And now it’s all this

I just said what I said and it was wrong
Or was taken wrong

A small probability correction

September 29, 2026 at 10:35 AM by Dr. Drang

This short video appeared in my YouTube feed last week. It’s from Hannah Fry, whom you probably know from her appearances on Numberphile and other STEM-oriented stuff. If you’re in the UK, you’ve may have seen her on the BBC, too.

The short presents a classic problem in conditional probability, but the answer she comes up with is wrong. It’s a good estimate, and it’s the same answer I got when I stopped the video and tried to work it out in my head, but it’s still wrong. By just a little bit.

Here’s the problem: There’s a disease that affects 1 in 1,000 people. A test for the disease is perfect in one sense but imperfect in another. If you have the disease, the test will return a positive result 100% of the time. If you don’t have the disease, the test will return a negative result 95% of the time but a positive result 5% of the time. If you have the test and the result is positive, what is the probability you have the disease?

What makes this a classic problem is that it presents you with conditional probability in one sense (the probability of a positive test given that you have the disease) and asks for a conditional probability in the opposite sense (the probability that you have the disease given that your test was positive). The solution combines the definition of conditional probability, the commutative property of intersections, and the law of total probability.

Let’s define some events. $D$ is you having the disease, and $T$ is you getting a positive test result. Putting a horizontal bar over these indicates not having the disease and not testing positive (i.e., testing negative), respectively. Therefore

$$\begin{array}{cc}\hfill P(D)& =0.001\hfill \\ \hfill P(\overline{D})& =0.999\hfill \\ \hfill P(T\mid D)& =1\hfill \\ \hfill P(\overline{T}\mid \overline{D})& =0.95\hfill \\ \hfill P(T\mid \overline{D})& =0.05\hfill \end{array}$$

The vertical bars are read as “given,” meaning the event after the bar is the condition. What we’ve been asked to find is $P(D\mid T)$. Let’s work it out.

By the definition of conditional probability, we can say

$$P(D\cap T)=P(D\mid T)P(T)$$

where $\cap$ means the intersection of the two events. So

$$P(D\mid T)=\frac{P(D\cap T)}{P(T)}$$

Because the intersection of events is commutative

$$P(D\cap T)=P(T\cap D)=P(T\mid D)P(D)$$

Both terms on the right-hand side of this equation are known, so we can say

$$P(D\mid T)=\frac{(1)(0.001)}{P(T)}$$

Since $D$ and $\overline{D}$ are mutually exclusive and collectively exhaustive, the law of total probability says

$$\begin{array}{cc}\hfill P(T)& =P(T\mid D)P(D)+P(T\mid \overline{D})P(\overline{D})\hfill \\ \hfill & =(1)(0.001)+(0.05)(0.999)\hfill \\ \hfill & =0.05095\hfill \end{array}$$

So

$$P(D\mid T)=\frac{0.001}{0.05095}=0.019627$$

which is, as I said, pretty close to the 2% answer in the video but not exactly.

This formal approach is how you’re taught to solve problems like this in an introductory probability class, but Dr. Fry and I used a more concrete method to get our nearly correct answers. Here’s what we did:

Imagine 1,000 typical people. Of these, 1 should have the disease (correct) and 50 should test positive (incorrect). That tells us that 1 in 50, or 2%, of the people who test positive will have the disease. Here’s a screenshot from the video that matches this calculation:

What makes this calculation wrong is that it implicitly assumes that 5% of everyone will test positive, not 5% of only those who don’t have the disease. The number who will test positive should be 5% of 999, which is 49.95, plus the 1 who does have the disease. So 1 in 50.95, or 1.9627%, of those who test positive will have the disease. This answer matches that of the formal approach.

This is somewhat unsatisfying, though, as the purpose of this “imagine a bunch of typical people” method is to have all the people counts be integers. Although the numbers work out when you get to the end, it’s distracting to litter the discussion with fractional people. You can get around this by imagining more people—a million, say—but then all the numbers get bigger: 1,000 people with the disease and 49,950 false positives. This isn’t a problem for the kind of people who read this blog but isn’t so great for the more general audience Dr. Fry is addressing.

Personally, I would have been OK with her saying that 5% of 999 is almost 50, so the number who test positive is nearly 51. And 1 out of 51 is just under 2%—call it 2% in round figures. You still make the point that it’s way less than 95% and that problems like this require some care.

