[FizMan]

I think we can all agree on the four possible combinations of children, right?

So imagine four houses, each having one different combination of children.

The question specifically states that a girl answers the door. I read that as a given. Leaving three houses left; two with a boy, and one with a girl. If you randomnly chose either house than it'd be a 33% chance.


IF you guys were going by the probability that a girl answered the door in the first place... then consider this. We can naturally exclude the boy/boy combination from calculations. This leaves FOUR girls and TWO boys in our last three possibilities.

So what are the chances that a girl will answer the door? 66%

If you choose randomnly between the houses, House A (has girl/girl) will yield a 100% girl answering door chance. Whereas Houses B,C (girl/boy x 2) have a 50% chance each. But the probability of having either possible combination is 33% (three possibilities)

1/3 x 100% + 1/3 x 50% + 1/3 x 50%
33% + 16.5% + 16.5% = 66% (2/3)

So you have a 66% chance that a girl will open the door, this is where your calculations should start. I agree with the idea that at this point, it can be a 50/50 chance for the second child being a boy or a girl (just like flipping a coin really), but since we already have the 66% chance of a girl opening the door in the first place, we multiply it by the 50% and we end up with a 33% chance of a girl/girl possibility through this method.