# Amazon poker

Investors are constantly checking the stock ticker, gamblers check the point spread, and everybody is forever checking their e-mail. For a writerly type like me, however, the unshakeable obsession is checking my Amazon sales rank. Amazon.com calculates a sales rank for every book listed on its Web site, and updates the ranking hourly. Here’s a graph of the hourly fluctuations in the ranking of my book Infrastructure: A Field Guide to the Industrial Landscape over the course of a single day last week:

At any given moment the current rank is listed in the “Product Details” section of the Amazon page for both the hardcover and the paperback editions. The graph above comes from a service called Rankforest, which also tracks both the hardcover and the paperback versions.

From an author’s point of view, there’s a lot of mystery in these numbers. How are the hourly rankings calculated, and what (if anything) do they mean? How do the rankings correlate with actual sales of the book? Amazon is not telling, and so authors and other interested parties have been left to speculate and experiment. The obvious experiment is to order a copy of the book and observe the effect of this purchase on the ranking. I welcome such experimentation with my book. (I would prefer that you do it with the hardcover edition, for which I get a more generous royalty.)

Morris Rosenthal of Foner Books seems to be the leading scryer of signs in this field. The interpretation of the rankings has also been discussed by Chris Anderson, the editor of Wired, in an article, a blog, and a book (current Amazon rank = 463, way above mine, dammit). It’s clear that the hourly ranking cannot simply reflect the number of copies sold within the past hour. After all, in any given hour the vast majority of books sell zero copies, and so there would be a gigantic tie for last place. The rankings also can’t be based in any simple way on total sales since publication, because the standings are much too volatile. The received wisdom is that each sale produces an uptick, followed by an exponential decay until the next sale.

I find all these speculations fascinating, but they are not what I want to write about today. My topic is even more trivial. I’ve been tracking my Amazon ranking for more than a year and a half now, ever since the hardcover edition was first listed in September 2005. (The paperback came out about a year later.) Here’s what the trend looks like:

The graph records the highest (i.e., numerically smallest) rank noted on each day, with rare gaps when I happened to be offline all day. Without question it would be fairer to use the daily average rather than the daily peak. But an author’s ego is a fragile thing, and so I have gone out of my way to make the outlook as rosy as I could.

Here are some of the actual numbers I recorded, for the month of April 2007:

   date          HB      PB
2007-04-01     138263   47878
2007-04-02     192152   22728
2007-04-03      29146   41862
2007-04-04      73628   57155
2007-04-05      37172   16948
2007-04-06     127858   12779
2007-04-07     171363   25212
2007-04-08      55256   20770
2007-04-09
2007-04-10      36333    7121
2007-04-11     112423   19015
2007-04-12     183063   40015
2007-04-13     225457   46781
2007-04-14     239879   29259
2007-04-15     142252   16030
2007-04-16      24803   39200
2007-04-17      93485   18939
2007-04-18     173691   26434
2007-04-19     217440   19360
2007-04-20      44426   17276
2007-04-21     213765   26652
2007-04-22      39014   21699
2007-04-23     150012   18598
2007-04-24      61301   46268
2007-04-25      33800   22474
2007-04-26      10603   39335
2007-04-27      19746   15984
2007-04-28      14027   19324
2007-04-29      16527   25999
2007-04-30       5844   63439


Notice anything out of the ordinary? What I’m looking at is not the overall pattern of rising and falling magnitudes but rather the inner patterns of digits within the numbers. As I’ve been writing down these rankings over the past 20 months, I have had the persistent impression of a peculiar overabundance of repeated digits. Just in this small sample we have 47878, 22728, 192152, 57155, 25212, 36333, 44426, 33800, 39335, 25999, and lots more. Is there something going on here? Is Jeff Bezos broadcasting secret signals hidden in the digit patterns of the Amazon sales rankings?

I consider myself a reasonably sophisticated probabilist. I know I’m not supposed to be shocked when I bring together 23 people and find that two of them share a birthday. And I know that when people try to generate a random sequence of digits by plucking numbers out of their imagination, the result is almost always too homogeneous, with a deficiency of repetitions and other patterns of the kind I’m calling attention to. Thus I was prepared to believe that the patterns I perceived were conjured up out of nothing—phantom regularities in purely random data. Still, day after day, I would note numbers like 55256 and 20770 (which in fact appeared on the same day, one for the hardcover, one for the paperback), and wonder about the odds of such coincidences. Maybe I wasn’t just letting my imagination run away with me.

