Maths problems (interesting ones)

[Alan Anderson]

Originally Posted by Kris Verdeyen

Actually, I'm the only one catching it - it's twice as likely that a girl will initally answer from a G/G family than from a B/G family, and there are twice as many B/G families. Therefore, the other sibling has a 50/50 shot of being girly.

No, you're counting families too soon. Only half of of the boy/girl families had a girl answer the door, so the others don't meet the criterion for this problem.

None of the boy/boy families are included. Half of the boy/girl families are included. All of the girl/girl families are included. But once a girl answers the door, you have to completely forget all the families that didn't get included. You can't count them, because they don't exist in the set of familes that did have a girl answer the door.

[FizMan]

Okay let's try this again...

Four possible combinations of children. Right? That's simple math too, you have two inputs, four outputs.

A girl answers the door... This is GIVEN.

We've now determined that it is impossible for the family to have a boy/boy combination.

Since a girl answering the door is GIVEN, it leaves us with three possibilities.

Two of those possibilites have boys. One possibility has a second girl.


I don't know how else to go aboot this, but I for one am going to trust my OAC finite before I trust logic that I feel is flawed.

[Joe Ross]

I think it comes down to, how are the kids selected to answer the door. I don't beleive the problem is clear enough, and that's why you have two different arguments for the answer.

When Kris reads the problem, he sees no method for picking which kid goes, and so assume random. That throws out the possibility of 2 boys, and also 1 of the 1 boy 1 girl possibilities. You get 1/2

When Denman and FizMan read the problem, they see that if a girl is in the family, she is sent out first. Thus, you can only eliminate the 2 boys possibility, and get 1/3.

Is one or the other the correct reading? Make your arguments to convince me.

[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.

[Denman]

If you know stats then think of it as permutations / combinations
i will write a full answer later, but its only 9am and i am tired lol