I decided the design was good enough to roll out, although it still requires a bit of tweaking. It currently looks lovely in OmniWeb and Safari (webkit) and not-ugly in Firefox, IE, and Safari Mobile. Fonts currently work in OmniWeb, Safari, and Firefox, but I've only got TTF so the others are backing off to a few standard fonts; and the rounded corners only work in the desktop webkit browsers and the gradients don't work on iPods. But in all these cases there is a good graceful degradation, so I'm happy enough to slide the CSS into place.

Some readers might be pleased to note that the background isn't quite white; it's just enough off-white to take the edge off, I think.

Enough fussing for now. Merry Christmas!

"Two approaches we could take here. The first is we just stick to the facts. Lotta fun that is. The second is we wave cheerily at the facts en route to a more entertaining sociopolitical perspective. This is the Fox News system, and you can see it works for them." --Cecil Adams

Posted by blahedo at 3:32am
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I'd read a while back about Tau Day and the idea that *τ=6.28...* is a better mathematical constant than *π=3.14...*, for a variety of reasons. (Go read the Tau Manifesto and learn several of them if you haven't already.)

One of them was the idea that far from being a strength of π, the area formula *A=πr ^{2}* is actually a weakness, because it camouflages the fact that there should naturally be a ½ in there, deriving from its integral relationship with the circumference formula. By contrast,

So anyway. I was thinking about the volumes of spheres, and I recalled that the formula was *V=⁴/₃πr ^{2}*; of course I knew that because I'd memorised it many years ago, not that it had any reason behind it:

A=πr^{2} |
V=⁴/₃πr^{3} |

A=½τr^{2} |
V=⅔τr^{3} |

V_{2} (area) |
=πr^{2} |
=(1/2)τr^{2} |

V_{3} (volume) |
=(4/3)πr^{3} |
=(2/3)τr^{3} |

V_{4} |
=(1/2!)π^{2}r^{4} |
=(1/2!∙4)τ^{2}r^{4} |

V_{5} |
=(8/5∙3)π^{2}r^{5} |
=(2/5∙3)τ^{2}r^{5} |

V_{6} |
=(1/3!)π^{3}r^{6} |
=(1/3!∙8)τ^{3}r^{6} |

V_{7} |
=(16/7∙5∙3)π^{3}r^{7} |
=(2/7∙5∙3)τ^{3}r^{7} |

V_{8} |
=(1/4!)π^{4}r^{8} |
=(1/4!∙16)τ^{4}r^{8} |

But wait! What if we take that awkward extra power of 2 in the even-dimension formulas and distribute it over the factorial?

V_{2} (area) |
=(1/2)τr^{2} |

V_{3} (volume) |
=(2/3∙1)τr^{3} |

V_{4} |
=(1/4∙2)τ^{2}r^{4} |

V_{5} |
=(2/5∙3∙1)τ^{2}r^{5} |

V_{6} |
=(1/6∙4∙2)τ^{3}r^{6} |

V_{7} |
=(2/7∙5∙3∙1)τ^{3}r^{7} |

V_{8} |
=(1/8∙6∙4∙2)τ^{4}r^{8} |

Check it out! Even if we don't have a deep understanding of what a double factorial is or how to compute the *Γ* function, we can clearly see the recurrence relation among the various dimensions, and the relationship between the even-numbered dimensions and the odd-numbered dimensions, and that they're much more closely related than might first appear from reading the Wikipedia article on *n*-spheres that I linked above.

So, chalk up one more success for the τists!

"When I go to get a new driver's license... or deal with the city inspector... or walk into a post office... I find public employees to be cheerful and competent and highly professional, and when I go for blood draws at Quest Diagnostics, a national for-profit chain of medical labs, I find myself in tiny, dingy offices run by low-wage immigrant health workers who speak incomprehensible English and are rud to customers and take forever to do a routine procedure." --Garrison Keillor

Posted by blahedo at 10:26pm
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It's been a loooong time since I've done a site upgrade. This place was already looking pretty dated a few years ago, and time has not treated it well. It had gotten to where I hesitated to post links because I knew I'd get the inevitable "Comic sans? Really?", among other snarky comments. And of course the web has long since moved away from coloured backgrounds, even light ones; pretty much anything other than white (or off-white) is hard to find among sharp-looking websites. But if I was changing things, I wanted to do more than just a trivial font change and a switch to a white background; other elements of the layout reflected an older web, and the last time I did a doc crawl on this stuff it was still the heady early days of CSS2, so probably more than ten years ago. You might be surprised to learn this, but a lot has changed in ten years.

So I'm working on a new design. A few minor changes will roll out early (where they show up directly in the HTML), but I want to wait to slide in the new CSS until I've tested it on multiple browsers. Soon, though.

"A human being should be able to change a diaper, plan an invasion, butcher a hog, conn a ship, design a building, write a sonnet, balance accounts, build a wall, set a bone, comfort the dying, take orders, give orders, cooperate, act alone, solve equations, analyze a new problem, pitch manure, program a computer, cook a tasty meal, fight efficiently, die gallantly. Specialization is for insects." --Robert Heinlein

Posted by blahedo at 2:05am
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Earlier today, on Hacker News someone posted a link to Tom Moertel's blog post "On the evidence of a single coin toss", where he poses a question about probabilities: if he claims he had a perfectly-biased always-heads coin, and you toss it once and it comes up heads, should that sway your beliefs about the claim?