Finally, this past weekend, I could stand it no longer; I had to find out.

I’m going to have to give away the punchline right now, lest it come as a disappointment later: There is nothing unusual about those numbers. The distribution of digits—the number of pairs and triples and what-not—is well within the range expected for randomly generated numbers of the same size. In other words, I have confirmed the null hypothesis. Often, such a negative result is considered unpublishable, but this one cost me a fair amount of effort, and so I’m determined to get a blog item out of it for better or worse. If the conclusion itself is not very interesting, perhaps I can find something to say about the techniques and technologies that led to it.

How do you decide whether a number like 36333 is too unusual to be a product of random processes? I decided to analyze the numbers in my sample as if they were poker hands (with the rules of poker adapted to a deck of cards with ten ranks and no suits). I confined my attention to the five-digit numbers, which are the most numerous in my sample; of the 850 rankings I had recorded, 458 were in the range between 10,000 and 99,999. Then I wrote a five-line program that takes any such number and classifies it as one of seven types of poker hands:

• five of a kind (e.g., 77777)
• four of a kind (36333, 11119)
• full house (44555, 28288)
• three of a kind (57155, 20900)
• two pairs (97097, 28002)
• one pair (36739, 14912)
• bust (53208, 16897)

I decided to ignore straights; the poker hand 56789 is simply a “bust” according to this scheme of classification. The concept of a flush—all cards of the same suit—doesn’t arise, since there are no suits.

When I ran my little program over the 458 five-digit Amazon sales rankings, here’s what I got:

   hand            number       frequency

five of a kind        0          0.0000
four of a kind        3          0.0066
full house            7          0.0153
three of a kind      34          0.0742
two pairs            51          0.1114
one pair            233          0.5087
bust                130          0.2838

TOTAL            458          1.0000


The question now, of course, is what I should expect to see in such a data set, on the hypothesis that the digit patterns are random rather than contrived for some secret, nefarious purpose. Should I find it remarkable that half of the hands have a single pair, or that more than 70 percent have a pair or better? What about those seven full-house hands—should that abundance arouse suspicion? To begin answering questions like these, we need to calculate the probabilities of the various hands.

Having just boasted of my sophistication as a probabilist, I must now confess that the main thing I’ve learned about calculating probabilities is that getting the right answer is highly improbable. I understand the principle of the thing. It’s just a matter of counting. You count the “success” cases and divide by the total number of cases. But counting—even though it tends to come early in the mathematical curriculum—is not always easy.

In the Amazon poker problem, the first trap for the unwary is in counting the total number of cases. You might think that with five decimal digits, there would be 105 possible arrangements, but in fact 104 of those arrangements are excluded from the sample, because numbers in this context cannot have a leading digit of zero. Thus the denominator in the probability calculation will be 90,000 rather than 100,000.

I think I can trust myself to calculate the probability of a five-of-a-kind hand. The first card dealt must not be a zero but can be any of the other nine digits; thereafter, each subsequent digit must be identical to the first one. Thus the number of ways of forming a five-of-a-kind hand is 9×1×1×1×1, and the probability of this outcome is 9/90,000, or 0.0001. You can expect five of a kind in one hand out of every 10,000, if the deal is fair.

I believe this answer is correct, and I am proud of having obtained it; on the other hand, the five-of-a-kind calculation is by far the easiest case. Allow me to try to work out a harder problem: the odds of a full house. I invite you to listen in on my so-called thought process:

