So you don't have the set of all primes after all... that's the point. The proof goes like this:
1) suppose you have a set of all primes, and the set is finite.
2) show that there's another prime not in your set - that contradicts (1).
3) therefore, there is no finite set that contains all primes.

All you've done is demonstrate one example of step 2. The original proof given by phantomfive gives a different example of a prime not in the set. Either works - the proof is valid.

Great description, thank you!

Oh great, now I have *another* thing to worry about!

This is fantastic. I think Julia has the right focus to be better for many purposes than MATLAB and C++, and if a compiler is coming it's even better. I'm familiar with MATLAB, Mathematica, Octave, C and C++, and the Julia language looks very easy to learn and use. I also applaud your choice of the MIT license - I hope this permissive license encourages widespread use and innovation - both for free and commercial software. Thank you!

It's not quite your stated ideal, but I'm happy with XBMC running on a rooted Apple TV. Cheap, easy and the videos look fantastic (without having to run iTunes or transcode anything!).

Why do you say ESR is more stable? It sounds like it's more for a controlled and long update cycle.

For example, from here:
ESR FAQ

"The ESR will not have the benefit of large scale testing by nightly and beta groups. As a result, the potential for the introduction of bugs which affect ESR users will be greater, and that risk needs to be understood and accepted by groups that deploy it"

HFA or Aspergers is really very different than the severe autism you are describing - the autistic spectrum is vast. And there are all kinds of problems people have - if you draw the line at autism (including HFA, which can be mild enough to go undetected), what else will be enough? Blindness? Risk of anger issues? A high likelihood of anxiety disorder?

And if six months is fine, how old is too old? Or does age not matter?

A society can be judged by how it takes care of its weak and vulnerable.

And when you do freak out it will be too late.

Is that 40 mSv absorbed dose, or is that measured outside your shielding?

It sounds like a lot, given that radiation workers have a dose limit of 50mSv per year, and 100mSv over a year is positively correlated with cancer... :/

Also 0.01 mrem is not 0.1 Sv (try 0.1 uSv). Given that 1 Sv starts to cause nausea, 0.1 Sv per banana would be bad. ;-)
Getting the prefix or scaling wrong seems to be happening all over the place. NHK said that the two firefighters standing in the water got between 2,000 and 6,000 mSv in one of their news stories, I'm hoping that was a misprint.

We were making the same comparison, but switched focus to RackSpace because of their simpler pricing model.

RackSpace Pricing

We also liked that they're releasing OpenStack under the Apache 2 license.

This has nothing to do with the Gambler's fallacy.

You're right that if the youngest is a boy, then the probability that the oldest is a boy is 1/2.

However if you only know one is a boy, and you don't know which that one could refer to, then the probability the other is also a boy is 1/3.

Take 1000 pairs - on average you get 500 pairs of one girl one boy, 250 of two boys, and 250 of two girls. If you know that one is a boy that means it must be one of the 750 made up of 250 boy/boy and 500 boy/girl. Since 250 of the 750 are two boys you have a probability of one in three that it's two boys.

You don't have to take my word for it - it's easy to write simulations to demonstrate that this is the case. There's even one posted in the comments to this story. :-)

Nope - assuming Tboy means the boy that's identified in the problem statement, then there are 125 Tboy boy and 125 boy Tboy pairs. (And 250 Tboy girl and 250 girl Tboy pairs).

So it's still 1/3.

The correct solution is counter-intuitive, but so is a fair bit of statistics.

OK, try this way then: the ratio of Tuesdays to all possible days is what matters. I.e. the probability that both boys meet the critera is much smaller (i.e. both born on Tuesday).

There are two extremes for this sort of problem:

1) You know one is a boy, but you have no information to say which. Then the probability the other is a boy is 1/3. (This is counter intuitive, but Devlin explains it well).

2) You know the youngest is a boy. Then the probability the oldest is a boy is 1/2.

When you have an extra piece of information, the chance that it might apply to both children affects the overall probability, and you get a value between 1/3 and 1/2. The day of the week is unlikely to be the same for both (ignoring twins), so it's close to 1/2.