millisecond, it would most likely take him considerably longer than the estimated age of the universe to reproduce these two sentences.

Now what about the first word? It has only three letters, so he should be able to get at least this part in a short time. The chance of randomly reproducing the word “two” is:

p(reproduce) = 1/26 · 1/26 · 1/26 = (1/26)^3

Note that I dropped the comma and space as a choice, so now there’s a 1 in 26 chance to get a character right. The result is 5.7 times ten to the power of -5, which is about a 1 in 17500 chance. Even a slower monkey could easily get that done within a year, but I guess it’s still best to stick to human writers.

.This was an excerpt from the ebook “Statistical Snacks. Liked the excerpt? Get the book here: http://www.amazon.com/Statistical-Snacks-ebook/dp/B00DWJZ9Z2. Want more excerpts? Check out The Probability of Becoming a Homicide Victim and Missile Accuracy (CEP).