Is James Dyson held back by the speed of sound?

I was intrigued by a story in the Daily Telegraph about a new electric motor created by Dyson. The DC motor apparently rotates at 104,000 RPM and is to be used in a portable vacuum cleaner.

The motor technology itself is switched reluctance. Essentially, the motor works by turning on and off electromagnets at just the right time to keep the rotor inside the motor spinning.

My immediate thought was 'how fast is the outside edge of the rotor moving if it's spinning at 104,000 RPM?' And shortly after that, 'how close is that to the speed of sound?'

In Electronic Weekly there's an article which states that the motor is 55.8mm across. Now, that's probably not the diameter of the rotor, but given that Dyson is attaching an impeller to the rotor anyway I'm going to take that as the diameter and work my calculations from there.

So the distance travelled in one rotation is π * 55.8mm and there are 104000 / 60 rotations per second. So, the outside is moving at 304ms^-1.

The speed of sound at sea level is 340ms^-1.

So the impeller is likely operating at near the speed of sound. I wonder if there are any nasty effects of rotating at that speed and if Dyson is close to the theoretical limit of what he can do.

There are two patent applications from Dyson that I believe cover this invention: 20070252551 and 20070278983. Neither mentions the speed of sound.

How to do customer service

I've previously complained about poor technical support that I received from Hewlett-Packard. That particular incident isn't over yet... the issue has been escalated a couple of times, HP has told me they are end-of-lifeing the product, ... I'll write that up when it comes to a resolution.

But it's not all moaning! Two companies that have provided excellent customer service recently are Apple and Bugaboo. I dealt directly with Apple myself, a friend with small children told me about the Bugaboo goodness.

First off, Apple. I own a MacBook Pro that I bought in mid-2007. Unfortunately, it suddenly started to suffer from the NVIDIA GeForce 8600M GT problem a couple of months ago. The upshot was that my machine would boot but couldn't find a display adapter (or at least it found the Intel display adapter, not the NVIDIA one).

I verified that I could ssh into the machine and ran System Profiler on the command-line. A quick search by serial number showed that my machine was susceptible to this problem and that Apple offered free service.

So, I called AppleCare. I never bought AppleCare for this machine and for this problem I didn't need it. I described my problem in detail to the technician including the steps that I'd taken to try to resolve it (including resetting the PRAM and SMC) and he did something great. He completely avoided going through any script, realized that I knew what I was talking about and immediately set the machine up for repair.

Next step was an appointment with the Genius Bar. This was the most annoying part because Apple's Concierge software is poorly designed. But once at the Genius Bar I got my appointment in about 10 minutes of the allotted time. The technician immediately verified that I had the NVIDIA problem and that I was eligible for a motherboard replacement.

While I was chatting with him I mentioned that my iPhone headphones had a fault and I wanted to buy some new ones. He asked me how long I'd had the iPhone (about 3 months) and simply went and got me a new pair, for free, just like that.

Then he told me to expect that my MacBook Pro would take about a week to repair. I left the Apple Store and went into work. That evening Apple called me to tell me the laptop was ready.

Nice.

Now Bugaboo. My friend Bill has two small kids and one of them is always in a Bugaboo Cameleon stroller. These are really high-end and expensive bits of kit. But they are very, very well made.

Now Bill's Bugaboo's brakes had developed a fault. They didn't always work and it was a minor annoyance. Little did Bill know that Bugaboo had identified this as a common fault and recalled the Cameleon.

Happily, Bill had filled out the warranty card for the stroller and sent it back when he bought it. One day a small package arrived unannounced containing a kit to fix the brakes. The kit worked perfectly.

Nice.

In both cases, Apple and Bugaboo, we were dealing with premium brands and got premium support. Apple's ability to just give me new headphones made my experience wonderful, and Bugaboo simply sending the repair kit to Bill made him a loyal customer for life (he just needs to have some more kids).

The 1944 US Presidential Election was fraudulent

OK, it wasn't really, but I thought I'd run the Scacco/Beber analysis on that election and see what it comes up with. Guess what.

If you look at the non-adjacent, non-repeated digits in the last two places in the votes counts by state for Roosevelt and Dewey you discover that 59.38% of the votes are non-adjacent, non-repeated. If the numbers were truly random you'd expect 70%. That's way worse than the 62.07% in the Iranian election.