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Equinox

September 22, 2026 at 7:05 PM by Dr. Drang

The episode of The West Wing that’s always bothered me is “Evidence of Things Not Seen,” which is Episode 20 of Season 4. It takes place on the day of the equinox (in March, not September, but I remembered it because today is an equinox), and one of the continuing subplots is C.J. trying to convince the others that you can stand an egg on its end at the exact moment of the equinox and only at that moment. Oh, and it has to be the vernal equinox, not the autumnal one.¹

That someone in the glorious Bartlet White House believes in nonsense isn’t what bothers me. Lots of smart people believe silly things. What’s wrong is that it’s C.J., the primary female character, who believes in superstition while the men around her are pooh-poohing it. I’m sure plenty of women will tell me that’s par for the course for Aaron Sorkin, but it still bothers me.

There are two small counters to the general sexism of the egg subplot. First, the men who tell C.J. she’s full of it are themselves wrong. They think you can’t balance an egg on its end, even though it’s not really that hard.

This photo was taken today (yes, on the equinox but not the vernal equinox and not at the exact moment) on my back patio. I didn’t do any tricks like slightly crack the shell or put it on a bit of salt and then blow the salt away. I just kept adjusting its position again and again until it stood. The plain fact is that it just takes a little time to balance an egg on its end, time that most people aren’t willing to spend. (And yes, it helps to have a rough surface.)

The second counter is that Jed seems willing to believe the egg story. He doesn’t defend C.J., but he does try to stand an egg on end in his office. In some ways, this is worse. For all his faults, Jed is supposed to know at least a bit about science. Remember the episode in which C.J. has to remind him to let the NASA experts handle the science questions coming in from schoolkids? If anyone should be instantly dismissive of the standing egg myth, it should be Jed.

At the very end of the show, C.J.’s alone and manages to balance an egg on its end, justifying her faith. The title comes from Hebrews 11:1, which she recites earlier in the episode. I guess this is supposed to get us to think that she hasn’t been spouting nonsense all this time, but what it really means is that she can do what anyone can do if they take their time.

What bothers me the most is the setup for her balancing act. Before trying, she looks pointedly at the clock, which shows midnight. Recall that the balancing is supposed to work only at the exact moment of the equinox. Does Sorkin think equinoxes always occur at midnight? Does he think the audience is dumb enough to believe that? The episode aired in April 2003, so I guess its events take place on March 20 of that year. In Washington, the 2003 vernal equinox occurred at about 8:00 pm, not four hours later. Was the show set a year or two earlier? No midnight spring equinoxes those years, either.

If you’re interested in the history of the egg balancing myth, Martin Gardner wrote an article about it in 1996. You can see the text on the Wayback Machine.

---

Cleaning up after Matt Parker

September 16, 2026 at 11:36 PM by Dr. Drang

This video Matt Parker posted a couple of days ago really pissed me off:

It’s cheating to say that a result arrived at through classical mechanics—in this case, that $\frac{1}{2}m{v}^2$ is the kinetic energy of a particle—is wrong because it’s only an approximation to relativistic physics. Of course that’s the case, and I swore at my phone when the “reveal” came.

By the end of the video, though, I had calmed down because I knew I could put together a quick post of my own by rewriting the infinite series equation he derived in a better form. Also, I could provide a general expression for the individual terms.

Let’s start with this result: that the kinetic energy of a particle (including relativistic effects) is

$$m{c}^2(\gamma -1)$$

where m is the mass of the particle, c is the speed of light, and

$$\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

is the Lorentz factor, with v as the velocity of the particle. The video gets to this result at about the 8-minute mark.

Matt then does a series expansion of the $\gamma$ term to get

$$\gamma =1+\frac{1}{2}\frac{v^2}{c^2}+\frac{3}{8}\frac{v^4}{c^4}+\frac{5}{16}\frac{v^6}{c^6}+\frac{35}{128}\frac{v^8}{c^8}+\dots$$

The leading 1 of this series gets canceled by the 1 that’s subtracted from γ, giving

$$\frac{1}{2}m{v}^2+\frac{3}{8}m\frac{v^4}{c^2}+\frac{5}{16}m\frac{v^6}{c^4}+\frac{35}{128}m\frac{v^8}{c^6}+\dots$$

for the kinetic energy. The leading term is the one we get from classical mechanics and the others are essentially zero unless you’re in a particle accelerator (which is discussed later in the video).