Well, a full house is a hand that matches the pattern aaabb. The first digit can be anything but zero, and so there are nine possibilities, but then the second and third digits have to match the first. The fourth digit must differ from the first three, and so there are eight candidates left…. No, wait…. This time zero is allowed, and so there are nine possibilities again. Then the fifth digit has to be identical to the fourth. That gives us the product 9×1×1×9×1, or 81 successes out of 90,000 total cases, for a probability of 0.0009…. Did I get that right?… Of course not. What I’ve calculated is the probability of seeing the pattern aaabb in that precise sequence; I am counting occurrences of 11100, 11122, 11133, …, 99988, but I am not including sequences such as 23233 or 45454. Okay. So one approach, now that I have the number of aaabb sequences, is to multiply by the number of ways of permuting that sequence. We can just take the aaa and intercalate the two b‘s in all possible positions: bbaaa, babaa, baaba,…. No, wait…. The b could be a zero, and so it’s not always allowed to appear in the first position. Hmmm. This is not going well. For the time being, let’s forget about the prohibition of leading zeros, and we’ll correct for it later. That way we can consider all possible permutations of aaabb. As a first step, take aaa, where a can be any of the ten decimal digits, and place a b is all possible positions, allowing b to assume any value that differs from a, so that there are nine choices. There are four places to put the b: baaa, abaa, aaba and aaab. Now, for each of these four sequences, the second b can be placed in any of five positions, and so there are 4×5 = 20 permutations overall. Which means that the total number of full houses is…. Hold on. No, no, no, no, no. Not all those 20 permutations are distinguishable. If we start with baaa and insert a b in either the first or the second slot, we wind up with bbaaa in either case; this sequence should not be counted twice. So how many permutations are there, really? Offhand, I don’t see any way to count them that’s easier than direct enumeration: bbaaa, babaa, baaba, baaab, abbaa, ababa, abaab, aabba, aabab, aaabb. That’s ten cases. Let’s sum up. We know that a can take any of ten values and b has nine possible values, and there are ten ways of arranging three a‘s and two b‘s. Thus the total number of combinations is 10×9×10 = 900. But, don’t forget, now we have to subtract away all those sequences that start with a zero. For this purpose it doesn’t matter whether the first symbol is an a or a b; exactly 10 percent of the sequences will have a leading zero. Thus the number of full houses is 900–90=810. This gives a probability of 810/90,000 = 0.009.

I happen to know, from an independent calculation, that this answer is correct, and so it’s time to stop. If I didn’t know, however, I might well go on to ask whether we need to interchange the a‘s and b‘s—that is, consider the case of sequences aaabb where a has only nine allowed values and b has ten. (Why don’t we have to take that into account?)

I am mildly embarrassed to put this lurching, stumbling, caricature of a probability calculation on public exhibition, and yet it is a fair description of how I often struggle with a problem of this kind. Am I the only one who suffers so? Not everyone does. I know people who could carry out the same computation quite deftly; they command the intuition, the spürkraft, to zero in immediately on the right approach, like a chess player who doesn’t waste time considering fruitless moves. I admire that kind of finesse, but I don’t possess it.

On the other hand, I know something else. I know another way to solve the problem, with less fuss. As I mentioned above, I have already cooked up a little program that can take any five-digit Amazon rank and classify it into one of the seven categories of poker hands. In milliseconds I can run that program on all possible five-digit numbers; after all, there are only 90,000 of them, and it’s quite easy to generate all of them in sequence. Then I can just count the number of hands in each category, and all the probabilities come tumbling out in a neatly formatted table:

   hand            number       frequency

five of a kind        9          0.0001
four of a kind      405          0.0045
full house          810          0.0090
three of a kind    6480          0.0720
two pairs          9720          0.1080
one pair          45360          0.5040
bust              27216          0.3024

TOTAL          90000          1.0000


If Fermat or Pascal or the Bernoullis were asked their opinion on this approach to probability, something tells me they would find it distasteful. I’m ambivalent myself. Returning to the chess analogy, this is the equivalent of the machine that beats a grandmaster by brute force, scanning millions of positions but knowing nothing of strategy. You can win that way, but you don’t get any style points. More important, the method is frustratingly opaque. I get the answers, and I have reasonable confidence that they’re correct, but the computation gives me no understanding of why they’re correct. Still another objection is that the method does not scale well. If my Amazon rankings had ten digits instead of five (perish the thought!), I’d have a hard time classifying the nine billion possible hands. (But then again I’m not sure the more analytic method scales all that well either.)

Setting aside these qualms, we can now take the predictions of theory and the observations of the Amazon sample and compare them side by side:

   hand            prediction       observation

five of a kind        0.0001          0.0000
four of a kind        0.0045          0.0066
full house            0.0090          0.0153
three of a kind       0.0720          0.0742
two pairs             0.1080          0.1114
one pair              0.5040          0.5087
bust                  0.3024          0.2838

TOTAL              1.0000          1.0000


Some of these frequencies match quite closely (three of a kind, one pair); others are a bit off the mark (full house, bust). In a finite sample, of course, you would never expect an exact match to the theoretical frequencies—but are the discrepancies we’re seeing significant or not? My personal instinct says that there’s nothing amiss here, that the observed frequencies are consistent with the null hypothesis. In other words, the rankings could just as well be random numbers. The old rule of thumb that the variation should be less than the square root of the observation leads to the same conclusion. We could quantify these intuitions by calculating variances or standard deviations, doing a Χ2 test, and so on. But if I can barely calculate a simple probability, can I be trusted to navigate all those treacherous subtleties such as choosing the correct number of degrees of freedom?

