No, Dr. O. Just choose whichever other identifying way you want of distinguishing them (but not one that's sex-related). Like the one with alphabetically earlier name, etc.

Read that link and you will see both justifications. The simple fact is, just as with cpin tosses, any given binary selection is not affected by the one before or the one after so it is a 50/50 shot that any given child will be a boy.

I have read the link, Draug. I was careful in the way I asked my question.

You are unjustifiedly assuming that a child has been singled out, which is not the case. (Perhaps this was the gist of spyman's initial concern).

twist: mr jones is krellin

Let me put it another way. BG GB are one in the same. Position being irrelavent you can't include both so it becomes just 2B 2G and B+G. Eliminate 2G and it becomes 50/50.

As for coins, I repeat: given only the information that a fair coin was tossed twice, and that at least one of the tosses was heads, the probability that the other toss was heads is 1/3.

If you disagree, I will be more than happy to co-design a random experiment with you and put money on the results.

no i saw the revised question
your bb-bg-gg-gb only works if there is a first and then second child
if both children are born at the same time do there is no first or second child only two children then the probability of each child is 50/50 so it would not matter whether one of them is a boy or not. both children have the same chance of being either male or female

BG and GB are not one and the same probabilistically, Draug. Together they account for 2/3 of the probability space.

Their being born at identical moments does not change a thing, dr. O (except via the probabilities that twins are fraternal or not).

Ok, I'll change my answer again. I agree that 1/3 is correct, based on the elimination of the double G possibility.

no it does change, as Draug says if position is irrelavent then you have either 1boy and 1girl, or you have 2 boys. out of two posible solutions then chance of having 2 boys is 50/50

My hamdbook of applied mathematics agrees with me. If you don't have a definitive position the all outcomes with the same division count as one outcome.

So let us extend this out further to three kids. One is a boy for certain. The odds of havimg at least one more boy become 2:1 (3B, 2B+1G:2G+1B with no chance of 3G).

#include <iostream>
#include <cstdlib>
using namespace std;

int main()
{

srand(time(0));

unsigned long cBoth = 0, cSplit = 0;

int r = 0;
for (int i = 0; i < 10000000; ++i)
{
r = rand();
if ((r & 0x01) || (r & 0x02))
{
if ((r & 0x01) && (r & 0x02)) ++cBoth;
else ++cSplit;
}
}

cout << static_cast<double>(cBoth) / static_cast<double>(cBoth + cSplit) << endl;

}


/* So what do you say, Draug? Run it and $100 to the site, whoever is closer? (You if the output is closer to 1/3, me if it's closer to 1/2. And I haven't compiled or run it yet, so there may be errors).*/

Wait a damn minute! When alderain asked me about simultaneous tosses of a coin, I thought he was just bringing back the synapse thread where poor synapse was hounded by draugnar across multiple threads because he thought the answer would be different if they were tossed at the same time versus consecutive. Now because they're babies it matters?

WTF? Toss two babies in the air at the same time or one at a time, it doesn't change the probability! I thigh everyone but synapse had been convinced of this months ago!

@dr. octagonopolous- The phenomenon that you are describing actually does happen with indistinguishable particles. If the question was talking about the spins of indistinguishable particles then the answer would in fact be 1/2. The thing about twins, though, is that they are distinguishable. For example, you can put a sticker on one of their foreheads and tell them apart that way.

@Draug- Maybe you should reread that wikipedia article. It doesn't say what you think it says.

Convert that out of PHp and into C# so I can.follow it on a stressed brain (been writing Linq queries and consuming web services and.parsing XML elements and interfacing with remote servers and mamaging memory and parallel processing threads all day - brain.is totally frazzled)and I'll look for any flaw then run it if I find it acceptable. But all money will be a site donation. I've already proven my word is my bond on debts.

And while were betting Draug, and bringing back prior threads, ill bet draug that if you throw in the following statement:

int sz = sizeof(cBoth);
count << sz << endl;

The answer will be 32, not 64

Draug,

It's in C++, not PHP. I don't know C#, so I'm going to post it in pseudocode. Then you can convert it to C#, post it again, and if I agree it looks right, then we run it. That seems more efficient than me looking up how to do things in C#. So give me 3 minutes.

Draug, you insult C++ programmers calling that php!

Are we seriously still arguing about this? Uclabb, next time you post a goddamn math answer, bring the damn evidence so that people don't fucking argue it otherwise. There isn't any damn "my feeling on this is" or "my applied math says" or "I've seen" crap. Just math.

You can answer this question in at least two ways.

One is by definition of the conditional: the question asks what is the probability of having both boys given at least one of the two kids is a boy? Translated to math, it's

P(both boys Given at least 1 boy)=P(both boys AND at least 1 boy) / P(at least one boy)=(1/4)/(3/4)=1/3

Or you can do it by looking at the sample space: For two kids with equal probability of either, the sample space is a 2x2 {GG, GB, BG, BB} **same as tossing two coins.**
If at least one kid is a boy, then the sample space is {BG, GB, BB}. Each event is equally likely, so the probability of having both boys is 1/3. Same as the probability of having two heads given at least one of them is heads.

If you were still stupid enough to not understand, please run a random program generating GB or HT say about 1 million times. Then after that, collect all the outcomes where there is at least 1 B or 1 H, and out of that, calculate the prob of getting BB or HH.

Ok, pseudo code will do. My brain is so frazzeled I wasn't even looking close at the code to.tell it was C++.

I thought you said you were cpp, back in the day draug, you never forget iostream if you've ever used it.

SeedRandomFromSystemTime;

Created unsigned integers cBoth, cSplit;

cBoth = 0;
cSplit = 0;

Create integers r1, r2, i.

r1 = 0;
r2 = 0;
i = 0;

for i = 1 to 10,000,000
{

r1 = RandomInteger() Mod 2;
r2 = RandomInteger() Mod 2;

if (r1 == 1 or r2 == 1) // One or the other is odd
{
if (r1 == 1 and r2 == 1) // both are odd
{
++cBoth;
}
else // only one is odd
{
++cSplit;
}
}

Print (float)(cBoth)/((float)(cBoth + cSplit)); // print the ratio that were both odd.
}

@uclabb, we have been given no information that the children are not indistinguishable, we are only given 2 facts, one: there are two children. two: one of these children is a boy
while in real life, children are yes easily differentiated the question does say using ONLY (in big shouty capital letters) these two facts
in response to the bet please note my first response was "what if i through a spanner in the works"

I did post the math answer. There are three possible microstates BG, GB, BB with eqaul probability, thus the answer is 1/3. Everything else you wrote is just wordiness. If my above explanation isn't enough, then my instinct is to refer people to the last paragraph of your post.

And yes, $100 to the site is fine. I'll trust you.

@dr octagonapus- First, I want to point out that I thought your point was a good one, and I addressed it as such. The thing, though, is that twins *can't* be indistinguishable by virtue of the fact that they are freaking huge.

@uclabb- i would have to agree, was just trying to make a point stick