I wrote it in VB because even the most casual coder hacker can read VB. although it was a challenge to write. I kept putting parenthesis, curly braces, and semicolons as well as initially saying "Long lSplit = 0;" for one good example.

You went to the wikipedia page out of exasperation too, semck? :-) I was just about to post that.

lol, I didn't!

But I did read it yesterday, so it's probably from there. :P

Anyway, Draug, since we're on a new page, let me ask again. Consider the set of all families in America with two children, one of whom is a boy. (Assume 50/50 birth rates, etc.)

What percentage of those families have two boys?

But semck isn't a casual hacker. His C++ version could have been converted to c# by changing about 5 lines. C# is readable by any C++ programmer.

@semck - In the case of all the familes... Because there are familes with GG, then there are twice as many with B+G (regardless of position) as there are BB. but as proposed, GG was impossible in the specific instance (the question was presented with a specific man and his specific set of two kids) which makes each child independent of the other making the odds 50/50 just as each door is independent of the other in Monty Hall.

@phil - I wanted everyone, not just semck, to be able to read and understand. You assume it was written just for him. Bad assumption.

@Draug, please read the question I asked. I said, consider the set of all families with two children, ONE OF WHOM IS A BOY.

So you're saying that, in that set, only 1/3 have two boys? Yes or no, please, keep this simple.

I guess what I am arguing is that in both the coin toss and the kids, we are talking about a specific instance where we know at least one object has that state set, so one object with that state must be set aside and not tossed or randomly chosen.

I wrote my routine in VBA because anyone with a copy of Excel can run it (and its the only language I know, apart from a little SQL).

Do I need Visual Studio to test Draugs?

Was that an answer to my question, Draug? If so, I can't tell. I'll take up other points later. For now, please answer my question.

Well you should go back and reread it, I did the conditional prob formula, I proposed an experiment where I flip 2 coins behind a blunder and give the home that one was a heads, and then took it to the point that I wrote the letter "K" on one. You should spend a little bit of time to see where I've taken it before trying from scratch.

Twice as many B+Gs as BBs. Yes, 1 in 3 of the reduced set is BB. 2 in 3 is B+G. But this is not a specific instance. Specific instances alter outcomes from over all. Mr. What'sHisName is a specific person. His two kids are specific people, not an amalgam of the population.

Visual Studio is free.

http://www.microsoft.com/visualstudio/eng/downloads

Download Visual Studio Express 2012. Completely free.

"You should spend a little bit of time to see where I've taken it before trying from scratch. "

The upshot, though, was that it didn't convince Draug. I figured I'd come try something new.+

I would very much like an answer to my question to. Draug, do you agree or disagree that the two questions I posed are different?

But surely the population is just a big group of specific people.

OK, Draug, thank you for your answer.

So you agree that in the set S, only 1/3 of families are BB.

Now would you agree that if we chose a random element of the set S, we would have only a 1/3 chance of getting a BB?

Fucking phone!

Repost as:

Well you should go back and reread it, I did the conditional prob formula, I proposed an experiment where I flip 2 coins behind a blinder and give the hint that one was a heads, and then took it to the point that I wrote the letter "K" on one. You should spend a little bit of time to see where I've taken it before trying from scratch.

Scratching the letter K after the fact is not the condition. The K is a constant and must be established *before* flipping. So scratch K, toss both, then make certain K is on heads even if it landed on tails. That is what would simulate the conditions for the experiment.

"The K is a constant and must be established *before* flipping."

Just like the Joneses spent millions of dollars with that genetic selection firm to make sure their first would be a boy.

I have it on my work laptop. I keep meaning to get around to learning VB. But I am on home computer which just has Excel (but your code is almost VBA is easy to convert).

@Semck - Yes, I would agree. But that was not the question. the question was "knowing that at least one is a boy, what are the odds of the second one or other one being a boy" again, we come to a specific concrete set of data. When pulled at random, the odds of a BB family are 2 to 1 against or 1 in 3 or 1/3 however you want to do it. But when dealing with concrete examples of a non-random family, the odds for each kid are independent of the other.

To put it another way:

Do a repeated coin toss and record the result. Discard the TT and the odss are 1 in 3 that you will randomly pick HH, but toss a coin and then toss another coin and the odds of the second coin being heads or tails are even so when one coin is a known, the odds of the second become even.

Draug,

OK, you've now agreed that if we choose a random family from the set S, then that family has BB with probability 1/3.

We happen to know the Joneses are in S, and we know NOTHING else about them. So therefore, we know that they have BB with probability 1/3.

You say that "for a concrete family, the odds for each kid are independent of each other." That is correct. It is also true of every family in S. Therefore, it doesn't change the analysis.

You are saying the following:

p(BB(Fam) | Fam in S) = 1/3

but

p(BB(Jones) | Jones in S) = 1/2.

This is unconvincing.

It's really a simple problem if you try to distill it. There are only 4 possible outcomes, and one assumption eliminates one of the possibilities. There are only three left, and only one is favorable.

Problem distilled in conditional probability language ...

What is the probability of two heads being thrown given that at least one head is thrown.

There are only three possible outcomes: all boys, all girls, or split. when you eliminate all girls, then, just as slput gets counted twice because the constant boy could be in either position, so too does double boy get counted twice because the constant boy could be in either position. But it is late and I have been up since before 7 this morning (11:45 now) and I have church tomorrow.

"There are only three possible outcomes: all boys, all girls, or split. "

True, but that doesn't mean they're equally probable. It's not like they are 1/3, 1/3, and 1/3. Rather, it's 1/4, 1/4, 1/2.

"when you eliminate all girls"

When you eliminate all girls, the remaining choices are scaled up with their current probability ratios preserved.

" just as slput gets counted twice because the constant boy could be in either position, so too does double boy get counted twice because the constant boy could be in either position."

There is no constant boy. That has been your mistake throughout. (Well, one of them). "There is at least one boy" does not give you a constant boy, or enable you to double-count BB.

Draug, you seriously need to read a book on probability. I'm sorry, but this is painful.

At least one boy means there is a constant boy... One of the two *will be* a boy. this is a constant. So you show the combinations with the established boy in as child 1 and the combinations with the established boy as 2.

"So you show the combinations with the established boy in as child 1 and the combinations with the established boy as 2. "

And you end up counting the freedom you had in choosing the B in the BB case as two distinct BB cases.

On other subjects -- how does one go about getting invited to this poker night you're always bragging of, Draug?