Again there’s another way to go about it, relying on lots of ignorant computation to replace a little smart mathematics. The question we want to answer is this: Given a set of 458 randomly generated five-digit numbers, what is the probability that the random set will differ from the predicted frequencies by at least as much as the observed Amazon set? Suppose we measure distance from the theoretical prediction in terms of the sum of the squared differences:

where the index i ranges over the seven types of poker hand, and the expression in parentheses is the difference between the observed and predicted frequency for each type of hand. (Technical note: The seven numbers entering into this statistic are not independent. For example, if full-house hands are in surfeit, there has to be a compensating deficiency somewhere else. How much should I worry about this?) For the actual Amazon results, S2 works out to 4.267×10–4. Now we can generate lots of batches of random Amazon poker hands, each batch consisting of 458 five-digit numbers, and calculate S2 for each batch. What proportion of them will have an S2 value exceeding 4.267×10–4? The answer, based on half a million batches, is 80 percent. Thus all the anomalies I thought I was seeing in those numbers are pure delusion.

Needless to say, I was rooting for another outcome. I would have enjoyed finding something spooky and inexplicable in the Amazon rankings. Instead, all I’ve proved is that my intuition about what random numbers look like is not to be trusted. This is a disappointment, but maybe I can salvage something from the ruins. Perhaps I can claim the discovery that the Amazon rankings are a fairly good source of random numbers.

Large-scale differences and movements in the rankings are surely not random. It’s not purely a matter of chance that my book’s current rank is 43,888 (what an interesting number!), while some preposterous tale about a pubescent wizard occupies the top of the list—even though that book hasn’t actually been published yet. (Do I sound bitter?) The rankings are nonrandom in another way as well: The first digits are not uniformly distributed but have a Benford or Zipf distribution, with an excess of ones and a shortage of nines. It’s interesting that even though the randomly generated numbers do not share this property—the first digits have a uniform distribution over the range 1 through 9—the poker-hand analysis shows that the frequency of pairs, triples, and other patterns is identical in the two data sets. Thus the digits to the right of the first digit do seem to have the statistical properties of random numbers. The source of the randomness is presumably the hourly reshuffling of the rankings by the actions of thousands of Amazon shoppers—actions that are all too predictable in the aggregate (curse you, Harry Potter) but quite random in detail.

Perhaps it’s worth noting that when the RAND Corporation prepared their famous book A Million Random Digits with 100,000 Normal Deviates in 1955, they also employed the poker test as a measure of randomness. I’m relieved to find that their calculation of the theoretical frequency of the seven types of hands agrees with mine. (They don’t say how they performed the calculation.) The current Amazon sales rank for the book is 2,668,928 (hardcover) and 281,270 (paperback).

This entry was posted in books, computing, mathematics, modern life.

### 9 Responses to Amazon poker

1. Carl Witty says:

Actually, the results using a Benford distribution are not the same as the results using a uniform distribution (although they are admittedly very very close). With a uniform distribution, the probabilities are given (as exact, terminating decimals) above; with a Benford distribution, the probabilities are:

hand prediction

five of a kind 0.0001105728747…
four of a kind 0.004763041358…
full house 0.009931354354…
three of a kind 0.0736751048…
two pairs 0.1092466737…
one pair 0.503990913…
bust 0.298900149…

Also, with a Benford distribution, the probability of a 0 as the second digit is 11.9679…%, and the probability of a 9 as the second digit is only 8.4997…%.

2. John S. says:

I had not heard about your book before, but it looks very interesting. My first job out of college was with a company that didn’t like to spend money on airfare, so I had to do a lot of cross-country driving with my boss. Whenever we would pass an interesting-looking factory or some other large piece of industrial infrastructure, he would point it out and ask us young engineers “What’s that?” or “What do you think they make there?” Guessing wasn’t good enough; we had to explain our answers. “How do you know?” he would ask, or “What makes you say that?”