If you then do the old Z-Test you get a Z value of -2.49 with a p-value of 0.013. That's well below the 0.05 critical value so you can reject the null hypothesis. The final digits are not random.

Is this fraud?

Is there any suggestion that the state-level numbers in the 1944 US election were invented by people?

If not, how can anyone claim that this test indicates fraud in the Iranian election?

Now run the other bit of their test looking at the frequencies of the last digit. You get 'too many' 7s (expected 10%, got 16%) and 'too few' 1s (expected 10%, got 5%).

I'm telling you, man, what's the chance of that happening, and the non-adjacent, non-repeating digits thing? (It's about 0.17% according to simulation) I mean, come on, that's gotta be fraud.

Oh, wait, it's not.

The Iranian Election Detector

OK, I thought I was done criticizing the Washington Post Op-Ed about how statistics leave 'little room for reasonable doubt' that the Iranian election was fraudulent. But then Hannah Devlin at The Times did her own analysis and it got me thinking about the errors in that article again.

Firstly, my previous post talks about the right way to determine whether the digits are random or not, I'm not going to go over that again, but I am going to go back over some of the actual figures that are presented in the article.

So begin with this quote:

But that's not all. Psychologists have also found that humans have trouble generating non-adjacent digits (such as 64 or 17, as opposed to 23) as frequently as one would expect in a sequence of random numbers. To check for deviations of this type, we examined the pairs of last and second-to-last digits in Iran's vote counts. On average, if the results had not been manipulated, 70 percent of these pairs should consist of distinct, non-adjacent digits.

Not so in the data from Iran: Only 62 percent of the pairs contain non-adjacent digits. This may not sound so different from 70 percent, but the probability that a fair election would produce a difference this large is less than 4.2 percent.

And there's a footnote:

This is a corollary of the fact that last digits should occur with equal frequency. For an arbitrary second-to-last numeral, there are seven out of ten equally likely last digits that will produce a non-adjacent pair. Note that we treat both 09 and 10 as adjacent.

Firstly, I believe they mean to say that they treat 09 and 90 as adjacent (not 09 and 10). That means that for any number there are two possible adjacent digits out of a ten, in other words 20% of digit pairs are adjacent, so 80% of digit pairs are non-adjacent.

In their article they say 70% 'distinct, non-adjacent'. OK, so their definition of non-adjacent means that you need to exclude repeats as well (so 23, 32 and 33 are all to be excluded).

They then present the argument that a figure of 62% or less will only happen in 4.2% of fair elections. Nowhere do they explain how they derived this figure, so I decided to run a simulation. (Hannah Devlin argues that this number is incorrect in her article, worth a read)

I ran a simulation of 1,000,000 elections that generate 116 counts of votes and I looked at the adjacent pairs of numbers in the vote counts and then I calculated the percentage of fair elections that would result in the same 62% or less as seen in the Iranian election. The figure is 2.66%. 2.66% of fair elections would produce the result (or 'worse') seen in Iran.

The difference, 4.2% vs 2.66%, comes about because the figure that they must have used is not 62%, but 62.07%. That is the actual number, to two decimal places, that comes from analyzing the digit distribution in the Iranian election results.

(Email me if you want my source code)

So, what does that tell you? That in almost 3 in 100 fair elections we would have seen the result in Iran. Or if you use their numbers 4 in 100. Either way that's pretty darn often. In the 20th century there were 26 general elections in the UK. Given their 4/100 number is 1/25 we shouldn't be at all surprised if one of those general elections looked fraudulent!

Now, we expect that the percentage of non-adjacent digits is normally distributed. And, in fact, my little simulation shows a nice little normal distribution centered on 70 with a standard deviation of 4.27.

So, we've got normally distributed data, a mean and a standard deviation and a sample (62.07%). Hey, time for a Z-test!

For this situation the Z value is -1.86 which yields a p-value of 0.063 for a two-tailed test (I'm doing two-tailed here because what I'm interested in is the deviation away from the mean, not the specific direction it went in). That's above the 0.05 value typically used for statistical significance and so we can't from this sample determine that there's statistical significance in the 62.07% figure.

So, I'd say that based on the figures given I can't find statistical significance. So I don't learn anything from that about the Iranian election.

Given that the Z-test on their 'non-adjacent, non-repeated' digits test doesn't find statistical significance, and my previous piece showed that the chi-squared test on the other claim in their paper didn't find statistical significance (that was on the randomness of the last two digits).

