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Despite FTC Settlement, Intel Can Ship Oak Trail Without PCIe 140

MojoKid writes "When the Federal Trade Commission settled their investigation of Intel, one of the stipulations of the agreement was that Intel would continue to support the PCI Express standard for the next six years. Intel agreed to all the FTC's demands, but Intel's upcoming Oak Trail Atom platform presented something of a conundrum. Oak Trail was finalized long before the FTC and Intel began negotiating, which means Santa Clara could've been banned from shipping the platform. However, the FTC and Intel have recently jointly announced an agreement covering Oak Trail that allows Intel to sell the platform without adding PCIe support — for now."

Comment Re:0.5 (Score 1) 981

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. :-)

Comment Re:Ordering and Convergence (Score 1) 981

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.

Comment Re:0.5 (Score 1) 981

You're implying in a random sample of 1000 pairs of children, 1/3 will be boy/boy, 1/3 girl/boy and 1/3 girl/girl...

That's incorrect - 1/2 of the pairs will be one girl and one boy. It's pretty easy to write some code to generate random pairs to convince yourself of this.

Comment Re:0.5 (Score 1) 981

But if I say that the youngest is a boy, then the probability that the oldest is also a boy is 1/2 (not 1/3). If I just say one of the children is a boy, with no other information, then I could be refering to either one in the boy/boy case, and the probability is 1/3 of both being boys.

Being younger or older is independent of the gender, but does affect the statistical outcome.

It's not easy to understand (and goes against instinct, especially as the pieces of information appear unrelated), but I wouldn't call it a trick.

Comment Re:The other problem posed in TFA (Score 3, Insightful) 981

That's incorrect - you've just skewed the population!

In 1000 pairs of children you'll have 250 girl/girl, 250 boy/boy, 500 girl/boy.

Of those, the ones that have at least one boy are the 250 boy/boy and 500 girl/boy pairs. So there's a 33% chance it's boy/boy if you know one is a boy.

The whole point is you could be talking about either of the boys in the 250 boy/boy pairs - it doesn't increase the probability that it's boy/boy instead of girl/boy (you're still twice as likely to have a girl/boy pair relative to a boy/boy pair). If you specify more about the boy you're talking about - for example (ironically) saying his name is Peter - then the boys are no longer interchangable and the probability tends towards 1/2.

It is tricky ;-)

Comment Re:He is wrong (Score 1) 981

The {known/unknown} bit is applied after the gender is fixed. By adding a fourth boy/boy option you're skewing the relative probabilities (rather than just removing the cases that no longer apply).

Maybe think of it this way: suppose there are 1000 pairs of children taken randomly from the population. On average, 250 of the pairs will be two boys, 250 two girls, and 500 a boy and a girl.

So if one pair is taken randomly from that group, and you are told at least one of the pair is a boy, then you know it's one of the 750 pairs other than girl/girl. 250 of those pairs are boy/boy, the remaining 500 are girl/boy - leaving you with a probability of 33% that it's boy/boy.

Comment Re:Ordering and Convergence (Score 1) 981

Nicely put!

So at one extreme you have BB, BG, GB (33%). At the other you have B*B, BB*, B*G, GB* (50%) where B* is the boy who fits the extra criteria and it is impossible for both boys to fit it. By making it improbable that both boys fit the criteria you're separating the BB into B*B and BB* to some degree, so the probabilities will tend towards 1/2.

Comment Re:Well? (Score 2, Informative) 981

If you ignore the Tuesday bit, then eliminating the option that you know is not possible (FF won't work since one child is a boy) you have three options remaining, each with equal probability, giving 33% chance.

Or to put it another way: If it was 25% chance like you say then so is FM and MF. There are no other possibilities so it should add up to 100%, but you only get 75%...

Comment Re:Ordering and Convergence (Score 1) 981

The number of days in the week does have 7 possibilities though... so saying it's a Tuesday narrows it down to one in seven children (on average) that fits that criteria.

That means it's less likely that both boys fit the critera (in the cases where there are two boys), which in turn alters the probabilities.

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