To me there’s no ambiguity. In both cases you know she has two children and at least one is a girl. This doesnt affect the probability that the second (whether it was chronologically second or not) was born male or female. Being told she has “at least one girl” doesnt affect the probability of the second child’s sex.

To me there’s no ambiguity. In both cases you know she has two children and at least one is a boy

Girl, but yes that’s how I see it we are not also trying to find out the order they were born so boy/girl is the same as girl/boy.

I asked the question in, what I hope, is an unambiguous way.

Here are two explanations. I find that typically people will click with one of them:

Let’s play a game. You toss two coins and don’t let me see the results.
I’ll ask you if one of them is heads, when you reply I’ll then guess what the other coin is.
Hopefully you can see that this isn’t a fair game. I’ll win most of the time. But it is exactly the same as the boy-girl problem.

-Or-

Okay take a group of 100 of these parents. If they all have 1 child then 50 of them will have a girl and 50 will have a boy.

B = 50, G = 50

When the 50 that have a Boy then have a second child, 25 of the will have a second Boy and 25 of them will have a Girl. Likewise for the 50 that had a Girl first.

So out of the 100 parents we now have a nice even spread:

BB = 25, BG = 25, GB = 25, GG = 25

With me?

If you take one of those parents at random, ask them if they have at least one girl and they say yes then you know they must be in BG, GB or GG.

So they will have a boy in 50 out of 75 cases… or 66%

Ah yes, because you’re not selecting from the original group. Makes sense. But I dont see the difference between that and the bonus question – the bonus question gave the same information didnt it?

I still don’t really get I almost did with the example of a hundred parents but were asking one she either has a boy and a girl or 2 boys, still can’t see why the order matters with one subject.

How about raising a suggestive eyebrow, swivelling one’s hips a little, adopting a louche tone of voice, and asking said lady if she’d like any more kids?

What’s the probability that she’d say ‘yes’?

The order doesn’t matter Drac, only the fact that there are twice as many “a boy and a girl” as there are “two girls”.

If you were asking “are your children mixed sex or same sex?” then it would be 50/50, but “mixed sex or two girls” is 50/25.

There’s always this little one that I heard on the radio the other day:-

A trader enters a bank where everyone else is doing exactly the same trade, and that trade has a 50% chance of making a profit. The trader knows of a trade that has a 75% chance of making a profit.

But in order for the individual trader to get his bonus, two things have to happen: the bank has to make a profit and the individual trader has to make a profit.

So, for the trader to get the best chance of getting his bonus, what should he do?

coffeeking: yeah that one is even subtler.

Basically if you say “Is at least one a girl?” then there is a possible negative response – she could say “No”. So when she says “Yes” you can eliminate the No cases.

Whereas if you just ask for the sex of one of them then you don’t actually gain any information.

(I think. To be honest that one still makes my head spin a bit)

I came into this and only saw the whole girl/boy, girl/girl thing. I thought I was on a very different thread.

This one will really mess with your minds.

Apparently the sex of one baby can influence the sex of another.

I’ve read somewhere thatif you have had two children of the same sex, then there is a statistically significant chance that if you have a third, it will be the same sex as the previous two.

The order doesn’t matter Drac, only the fact that there are twice as many “a boy and a girl” as there are “two girls”.

I still same that combination as the same girl/boy is no different to boy/girl.

Drac.. order doesn’t matter as long as you take into account that it it is twice as likely that two children are boy/girl as it is for them to be girl/girl.

So, if:-

the probability of having girl/girl = x

then

the probability of having boy/girl = 2x

total sum of the probabilies = 3x

therefore

probability of the woman having boy/girl = 2x/3x = 2/3 = 66%

Why the **** didn’t you write that in the first place I’ve got it now. 😳

Mind I never said I was right just couldn’t grasp it in writing, maybe I should have wrote an equation down.

Drac.. I know it’s hard to believe, what with arguing maths on an internet forum, but lets assume that I have a child X, that child could either be a boy or a girl. Now assume I have another child Y, and that child could be a boy or a girl.