You might be scratching your head wondering how the authors made the claim that this was definitely fraud (their words: 'But taken together, they leave very little room for reasonable doubt.')

Well, what they do is take the probability of seeing the 62% or less number in a fair election (4.2%) and multiply it by the probability of seeing the specific variance they see in the digits 7 and 5 in a fair election (4%) to come up with 1.4% likelihood of this happening in a fair election:

More specifically, the probability is .0014 that a fair election (with 116 vote counts) has the characteristics that (a) 62% or fewer of last and second-to-last digits are non-adjacent, and (b) has at least one numeral occurring in 17% or more of last digits and another numeral occurring in 4% or fewer of last digits.

That's a very specific test. In fact, it's so specific that I'm going to name it the "Iranian Election Detector". It's a test that's been crafted from the data in the Iranian election results, it's not the test that they started with (which is all about randomness of digits, and adjacency).

So, let's accept their 1.4% figure and delve into it... that's 1.4 in 100 elections. That's roughly 1 in 71. So, they are saying that their test would give a false positive in 1 in 71 elections.

How is that 'leaving little room for reasonable doubt'?

Running the numbers on the BBC executives expenses

So, another lovely data set appeared, the expenses of senior BBC executives. And the papers went a little wild highlighting the spending that they don't like.

Me, I just punched the numbers into a little program and took a look at how well they fit Benford's Law. I wanted to see if there were any interesting anomalies to look at. And there are. No smoking guns, though. Just some fun on a Number 22 bus with some numbers.

First, here are the chi-squared values for the fit of first digits of the expenses for 2007/2008 by executive. The critical value (for p = 0.05) is 15.51, so most of the expenses do not fit Benford's Law. The fun is in finding out why.

Ashley Highfield,3.06874517880956
John Smith,4.61752636949634
Mark Byford,15.5817014457229
Jana Bennett,15.6178982350545
Zarin Patel,17.5034731417114
Jennifer Abramsky,20.8803339214804
Mark Thompson,22.4588511455346
Caroline Thomson,37.666988157616
Timothy Davie,143.433388695613
Stephen Kelly,178.662639451409

The best fit is Ashley Highfield's lovely curve:



We'll come back to Mr Highfield later, but let's go to the other end of the spectrum and look at the extremes that don't match the expected. The 'worst' offender is Stephen Kelly (he's not with the BBC anymore). Here's his curve.



Whoa. What happened with all those 8s? Delve into the data and you find lots of £8.00 claims for "Road/Bridge Tolls". My guess is that Mr Kelly passed through the London Congestion Charging Zone in his own car. That's enough to skew the data. And if you match up his £8.00 charges and his mileage claims it all makes sense.

Now to Timothy Davie and here's his curve:



So, he's like Stephen Kelly and sure enough there are lots of £8.00 charges for the same "Road/Bridge Tolls".

Next on ths list comes a different pattern created by Caroline Thomson. An excess of 1s:



She's got a ton of taxi trips in the £10 to £19 range. According to Transport for London you'd see those fares on a weekday when traveling around 4 miles in central London. Given the location of BBC Television Centre it's pretty easy to imagine the need for these trips. Also, she doesn't claim any mileage or congestion charge so she's not using her own car.

Next up is the Big Kahuna Mark Thompson. His curve shows an excess of numbers 6, 7 and 8.



Why is that? Well, if you look at his expense claims line by line (and if you do you're a total nerd) you'll see that Mr Thompson takes people out to lunch a lot and spends a lot of money on lunches under £100. You can imagine this being totally legitimate. He probably has to do that for his job, there could be a BBC guideline about how much to spend on lunch, or Mr Thompson could simply have a moral compass that says he shouldn't go totally wild on lunch costs.

He doesn't have lots of taxis or tolls, but then again there's a note in his expense report where he did take a taxi that says "Driver not available" so I'm guessing he has a chauffeur.

And so it goes on. You can carry on down the list and look for little anomalies, but there's nothing glaring.

So, how come Ashley Highfield has such a perfect curve? Well, he doesn't take a lot of cabs (so no excess of 1s), drives his own car (lots of little mileage claims) and doesn't seem to claim the congestion charge (no excess of 8s). Did he forget to claim the congestion charge, or does he drive an electric car?

He was, after all, the BBC's Director of the Future Media and Technology.

