Slight inaccuracy, you can't DM a user on Twitter unless you're following that user and they also follow you back. You can still @reply to them, and they'll see it in their Mentions tab, which I don't know how many people check and how often.

You're entirely right about it being easy to search for keywords the company is interested in, so that it can know who is talking about it, though. Even more so with the new Streaming API... :)

500 KB Web fonts on a Slashdotted page... I guess one thing you can do now is replace that declaration with "font-face: Arial, Helvetica, sans-serif;" everywhere.

Just for future reference,

33,000 USD of hard drives, currently at about 1.5 TiB for 80 USD, is 633,600 GiB.

633,600 GiB can store 158,400,000 songs, at 4 MiB apiece.

The second trial of Jammie Thomas awarded the RIAA 1,920,000 USD for 24 songs, which comes out to 80,000 USD apiece.

For 158,400,000 songs, the RIAA would be awarded 12,672,000,000,000 USD (12 trillion short scale). That's only a bit less than the national US debt, which is 13,208,593,598,669 USD (13 trillion short scale) as of this comment!

It is given. "At least one of whom is a boy" means that the number of boys is guaranteed to be greater than 0. Therefore, one child is a boy, and I pinpointed him as being the given of the problem, and gave him a name to differentiate him from the other unspecified child. If the problem had stated "Exactly one of whom is a boy", then the probability of both children being boys is 0 (0%), because the number of boys is guaranteed to be greater than 0 and less than 2.

You gave the right probability here, but for the wrong reason. "Is at least one of your children a boy?" answered in the affirmative means that you can now answer the question as if one boy was a given. The question now is, "Given that I have at least one son, what is the probability that I have 2 sons?"

Per this page, this can be written as P(2 sons | at least 1 son) = P(2 sons and at least 1 son) / P(at least 1 son) = (1/4) / (3/4) = 1/3.

And I just invalidated all of my other comments on this thread... Ouch!

*takes a huge bite of humble pie*

And now you're saying that it's twice as likely to have a girl as it is to have a boy?

You say:
P(Peter, Boy) + P(Boy, Peter) = 1/3
P(Peter, Girl) + P(Girl, Peter) = 2/3

Let's try to reverse-engineer this problem.

Would you agree that "I rolled a die and it landed on 6. What's the probability that it landed on 6?" yields a probability of 1 (100%)? That's because it's a given of the problem. We don't even have to know if it's unweighted, or 6-sided; it could be 100-sided, and it still wouldn't change the fact that it landed on 6. If you don't agree with this, then you say "the probability is undefined, because I have insufficient information about your die".

Would you agree that, in "I tossed a coin twice, and at least one of these landed heads. What's the probability of both having landed heads?", a given of the problem is that there was 1 Heads, and therefore that the probability we're looking for is 0.5 (50%) for the unspecified coin? If you didn't agree with saying that a given of a problem has P = 1, then the universe of the problem is {(H,H), (H,T), (T,H), (T,T)}, each occurring with equal probability, and the answer is 0.25 (25%). Note that, if you don't accept givens to problems, (T,T) is not impossible, because you ignore "and at least one of these landed heads".

The problem posed in TFA amounts to "My wife gave birth twice, and at least one of the children was a boy. That boy was born on a Tuesday. What's the probability of both being boys?" Therefore the problem is exactly like the two coin problem, with 1 boy being a given, and the birth weekday being extra information that isn't used in the problem's question therefore doesn't affect the resulting probability. What's the probability now?

This is the question posed without the birth weekday specified. TFA actually tries to say that there are 4 outcomes for the pair of children, one of which is impossible, so they remove it. Since "boy, boy" is only one of the 3 outcomes, then the probability must be 1/3. Right?

Wrong.

The boy (let's call him Peter) being a boy is a given of the problem, so it has P = 1. The other child -- we don't care about it being born before or after Peter -- is independent, so the probability that it's a boy is 0.5*. The 4 outcomes are as follows:

Peter, Boy = 0.25*
Peter, Girl = 0.25*
Boy, Peter = 0.25*
Girl, Peter = 0.25*

So, whichever way we slice this problem, the solution is 0.5*.

P(Peter, Boy) + P(Boy, Peter) = 0.5*
1 * P(Other is Boy) = 0.5*

- - - - - -
* May slightly differ due to the male:female ratio at birth. It is assumed here to be 1:1.

I don't know about the chance of it landing heads up, but if you threw it up, it has a rather high chance of being damaged by your stomach acid...

The problem is stated thus:

One of whom is not "exactly one of whom", so 'one' might be opposed to 'the other' or 'two'. All we know is that one of the problem poser's children is a boy born on a Tuesday. It states nothing about the relationship between the two children in time or space, so the probabilities are independent.

Further, the problem doesn't ask about any probability related to the second boy's birthday. The problem doesn't ask, e.g., What is the probability that my other child is a boy not born on Tuesday?. That makes the birth weekday completely irrelevant.

What if the second child was born on Tuesday but was not a twin of the Tuesday boy?

Say, the Tuesday 75 weeks later. It's still a Tuesday...

I bet you would freak out if there was snow in California. Hell, the UK was like a bunch of pussies last winter because of about 15 cm of snow (half a foot), using up all of their grit, and Canada gets more than that every winter.

It just depends on what you're used to.

(Disclaimer: I am Canadian.)