Now, that gives us four combinations:

X B B G G
Y B G B G

As you can see:

the chances of having boy/boy = 25%
the chances of having girl/girl = 25%

Therefore

The chance of having boy/girl (irrespective of order) has to be 50%

If I ever have kids they are gonna hate me for just giving them letters for names though…

😀

See don’t write it like that use numbers I then get it. 🙂

Glad you got there Drac. Don’t worry, like the Monty Hall problem it is a counter-intuitive headf**k for most people.

http://en.wikipedia.org/wiki/Boy_or_Girl_paradox

The Monty Hall problem I never did get for ages then it sunk in why when watching 21.

What are the odds o nnumber 8 being a certain sex when you were only expecting 7 though??

50%… :o)

Guys the chances of you EVER getting girl/girl are 0%.

I know we all harbour less than discrete fantasies about getting in on a bit of girl/girl lezbo action, but seriously, it’s never going to happen.

funkynick

Here’s one you should like.

In my pocket, I have a die that I use to play board games with.

What do the sides of the die add up to in total?

miketually – Member

Oh, and as for the plane on a conveyor, the wheels on a plane have nothing to do with its ability to take off or not. Air speed over the wings is what matters!

You’re correct on both of those. But that still doesn’t answer whether or not it takes off. I want to know what, because I’d love it to turn out that you’re useless with physics as well as verbal reasoning after your rant about people not understanding probability above [:)]

Your point being?

It was late, I read the question slightly wrong (I read it that the lady had announced the question of the first child, rather than just “one of them”), realised my mistake, then apologised.

Regarding the plane, there are WAY too many variables that were not discussed in the thread to dictate whether or not the plane would take off. Speed of the conveyor belt, wind speed, wind direction, are the engines running or not, required take off speed for the plane etc etc. That’s as well as needing to know if the wheel brakes are on or not, and if they’re not, the friction coefficient of the conveyor belt, the friction coefficient in the wheel bearings, the drag coefficient of the plane etc etc etc.

Huh? That could be anything. A die is polyhedron, so it depends how many sides it has and how they are numbered surely?

What do the sides of the die add up to in total?

How many sides?

GrahamS and Drac, fair do’s. Most people usually just rush in with “21” as that’s what the sides of a “normal” 6 sided die add up to.

There is an equation to work it out though, regardless of the number of sides! Prize to the first person to work it out 😉

Nice try Mboy but I use to play AD&D so was well aware.

n/2 * (n+1)?

Assuming of course that the die is sequentially numbered from 1.

On the ball GrahamS!

Again, everybody forgets to mention that the die should be sequentially numbered too at this point.

Bonus point for working out the formula for a die that does not necessarily go up sequentially by 1 on each side (though assuming that each increment is equal).

Regarding the plane, there are WAY too many variables that were not discussed in the thread to dictate whether or not the plane would take off. Speed of the conveyor belt, wind speed, wind direction, are the engines running or not, required take off speed for the plane etc etc. That’s as well as needing to know if the wheel brakes are on or not, and if they’re not, the friction coefficient of the conveyor belt, the friction coefficient in the wheel bearings, the drag coefficient of the plane etc etc etc.

It’s a normal plane, taking off as normal on a runway that happens to be moving in the opposite direction to the plane. The runway’s not made of glue or treacle and the pilot’s not an idiot, so he took the brakes off before trying to take off. The wheel bearings have not been replaced with kryptonite and Superman isn’t holding the plan back. Nobody is exceeding the excess baggage allowance and the air traffic controllers are not on strike.

It. Takes. Off.

n(2+d(n-1))/2

n = number of sides
d = increment

Sockpuppet, nearly. Brackets aren’t quite all there though.

It’s n(2+(d(n-1)))/2

The extra brackets do make the difference.

It is still 50:50. Any 2/3 is flawed maths.

If at least one is a girl you must take the first one as being that girl thereby removing them from the equation. Leaving a 50:50 chance of the next one being a girl.

not sure they do tbh…

does that mean no bonus point? 🙁

Is it just me misreading the brackets or is n(2+d(n-1))/2 not exactly the same as n(2+(d(n-1)))/2?

And it’s 2/3, not 1/2.

In the second you times it by d then add 2 where as in the first you times it by d+2

[teams up with miketually]

yeah!

(elbows owenfracknell out of the way)

[/teams up]

I’m with mike on this one too, there is definitely a redundant set of brackets in there…

The correct answer is C

End of thread.

HTH

Yay smee! Fight the logic. Keep this thread going 🙂