This was an excitingly interesting question on which I spent far too much time working out an answer, so to sort of justify the time I wrote up the answer and posted it on HN. I figured I should clean it up a little and post here. The tl;dr is that "it depends", first on your formalism (and whether you buy into Bayesian analysis), and second on how much you trust Tom in the first place.

There are at least three different lines of inquiry here:

- Hypothesis testing.
- If the [null] hypothesis is that
*p(heads)*is 1, you can't prove this, only disprove it. So: "doesn't sway". Not very interesting, but there it is. - Simple Bayesian.
- The probability of his claim given that it
comes up heads,
*p(C|H)*, can be understood as[3] the prior of his claim,*p(C)*, times*p(H|C)*, divided by*p(H)*. Well,*p(H|C)*is 1 (that is the claim), and*p(H)*, if I fudge things a little bit, is about 1/2, so*p(C|H)*should be about double*p(C)*—assuming*p(C)*is very low to start with.[0][2] - Complex Bayesian.
There's a hidden probability in the simple case, because

*p(C)*is encompassing both my belief in coins generally and also my belief about Tom's truthtelling. So really I have*p(C)*"p that the claim is true" but also*p(S)*"p that Tom stated the claim to me". Thus also*p(S|C)*"p that if the claim were true, Tom would state this claim to me" and*p(C|S)*"p of the claim being true given that Tom stated it to me"; but also the highly relevant*p(S|¬C)*"p of that if the claim were NOT true, Tom would state this claim to me ANYWAY" and a few other variants. When you start doing Bayesian analysis with more than two variables you nearly always need to account for both*p(A|B)*and*p(A|¬B)*for at least some of the cases, even where you could sometimes fudge this in the simpler problems.SO this brings us to a formulation of the original question as: what is the relationship between

*p(C|S,H)*and*p(C|S)*? The former as

and then*p(H|C,S)p(C,S)/(p(C,S,H) + p(¬C,S,H))*

and if I take*p(H|C,S)p(C,S)/(p(H|C,S)p(C,S) + p(H|¬C,S)p(¬C,S))**p(H|C,S)*as 1 (given) and*p(H|¬C,S)*as 1/2 (approximate), I'm left with

For the prior quantity*p(C,S)/(p(C,S) + 0.5p(¬C,S))**p(C|S)*, a similar set of rewrites gives me

Now I'm in the home stretch, but I'm not done.*p(C,S)/(p(C,S) + p(¬C,S))*Here we have to break down

*p(C,S)*and*p(¬C,S)*. For*p(C,S)*we can use*p(C)p(S|C)*, which is "very small" times "near 1", assuming Tom would be really likely to state that claim if it were true (wouldn't you want to show off your magic coin?). The other one's more interesting. We rewrite*p(¬C,S)*as*p(¬C)p(S|¬C)*, which is "near 1" times "is Tom just messing with me?".Because a

*crucial*part of this analysis, which is missing in the hypothesis-test version or in the simpler Bayesian model, but "obvious" to anyone who approaches it from a more intuitive standpoint, is that it matters a lot whether you think Tom might be lying in the first place, and whether he's the sort that would state a claim like this just to get a reaction or whatever. In the case where you basically trust Tom ("he wouldn't say that unless he at least thought it to be true") then the terms of*p(C,S) + p(¬C,S)*might be of comparable magnitude, and multiplying the second of them by 1/2 will have a noticeable effect. (Specifically, if*p(C,S)*and*p(¬C,S)*turned out to be exactly equal, then flipping once would make us about 4/3 as likely to believe the claim as before.) But if you think Tom likely to state a claim like this, even if false, just for effect (or any other reason), then*p(C,S) + p(¬C,S)*is*hugely*dominated by that second term, which would be many orders of magnitude larger than the first, and so multiplying that second term by 1/2 is still going to leave it orders of magnitude larger, and the overall probability—even with the extra evidence—remains negligible, with a*very*slight increase to the belief in Tom's claim.

[0] This clearly breaks if *p(C)* is higher than 1/2, because
twice that is more than 1. If we assume that the prior *p(H)* is
a distribution over coins, centred on the fair ones and with a long tail
going out to near-certainty at both ends, the claim "this coin is an
always-heads coin"[1] is removing a chunk of that distribution in the H
direction, meaning that *p(H|¬C)* is actually slightly,
very slightly, greater than 1/2. This is the "fudge" I refer to above
that lets me put the *p(H)* as 1/2. Clearly if my prior *p(C)*
is higher than "very small" this would be inconsistent with the prior
*p(H)* I've described.

[1] I'm further assuming that "always" means "reallllllly close to always", because otherwise the claim is trivially false and the problem isn't very interesting.

[2] Note that this is not actually a "naive Bayesian" approach—that's a technical term that means something more complicated.

[3] This is what I meant about buying into the Bayesian approach. I'm going to continue the post under the assumption that Bayesian reasoning is valid (even if not what is traditionally called a "probability"), and I'm going to use the language and notation of probability to do it. If that doesn't sit well, imagine that I am quantifying something like belief or confidence rather than probability per se.

"How would Apple like it if when they discovered a serious bug in OS X, instead of releasing a software update immediately, they had to submit their code to an intermediary who sat on it for a month and then rejected it because it contained an icon they didn't like?" --Paul Graham

Posted by blahedo at 5:05pm
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