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.
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!).
"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.
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.
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.
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.
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.
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.
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.
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%...
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.
I'm trying to understand this in laymans terms. Is this a valid way to interpret it?
If you know that a specific child is a boy (e.g. the youngest), then the probability the other one is a boy is 1/2.
If you only know that one of the two children is a boy, and you don't know which, then the probability they are both boys is 1/3.
So the more you pin down details towards identifying a specific child is a boy the closer the probability tends towards 1/2 for both boys. In this case you're specifying a day, and as that reduces the ambiguity a lot it gets you quite close to 1/2.
Slashdot Mirror
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!
unless Godzilla will come to the site, eat all the Cesium and starts farting.
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.
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.
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.
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.
That Bridge article is harder to understand than the answer to this pair of children problem ;-)
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
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.
... and that "common sense" leads to an incorrect answer.
The first step is to see that if one of the children is a boy, then the probability that they both are is 1/3 not 1/2.
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.
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%...
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.
I meant "the more specific you get the closer the probability tends towards one HALF."
Gah!
Just because it's an arbitrary division of possibilities doesn't mean it isn't useful.
Saying one of the children is a boy who's favourite number between 1 and 7 (inclusive) is 3 would give the same probability of 13/27.
You're just narrowing down which boy you're talking about - the more specific you get the closer the probability tends towards one.
"The general rule is that for a set with N possibilities you would have (N*2 - 1)/(N*4 - 1)."
The wisdom of EldavoJohn
Yes.
It's P=(4x-1)/(2x-1) where x is the number of options for the extra information.
So first day of year, x=365, (ignoring leap years),
born in January x=12,
winter x=4.
I'm trying to understand this in laymans terms. Is this a valid way to interpret it? If you know that a specific child is a boy (e.g. the youngest), then the probability the other one is a boy is 1/2. If you only know that one of the two children is a boy, and you don't know which, then the probability they are both boys is 1/3. So the more you pin down details towards identifying a specific child is a boy the closer the probability tends towards 1/2 for both boys. In this case you're specifying a day, and as that reduces the ambiguity a lot it gets you quite close to 1/2.
Thank goodness I'm smart enough to leap ahead without thinking!