i was in the 50% column until I ran
double boys = 0;
double girls = 0;
Random r = new Random();
while(girls 1000000 )
{
int first = r.Next(0, 2);
int second = r.Next(0, 2);
if ( first == 1 || second == 1 )
{
if (first == 0 || second == 0) boys++;
girls++;
}
}
double percentage = boys/girls;
and got .667199
I guess knowing the ordering matters. If you know the ordering it changes the percentage to .50 Weird
50%
The fact that one is a girl has nothing to do with the probability of the event of the other being a boy.
Anyone saying odds are better than that, I live in chandler Arizona and would love to play poker with you.
33%
True, this is not exactly monty hall, but it can be looked at the same way. There are essentially 4 ‘doors’ in this problem: gg gb bg bb. By saying that at least one is a girl they essentially ‘open’ the bb door. This leaves 3 doors, two of which have one boy and one girl behind them. probability 2/3.
Could also be solved as follows:
P(girl then girl) = .5*.5 = .25
P(girl then boy) = .5 * .5 = .25
P(boy then girl) = .5 * .5 = .25
P(boy then boy) = .5 * .5 = .25
thus
P(girl and boy) = P(boy then girl) + P(girl then boy) = .5
so gg = .25 bg = .5 bb = .25
eliminating the possibility of bb, you need to get the probability of bg from the remaining choices:
bg/(gg+bg) = .5/.75 = .66
Actually, 100%. They say one of them is a girl. Therefore the other must be a boy.
quick thought: 45% and I’m not ashamed of being wrong.
I read once somewhere that 55% of humans are female.
if(numberOfGirls == 1 numberOfBoys == 1)…
else if(numberOfBoys == 1 numberOfGirls == 1)…
There is a very clear difference between those two cases - Where you’d crash with the following definitions:
numberOfGirls=0
numberOfBoys=one
See? Order does matter 
We already know numberOfGirls 0 so that one is out from the start.
I wasn’t trying to show all the possibilities, just my opinion that saying BG != GB is silly.
And what abou hermaphrodites???
It’s 50%. You already know that one of the two kids is a girl, so the problem is reduced to What is the probability that the other one is a boy?
I’m going for 2/3
The combinations could be:
Boy/Girl
Boy/Boy
Girl/Girl
Girl/Boy
We know one of the children is a girl so we can drop the boy/boy combination. This leaves three combinations, two of which include boys.
Toot toot.
50% or .5 probability.
Two thirds.
My explanation:
Since this is hypothetical, we may assume we are talking about the expected value. We may further assume that in an independent experiment (i.e. birth), the odds for a boy or a girl are identical at 50%.
Now, for two children there are four possibilities, with equal probabilities:
Options are BB BG GB GG
Comment rules out BB so the answer is 2 to 1 (assuming B/G 50/50)
But realistically, anyone who says one is a girl implies the other is a boy
2/3
If they have 2 children, there are four possible permutations, all other things being equal:
BB
BG
GB
GG
If you know that one of them is a girl (ignoring the very valid semantic point that Isaac raised) that eliminates the ‘BB’ possibility. Of the three remaining possibilities, two satisfy the ‘boy and girl’ condition. Therefore, 2/3.
@Isaac - that’s brilliant reasoning.
I’d say that there are four possible combinations for two children:
a) boy + boy
b) boy + girl
c) girl + boy
d) girl + girl
and that we could naively assume 25% probability for each.
(a) is ruled out, therefore the person has one of a,b,or c, with 33% probability each. The odds of boy and girl are therefore 66%.
i have to revise my answer to 1/3 because my answer was for probability of both children being girls.
Same cards analogy, let A seeing green on first side, B = seeing RED on other side (not green as my original post had it).
P(A and B) = 1/6, not 1/3 because there are six possible ways of drawing one card (3 cards * 2 sides per each card). So
P(B|A) = P(A and B)/P(A) = (1/6)/(1/2) = 1/3
Of course, the closer answer is 68.75%.
The natural birth sex ratio being 1.1 boy/girl, probabilities for pairs are:
bb 27.44%
bg 24.94%
gb 24.94%
gg 22.68%
Discarding bb and considering the remaining population:
bg 34.38%
gb 34.38%
gg 31.25%
so bg|gb = 68.75%
Ok, I cheated: I did a web search to find the sex ratio there:
http://en.wikipedia.org/wiki/Human_sex_ratio#Natural_ratio
There are a ton of very different, very complicated answers to a question that a 4th grader can figure out.
They tell you a person has two children, and they ask you the odds of the person having one boy and one girl. Without any other information, the answer would be 50%(combos of: GG, GB, BG, BB).
They tell you that one of the children is a girl, so all they are asking for is the probability that the other child is a boy. There are only two choices to choose from(boy/girl), meaning you have a 50% chance of getting it right. Nothing else matters, probability does not change of the second child being a boy/girl no matter what the first child is.
Simple Example: I flip a coin, it comes up heads. I flip the coin again, what are the odds it comes up heads again?
It is completely irrelevant what happened the first time you flipped the coin, the odds don’t change, it is still 50%. It’s the same with children, you have a 50% chance of a baby being a boy or a girl.
This problem is well documented:
See http://en.wikipedia.org/wiki/Boy_or_Girl
for the answer being 66% Have a good day
This problem is well documented:
See http://en.wikipedia.org/wiki/Boy_or_Girl
for the answer being 66% Have a good day