I don’t like this form for the equation. It’s cleaner if you factor out all the terms that give the expression the units of energy and then have a nondimensional expression afterward. Like this:

$$\frac{1}{2}m{v}^2(1+\frac{3}{4}\frac{v^2}{c^2}+\frac{5}{8}\frac{v^4}{c^4}+\frac{35}{64}\frac{v^6}{c^6}+\dots )$$

Isn’t this nicer? The expression in the parentheses is a function of the ratio of the velocity of the particle to the speed of light. If we call that

$$\varphi =\frac{v}{c}$$

then the kinetic energy is

$$\frac{1}{2}m{v}^2(1+\frac{3}{4}{\varphi }^2+\frac{5}{8}{\varphi }^4+\frac{35}{64}{\varphi }^6+\dots )$$

and it’s much easier to see why the terms after the 1 are vanishingly small for most situations—all the situations for which classical mechanics applies.

One last thing. Matt sort of explained how to calculate the terms of the expansion of γ in the companion video, but there was a lot of handwaving and he bailed out after the second term. It doesn’t take too much effort to show that the series expansion of γ can be written like this:

$$\gamma =\sum _{n=0,2,4,\dots }^{\infty }\frac{n!}{2^n{\left[\left(n/2\right)!\right]}^2}{\varphi }^n$$

That means the kinetic energy, $m{c}^2(\gamma -1)$, is

$$\frac{1}{2}m{v}^2\sum _{n=2,4,6,\dots }^{\infty }\frac{n!}{2^{n-1}{\left[\left(n/2\right)!\right]}^2}{\varphi }^{n-2}$$

You can confirm that these terms match the equation we saw earlier by plugging values of n from 2 through 8 into this expression. And now we can extend the series as far as we like, even though the additional terms add essentially nothing.

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Apple drawings

September 14, 2026 at 8:09 PM by Dr. Drang

Everyone’s favorite Apple archivist, Stephen Hackett, wrote a short post last week in which he linked to Apple’s page with dimensional drawings of its products. He included this drawing of the iPhone 17 Pro:

You can click on it to see a 300 dpi version. The Apple page linked above will give you infinite-resolution PDFs.

When I zoomed in on Stephen’s page, I was initially confused, but then I realized what seemed odd to me. You see, I’ve spent an awful lot of the past 40 years looking at engineering drawings, and I can’t think of any that looked like this. That’s because the drawings I’m used to are for making the depicted product, but the drawings Apple’s offering up here aren’t for that.¹ These are for others—not Apple and not its suppliers—to use for making cases and other accessories. I figured it was worth a quick post on what makes a drawing like the one above so different from what I’m used to.

First off, there are so many dimensions. Drawings of full products are typically called assembly drawings, and they have almost no dimensions. What they do is show how the various component pieces (subassemblies) are put together. The subassemblies have their own drawings that show how they are put together from their components, which are themselves typically subassemblies. Eventually, you get down to the individual part drawings, which depict a single piece of metal or glass or plastic. At each level in this hierarchy, the drawings tend to include only the dimensions necessary to build the object depicted. You don’t include the length and diameter of a screw on an assembly drawing; you simply call out its part number (which is also its drawing number).

Speaking of drawing numbers, if you look in the title block you’ll see that there is no drawing number. Design drawings always have these. They also have the names or initials of the draftsmen and checkers and, usually, a list of revisions. These public drawings don’t reveal any of that internal information.

There are names for many of the items and dimensions: product length, display active area, volume button, rear sensor, etc. These are not common in design drawings because they tend to be of no value to those doing the manufacturing. You don’t need to know what this hole is for; just make it this big and put it here.

Which leads us to tolerances, which are absent. Dimensions and positions are given to the nearest hundredth of a millimeter and there is no give or take. To the outside world, Apple’s dimensions are absolute and invariable. Inside Apple and its suppliers, people know better.

Also absent is any specification of material, other than a few generic mentions of glass. While assembly drawings don’t usually include materials, parts drawings always refer to material specs, usually included in another company document.

The last thing I’ll mention is the Notes section in the upper left corner. It has instructions clearly meant for people outside of Apple and making accessories. If I’d zoomed in on this section first, I probably wouldn’t have been even momentarily confused.

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