The Scacco/Beber analysis of the Iranian election is bogus

OK, I wasn't going to write another blog entry about the 2009 Iranian election, but the article in the Washington Post that supposedly gives statistical evidence for vote fraud just won't die in the blogosphere and just got a boost from a tweet by Tim O'Reilly.

The trouble is the analysis is bogus.

The authors propose a simple hypothesis: the last and second-to-last digits of vote counts should be random. In statistical terms this is often called uniformly distributed, which just means that they are each equally likely. So you'd expect to see 10% 0s, 10% 1s, 10% 2s, and so on.

Of course, you only expect to see that if you had an infinite number of vote counts because the point about random processes is that they only 'even out' to the expected probabilities in the long run. So if you've got a short run of numbers you have to be careful because they won't actually be exactly uniform.

To confirm that try tossing a coin six times. Did it come up with exactly 3 heads and 3 tails? Probably not, but that doesn't mean it's unfair.

Now, given some run of numbers (vote counts for example), the right thing to do is ask the statistical question "Could these numbers have occurred from a random process?" If they couldn't then you can go looking for some other reason (e.g. fraud).

The question "Could these numbers have occurred from a random process?" is given the ugly name the 'null hypothesis' by stats-heads. That just means that thing you are testing.

More concretely, the Scacco/Beber null hypothesis is "the last and second-to-last digits in the vote counts are random". What you want to know is with what confidence can you reject this, and for Scacco/Beber rejecting means fraud.

Now, what you don't do is go count the last and second-to-last digits, look for some that have counts that deviate from what you expect (the exactly 10% figure) and then try to work out how often that happens. That's like tossing a coin a few times, noticing that heads has come up more than 50% of the time and then starting to think the coin is biased.

Unfortunately, that's essentially what Scacco/Beber did. They picked on two numbers that lay outside their expected value and went off to calculate how frequently that would occur. That's cherrypicking the data.

What you do do is apply a chi-square test to figure out whether the numbers you are seeing could have been generated by a random process. And you use that test because it gives you the probability with which you can reject your null hypothesis.

To prevent you, dear reader, from having to run the test I've done it for you. I took their data and wrote a little program to do the calculation against the last and second-to-last digits. Here's the program:

use strict;
use warnings;

use Text::CSV;
my $csv = Text::CSV->new();

my %la;
my %sl;

foreach my $i (0..9) {
  $la{$i} = 0;
  $sl{$i} = 0;
}

my $count = 0;

open I, "<i.csv";
while (<I>) {
  chomp;
  $csv->parse($_);
  my @cols = $csv->fields();
  for my $i (@cols[1..4]) {
    my @d = reverse split( //, $i );
    $la{$d[0]}++;
    $sl{$d[1]}++;
    $count++;
  }
}
close I;

print "Count: $count\n";

my $e = $count/10;

my $slchi = 0;
my $lachi = 0;

foreach my $i (0..9) {
  print "$i,$e,$sl{$i},$la{$i}\n";

  $slchi += ( $sl{$i} - $e ) * ( $sl{$i} - $e ) / $e;
  $lachi += ( $la{$i} - $e ) * ( $la{$i} - $e ) / $e;
}

print "slchi: $slchi\n";
print "lachi: $lachi\n";

Here's a little CSV table that you can steal to do your own analysis:

Digit,Expected Count,Second-to-last Count,Last Count
0,11.6,10,9
1,11.6,9,11
2,11.6,15,8
3,11.6,6,9
4,11.6,11,10
5,11.6,11,5
6,11.6,14,14
7,11.6,18,20
8,11.6,13,17
9,11.6,9,13

And true enough I get the same figures as Scacco/Beber. The number 7 does occur 17% of the time in the last digit, and the number 5 only occurs 4% of the time. But, I don't care. What I want to know is, is the null hypothesis wrong. Could these results have occurred from a random process? And with what likelihood.

So here's where I avoid staring at the numbers (which can get to be borderline numerology) and do the chi-square test.

For the last digit the magic chi-square number is (drum roll, please): 15.55 and for the second-to-last digit it's 9.33. Then I go to my chi-square table and I look at the row for 9 degrees of freedom (that corresponds to the 10 possible digits; if you want to know why it's 9 and not 10 go read up on the subject) and I see that the critical value is 16.92.