I learned a lot from that guy!

3. Jonathan Katz says:

It seems an easier way to calculate the probability of a full house is as follows: there are 10=(5 choose 3) ways that 3 ‘a’s and 2 ‘b’s can appear in a sequence of 5 items. We then have 9 choices for whatever letter appears in the first position (regardless of whether it’s an ‘a’ or a ‘b’) and are then left with 9 choices for the other letter. This gives a total of 10*9*9=810 ways to make a full house, in agreement with your calculation.

4. brian says:

To Carl Witty:

Many thanks for the correction. Can you say a word about how those numbers were calculated?

5. brian says:

To John S.:

Far be it from me to use my blog to promote my book! :)

As it happens, the genesis of the book was very much like the highway experience you describe. The difference is that the inquisitor continually asking “What’s that?” was my young daughter.

6. brian says:

To Jonathan Katz:

As I said, some people can carry out such a calculation quite deftly. It all looks easy — almost inevitable — once you see it done.

7. Carl Witty says:

I computed the probabilities with a short SAGE program (SAGE is a computer algebra system using Python as a programming language, and providing bindings for many many other computer algebra programs: http://www.sagemath.org/).

I put the code I used here: http://sage.math.washington.edu/home/cwitty/bitplayer.py

The code is short, rather than fast or readable. Note that the computations use interval arithmetic, giving guaranteed (barring bugs) bounds on the probabilities in question; it would be trivial to use higher-precision interval arithmetic, providing as many digits of the correct answer as were desired.

To run the code, install SAGE and run it; then type:

sage: hands()
sage: dig2()

8. Barry Cipra says:

Here’s another fun statistical test you can run on your data. Keep track of the sequence of ups and downs in your rankings. For example, your April hardback sequence goes

ududuudduuuudduuudududddudud

(i.e., it goes up from 138263 to 192152, then down to 29146, etc.) Note there are 15 ups and 13 downs. Let’s compare this to a “random” string of 28 binary digits, say

0010010000111111011010101000

(the astute bit-player will notice the fractional part of pi written in binary…). By happenstance (or perhaps Jeff Bezos has a hand in this as well), this string has 15 zeroes and 13 ones, so it’s quite comparable to your ups and downs. Offhand they look a lot alike. One way they differ, though (aside from u’s and d’s versus 0′s and 1′s), is in how often consecutive entries are the same as opposed to different. For the random string, the entries stay the same 13 out of 27 times (thanks in part to that long, non-random-looking string of 1′s). For your ups and downs, it happens only 10 times.

Is this significant? I’ll let someone else run a chi-square test on it, but I’ll argue here that, if anything, your sequence should have stayed the same 9 out of 27 times, not 10, and changed the other 18. That’s because your ups and downs are not coin flips. For the random string, each entry could care less what came before it, so it should agree with it half the time and disagree half the time, which 13 out of 27 tries its best to do. But your ups and downs are based on numbers which, let’s assume, are drawn at random from within some range. A little analysis, based on independent draws of random numbers in the interval [0,1], shows one expects a reversal of fortune not half the time, but two-thirds of the time. This makes sense, at least qualitatively, because going up means you probably picked a big number, which makes it more likely you’ll pick something smaller on the next round, and vice versa.

I became aware of this (and did the simple analysis) about 15 years ago, when I was still working with my first computer, a Macintosh Plus, the one with a full megabyte of RAM. The screensaver it came with was the old happy Macintosh icon, which jumped around an otherwise dark screen. I found myself staring at it one day while I was supposedly thinking, and noticed that it tended to reverse its left-right direction of motion much more often than not. (It did the same in the up-down direction as well.) I won’t confess to how long I spent staring in fascination at that screensaver. Let’s just say I collected a lot of data before I did the analysis.

It would be of idle interest to see what your complete sequence reveals. One final note: I elided the gap in your April data, giving a down trend from the 7th to the 8th to the 10th before the uptick on the 11th — i.e., …ddu… If the missing ranking bounced up from 55256, the sequence would go …dudu… — for 9 out of 28 consecutive entries staying the same. On the other hand, if the missing data were less than 36333….

9. John S. says:

This post recalls the recent “controversy” over whether the iPod’s shuffle algorithm was truly random or not. See for example
http://www.msnbc.msn.com/id/6854309/site/newsweek/