Boy girl paradox and information I was watching a TED video about the boy/girl paradox and several questions come up and there are several things I dont understand.
Why is there a paradox? In the TED video they say its 1/3 invariabily, But the ODDs for a frog croaking wouldnt be doubled on the Male/Male subgroup? Meaning the correct answer would be 1/2?
Isnt the same with the original boy/girl paradox?


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*John Smith have two children, he introduces one of them as a boy. Wich are the odds the other is also a boy?


If he is introducing at random isnt introducing a boy the double the odds in the boy/boy subgroup? [BB; BG; GB; GG] arent the odds: [50;25;25;0]?
Isnt it showing exactly this? When you take the probabilities of the event happening into account, the odds are always 1/2.
Another example: You are playing poker, you have a very good hand with 80% chance of winning. If someone increases the bet, folds or pay, it makes you wonder whether your chances really are higher or lower than 80%.
Now when adding irrelevant information, as in the day of the week boy/girl paradox. I quite dont understand why it changes the probabilities addin irrelevant information, isnt it illogical?


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*You pick at random all families with 2 children, with at leat one boy, born on tuesday. The chances of the family consisting of a boy and a girl is 52%.


The same can be said about every other day of the week, it cant? for Monday 52%, Wednesday 52%, Thursday 52%... and so on. If the odds of every day are 52% how can it be 1/2 or even 1/3 if you disconsider the days of the week??
The way I see it is if you have multiple answers for the same problem one of them is wrong.
Cant you just make up information and achieve the same result?
 A: 
If he is introducing at random isnt introducing a boy the double the odds in the boy/boy subgroup? [BB; BG; GB; GG] arent the odds: [50;25;25;0]?

There's no "double the odds" happening here.   What you are measuring is the ratio of outcomes for "boy and girl" among those of the given criteria; when all outcomes have equal probability.
It comes down to the difference between "at least one" and "this particular one".

I: Bob Smith brings his two new babies to the park in a stroller on a cold day. He tells some anecdotes, and thus reveals that at least one of them is a boy. The probability that the other is a girl is $2/3$.
$$\require{enclose} \rm \enclose{circle}{\bbox[1ex]{BB, \underbrace{BG, GB}}}, GG$$

II: Bob Smith brings his two new babies to the park in a stroller on a cold day. He tells some anecdotes, and points to reveal that a particular one of them is a boy. The probability that the other is a girl is $1/2$.
$$\require{enclose} \rm \enclose{circle}{\bbox[1ex]{BB, \underbrace{BG}}}, GB, GG$$

III: Bob Smith brings his two new babies to the park in a stroller on a cold day. He tells some anecdotes, to reveal that at least one of them is a boy that was born on Tuesday. 


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*In this case the sample space consists of $14^2$ outcomes


$$\begin{array}{|l|c{:c}|} \hline L\backslash R & \rm B_{M} & \rm B_{Tu} & \rm B_{W} & \rm B_{Th} & \rm B_{F} & \rm B_{Sa} & \rm B_{Su} & \rm G_{M} & \rm G_{Tu} & \rm G_{W} & \rm G_{Th} & \rm G_{F} & \rm G_{Sa} & \rm G_{Su}\\ \hline 
 \rm B_{M} & \times& \circ & \times& \times& \times& \times& \times& \times& \times& \times& \times& \times& \times& \times\\ \hdashline 
 \rm B_{Tu} & \circ & \circ & \circ & \circ & \circ & \circ & \circ & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet \\ \hdashline
 \rm B_{W} & \times& \circ & \times& \times& \times& \times& \times& \times& \times& \times& \times& \times& \times& \times\\ \hdashline
 \rm B_{Th} & \times& \circ & \times& \times& \times& \times& \times& \times& \times& \times& \times& \times& \times& \times\\ \hdashline
 \rm B_{F} & \times& \circ & \times& \times& \times& \times& \times& \times& \times& \times& \times& \times& \times& \times\\ \hdashline
 \rm B_{Sa} & \times& \circ & \times& \times& \times& \times& \times& \times& \times& \times& \times& \times& \times& \times\\ \hdashline
 \rm B_{Su} & \times& \circ & \times& \times& \times& \times& \times& \times& \times& \times& \times& \times& \times& \times\\ \hdashline
 \rm G_{M} & \times& \bullet & \times& \times& \times& \times& \times& \times& \times& \times& \times& \times& \times& \times\\ \hdashline
 \rm G_{Tu} & \times& \bullet & \times& \times& \times& \times& \times& \times& \times& \times& \times& \times& \times& \times\\ \hdashline
 \rm G_{W} & \times& \bullet & \times& \times& \times& \times& \times& \times& \times& \times& \times& \times& \times& \times\\ \hdashline
 \rm G_{Th} & \times& \bullet & \times& \times& \times& \times& \times& \times& \times& \times& \times& \times& \times& \times\\ \hdashline
 \rm G_{F} & \times& \bullet & \times& \times& \times& \times& \times& \times& \times& \times& \times& \times& \times& \times\\ \hdashline
 \rm G_{Sa} & \times& \bullet & \times& \times& \times& \times& \times& \times& \times& \times& \times& \times& \times& \times\\ \hdashline
 \rm G_{Su}& \times& \bullet & \times& \times& \times& \times& \times& \times& \times& \times& \times& \times& \times& \times \\ \hline
\end{array}$$