If either of my numbers exceeded 16.92 then I'd have high confidence (greater than 95%) that the digit counts were not random. But neither do. I cannot with confidence reject the null hypothesis, I cannot with confidence say that these numbers are not random, and I cannot with confidence, therefore, conclude that the vote counts are fraudulent.

What this means is, is that there is no 'statistically significant' difference between the Iranian results and randomness. So, what we learn is that this statistical analysis tells us nothing.

It doesn't mean that the numbers weren't fiddled, it just means that we haven't found evidence fiddling.

PS In the notes added to their annotated version of the article Scacco/Beber mention that they did the chi-square test and got a p-value of 0.077. This is below the 'statistical significance' cut off of 0.05 and so their results are (as I find) not statistically significant.

To put 0.077 in context it means that there's a 7.7% chance that the digits are random. Sounds small but 7.7 is approximately 8 in 100 or 4 in 50 or 2 in 25 or ... 1 in 12.5. i.e. in 1 in every 12.5 fair elections we shouldn't be surprised to see the sort of figures we saw in Iran. That's pretty often! That's why chi-square tells us not to find non-randomness in the Iranian results.

30 June 2009 Update: I've removed that paragraph because that interpretation of the p-value is arguably inaccurate and if you are a statistician you'd probably shout at me about it. Doesn't change the fact that the data says the Iranian result is not statistically significant; it just says that my attempt to do a 'layman's version' is faulty.

To come up with better layman's version I ran a little simulation to find out how often you'd expect to see one digit occurring more than 17% of the time with another occurring less than 4% of the time (as in the Iranian election). The answer is about 1.48% of the time, or in about 1 in 67 fair elections.

Britannica.com makes me want to weep

I got a marketing mail from Britannica.com trying to entice me back after I canceled my subscription. So, I figured I'd just go take a quick look at a random Britannica entry and remind myself of what I was missing. Nightmare.

On the Britannica.com home they were mentioning that their article about the US Voyager program was featured and I could see it for free. So I clicked.

This featured article contains 503 words that give the briefest of introductions to Voyager. The related articles are all about the planets that Voyager passed, and there's a connection to a general article about space exploration. There's absolutely no drill down to explore Voyager in any depth.

Of course, I whizzed over to Wikipedia and looked up the same subject. The main article contains 2,009 words and links to in-depth articles about Voyager 1 and Voyager 2. And there are links to interesting articles about their voyages, their power systems, the Voyager Golden Record and more.

And Wikipedia links you straight to the definitive source for Voyager information: NASA's Voyager Program page. Britannica doesn't link; they choose to link to a small collection of images of the Voyager craft from NASA's web site.

So, basically Britannica.com's article is close to useless because it's a dead-end and a short dead-end at that. In contrast, Wikipedia's article is rich, links to even more information and lets me get to source material.

And if that's not enough Britannica.com's page is infested with distracting ads. The worst of these are the weird keyword-linked ads buried right inside the article itself.

It looks like you might be able to click on, say, solar system in the article to drill down. Far from it! Hover over solar system and you get the following irrelevant, useless, pop-up ad.



Pure genius, Britannica.com. Pure, pure genius.

Now, Britannica.com's article does contain some drill down, but some of it is useless. For example, the Voyagers each contain a phonograph record with a recording of sounds from Earth (language, music, etc.). On the Britannica.com page the words phonograph record are a link. Click through and they will tell you what a phonograph record is, not about the ones on board the Voyagers. Thanks, I'm old enough to know what a phonograph record is.

So, Britannica.com, now you know why I donate money to Wikipedia, and don't buy your service.

Michael Faraday criticizes 'security theatre' from beyond the grave

I was reading David Knight's book Humphry Davy and at one point he describes the arrival of Davy and Faraday in France in 1813:

On arrival, Faraday reported, they were searched, an unusual experience for a true-born Englishman: 'he then felt in my pockets, my breast, my clothes, and lastly, desired to look into my shoes; after which I was permitted to pass', and could hardly help 'laughing at the ridiculous nature of their precautions'.

Lucky he doesn't have to fly anywhere.

The Turing Test and Prejudice

Yesterday, I blogged about how I believed that Alan Turing deserves an apology for the way he was treated. A few people asked how we should apologize. Here's what I would say.

Alan,

I am sorry for the way you were treated. I am sorry that Britain treated a man of your genius, a hero of the Second World War, so despicably.

You laid the foundations of computer science, you helped break the Nazi Enigma code, and you were cut down in your prime because of prejudice. In your death we lost a great man who no doubt had much more to offer the world.

