petrogradphilos

I like what you've written here. If you write stuff elsewhere, please put a pointer in your HN profile.

Judging by your influences as an architect, I think you would enjoy these books:

- Ronald Ross, Principles of the Business Rule Approach.

- David B. Black, Wartime Software. If you like this, you'd probably like his other books as well. They each cover a different facet of software development, including QA and "project management" (which Black refers to as a disease).

You might also enjoy the essay by David Black in which he proposes "Occamlity" as the metric of software "goodness" (https://www.blackliszt.com/2020/03/william-occam-inventor-me...):

> I propose that a piece of software can be measured by its “Occamality.” The more “Occamal” it is, the better it is. And I propose that Occamality is strictly correlated with the extent to which there is no redundancy of any kind in a program...

> You will never understand every detail of the implementation of every level on that stack; but you can understand all of them to some level of abstraction, and any specific layer to essentially any depth necessary for any purpose.

As David Deutsch argues in The Beginning of Infinity, this is true not just for computers, but for everything.

> (1 - 2e-6)^(3e6) ≈ 0.002

Using the binomial directly is a good way to get the probability of 0 heads. Note, though, that the U.S. population is in the neighborhood of 300 million, not 3 million (as you seem to have used).

(1 - 2⋅10⁻⁶)^(3⋅10⁸) ≈ 10⁻²⁶¹

https://www.wolframalpha.com/input/?i=%281+-+2+*+10%5E-6%29%...

> Since 0 is about 25 standard deviations from the mean, the probability of seeing 0 heads is on the order of 10⁻¹³⁸.

Note: the above figure comes from the normal approximation to the binomial, which loses accuracy towards the tails. The exact probability of seeing 0 heads is (1 - p)^n = (1 - 2⋅10⁻⁶)^(311⋅10⁶) ≈ 10⁻²⁷⁰ [1].

[1] https://www.wolframalpha.com/input/?i=%281+-+2+*+10%5E-6%29%...

This is like having a weighted coin that comes up heads with probability 2⋅10⁻⁶, flipping it 311 million times, and seeing 0 heads. That's astronomically unlikely.

To see this, observe that the number of heads follows a binomial distribution with n = 311 million and p = 2⋅10⁻⁶. This can be well approximated¹ by a normal distribution with mean μ = np = 622 and standard deviation σ = Sqrt[np(1 - p)] = 25.

99.7% of the time², when you sample from this distribution, the sampled value will be within 3 standard deviations of the mean, i.e., between μ - 3σ = 547 and μ + 3σ = 697. Results further from the mean are more unlikely. For example, seeing a value more than 7 standard deviations from the mean (i.e., less than 447 or more than 797) is about a 1 in 2 trillion event³. Since 0 is about 25 standard deviations from the mean, the probability of seeing 0 heads is on the order of 10⁻¹³⁸.

[1] https://math.stackexchange.com/questions/2021801/conditions-...

[2] https://en.wikipedia.org/wiki/68–95–99.7_rule

[3] https://www.johndcook.com/blog/table-of-normal-tail-probabil...

You could also replace % with 10⁻² and use scientific notation:

      0.72% ⋅ 0.03%
    = 0.72 ⋅ 10⁻² ⋅ 0.03 ⋅ 10⁻²
    = 7.2 ⋅ 10⁻³ ⋅ 3 ⋅ 10⁻⁴
    = 7.2 ⋅ 3 ⋅ 10⁻³ ⋅ 10⁻⁴
    = 21.6 ⋅ 10⁻⁷
    = 2.16 ⋅ 10⁻⁶
    ≈ 2 in 1 million

Never use a macro to do an inline function’s job.

It’s correct to regard trees, but not people, as products of their environment. Trees don’t make choices. People do.

> 1 Introduction

> The increased availability of data linking students to teachers has made it possible to estimate the contribution teachers make to student achievement.

There was some data available before (or the sentence would not have used the word "increased"). Why wasn't it possible to estimate with that?

> By nearly all accounts, this contribution is large.

It goes on to talk about what "large" means:

> Estimates of the impact of a one standard deviation (σ) increase in teacher “value-added” on math and reading achievement typically range from 0.10 to 0.30σ, which suggest that a student assigned to a more effective teacher will experience nearly a year's more learning than a student assigned to an less effective teacher (Hanushek & Rivkin 2010;...).

(Typo: "an less effective" should be "a less effective".)

A "range from 0.10 to 0.30σ" doesn't make sense. A Greek lowercase sigma (σ) is used to represent one standard deviation, but the sigma is used only on the upper end of the range. Should it have been from 0.10σ to 0.30σ?

And how are they measuring the impact on achievement of an increase in teacher "value-added", anyway? It says that estimates of the impact "typically range from 0.10 to 0.30σ", but it doesn't say what units those figures are in.

The sentence goes on to say that those unit-less estimates "suggest" that "a student assigned to a more effective teacher will experience nearly a year's more learning than a student assigned to an less effective teacher". Over what time period? That is, how long does a student have to study under a "more effective teacher" to get "a year's more learning"? 1 week? 12 years? It doesn't say.

And finally, how do those unit-less estimates "suggest" an impact measured in learning time? It doesn't say.