*

*$14$ of these outcomes have the left child a boy born on a Tuesday, and $14$ of these have a right child a boy born on a Tuesday, however there is a common outcome that has both children a boy born on a Tuesday (so don't over count this).   In total $14+14-1$ outcomes of the sample space have at least one child that is a boy born on a Tuesday.   Indicated above as black and white circles.

*Among these, $7$ elements consist of a boy born on Tuesday then a girl, and $7$ are a girl then a boy born on a Tuesday.   Indicated above as the black circles.

*So the (conditional)probability that the family consists of a boy and a girl when given that at least one child is a boy borne on a Tuesday is: the ratio of black circles to all circles. $$\dfrac{7+7}{14+14-1}=\dfrac{14}{27}\approx 0.51{\small 8}$$



Assuming that: There is a equal chance of each child being a boy or a girl, an equal chance of each child being born on any particular day, and these events are mutually independent. (IE: all outcomes in the sample space have equal probability)
A: The key issue for the first example is this "If he is introducing at random..."
He isn't. You have already been told that he introduced a boy. If I came to you before any of this happened and said "Mr. Smith will introduce a random child to you. What do you think his other child will be?" Then it would be correct for you to say 50/50 boy/girl.
If he hadn't introduced a child yet, you could say that there is a 2/3 chance that the second child will be the opposite gender of the first. Once the first child is introduced, you know which 2/3-1/3 split you are looking at.
I do not understand what you are asking with the day of the week paradox.
A: The origin of these questions ultimately traces back to Martin Gardner's May, 1959 column in Scientific American. He asked: "Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?" Most people only remember that he said the answer was 1/3. What they don't remember, is that he retracted that answer the next October. The answer depends on how you learned the information "at least one is a boy." If you learned it by a process that would tell you that answer for every family that included a boy, then 1/3 is indeed correct.
But if you could have learned about a girl just as easily - for example, a father telling anecdotes about his two new babies is just as likely to reveal that at least one is a girl, as to reveal that at least one is a boy - then the answer is 1/2. Note that I'm not saying that the anecdote reveals the gender of a particular, just that you can learn about a girl or a boy when there is one of each, and the 1/3 answer requires that you can only learn about a boy from an anecdote.
Now, suppose it was Sherlock Holmes who hears the anecdotes. He says "Interesting, Watson, from those stories I was able to deduce the day of the week that a boy from that family was born. Can you guess what day that is?" If you are Watson, you probably learned nothing that would let you favor one day over another, so they all must be the same. Since they must add up to 1, you can only think that there is a 1/7 chance for each day. 
But if you look at Graham Kemp's table, there are 196 combinations. And no matter what day it is that Holmes deduced, there are 27 combinations that include a boy born on that day. So it seems the chances should be 27/196, which is a little less than 1/7. Unless we can find an error in one of these answers, this is a paradox.
It is actually a form of what is known as Bertrand's Box Paradox. Joseph Bertrand did not use the word "paradox" to refer the puzzle itself, but to how to tell which answer has the error. The chances that Holmes deduced "Tuesday" can't change from 1/7 to 27/196 just because he told you about the deduction, to the error lies in that calculation.
In fact, if the family includes a boy born on a Tuesday, and (say) a girl born on a Thursday, then there should be a 50% chance Holmes would have made a different deduction. There can only be a 100% chance if both are Tuesday Boys. Using this method, there is one combination where Holmes could only deduce "Tuesday Boy," and 26 where there was only a 50% chance. The answer to Graham Kemp question isn't (7+7)/(14+14-1)=14/27, it is (7/2+7/2)/(13/2+13/2+1)=7/14=1/2.