In 1950 you published a paper in which you proposed a way to determine if a machine was intelligent or not. This has become known as the Turing Test, and it has wider implications that just machine intelligence.

Your test involved asking a person to distinguish between a human intelligence and a machine intelligence by removing prejudice. You limit the judge to communicating with the human and machine via an intermediary (you proposed a teleprinter) so that the judge is unable to see or hear who they are communicating with. The judge is limited to judging intelligence alone. If the judge cannot tell the difference then the machine is deemed intelligent.

But replace the machine in your test with another human with some supposedly undesirable characteristic. Your test can pit a straight man against a gay man, a white man against black man, a man against a woman. I imagine if you and I were hidden behind the teleprinters, that you would be determined to be more intelligent.

Without the prejudice of knowing your sexuality, skin colour or sex, only your true values come through in the Turing Test.

At the end of that paper you write "We can only see a short distance ahead, but we can see plenty there that needs to be done." You took your own life four years later, after being prosecuted for homosexuality. I cannot fathom how much we lost when we lost you.

I am sorry that these things happened to you. But you may ask me, "What good is being sorry?"

An apology is really an atonement for the past, wrapped around a promise for the future. My promise (and I know there are others who will agree with me) is that we won't let prejudice prevent us from applying our own Turing Test to the people we deal with.

You may also ask me whether I write this because of an agenda. Because I want to take your death and use it to promote gay-rights, to hold you up as an example of how prejudice against gays harms the world.

I would be lying to you if I didn't tell you that I am personally uncomfortable with the implications of the acceptance of homosexuality. I suffer great internal conflict about questions such as "Should gay couples be allowed to have or adopt children?". I cannot overcome these feelings because they are grounded in an upbringing and they are in conflict with rationality and my own experiences.

But I see clearly that my feelings conflict with a simple truth: if I allow irrational opinions to guide my actions I lose my way. In allowing irrationality around homosexuality to guide our collective actions we lost you.

What we all need is to apply the Turing Test daily. I know that despite my own misguided feelings, I apply your test to those I encounter, and that is my way of apologizing to you.

John.

Alan Turing deserves an apology from the British Government

When I started writing The Geek Atlas there was one name that was getting in the book no matter what: Alan Turing.

Alan Turing matters on many levels because he was, in the words of the memorial in Manchester:

Father of computer science, mathematician, logician, wartime codebreaker, victim of prejudice

Turing's work has affected us all. He's best know for his involvement in Second World War code breaking (especially for helping to break Engima) and if all he had done was that we would be grateful.

But Turing was also a critical pioneer of computer science. He defined a theoretical model of computers (at a time when 'computer' meant a person, often a woman, who computed numbers) that holds true today. He suggested how we might determine whether a computer was sentient (with the Turing Test).

Turing's death should remind us how prejudice ruins and degrades.

Alan Turing was gay. And he was prosecuted for 'indecent acts' and eventually took his own life aged 41. This man, younger than me, killed himself because at the time homosexuality was illegal and having been prosecuted he was chemically castrated in an attempt to 'cure' him. He had been stripped of his security clearance.

For years, his legacy was largely ignored outside the computer community. To quote Wikipedia:

In 1994 a stretch of the A6010 road (the Manchester city intermediate ring road) was named Alan Turing Way. A bridge carrying this road was widened, and carries the name 'Alan Turing Bridge'.

A frikkin' Ring Road!

It wasn't until 2001 that a statue was erected.

Today is Alan Turing's 97th birthday. Or at least it could have been if it were not for his prosecution and untimely death.

Isn't it time the British Government apologized for the way he was treated? We shouldn't let this anniversary of his death go by without recognizing the great works this man did and the ignominious way in which he was treated.

I need your help: I need iphone 3GS serial numbers

Do you own an iPhone 3GS? Would you be willing to give me its serial number?

I don't need to know who you are, what your phone number or IMEI number is, just the serial number.

"Why?", you ask.

Because I want to do an order statistics analysis of the serial number to attempt to calculate the number of iPhone 3GS models being sold.

If you click on Settings then General and then About you can find the serial number. Just email it to me. I'll publish the results later.

"How are you going to estimate the number of iPhone 3GS phones sold from serial numbers?", you cry.

Just like the Allies estimated the number of Nazi German tanks being produced.