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Mike - nonsense. see above.
Sorry smee, that reasoning is entirely wrong. If you know one is a girl it clearly includes both options as both contain a girl. The only one it excludes is two boys, as neither is a girl?
B/G, G/B are two distinct possibilities and both meet the criteria of 'at least one girl'
[i]I then ask what is the gender of your second child (if you have one)[/i]
That's not the same question.
Wiredchops - you know that one IS a girl, so you cant have both options. It must either be one or the other, it cant be both.
It can't be both, but being either one is possible. How possible? 1 in 3...
Apart from repeating what has already been said, I can't see another way of changing your mind. The proof is all here, and it's been explained several times.
(Am I right in thinking Smee is Glupton? If so, there's a 100% chance that convincing him he's wrong is impossible.)
Yup Smee is Glupton.
miketually, I remember reasing something about the monty hall program and big arguments ensuing about it. One person was so massively convinced that he made some gigantic spreadsheet with 10,000 trials or something, relishing the thought of smashing it into the face of the smug maths guy. Only to see the probability drop out at 66%.
Then began the fiery e-mails to microsoft regarding the massive bugs in Excel....
Sometimes you just have to walk away.
If you choose B/G - how can you possibly also include G/B?
Smee,
Ok look at it this way, (I'm waiting on stuff to process on my PC so have time to kill)
There are three distinct outcomes, all boys, all girls, mixed.
For the last option (mixed) there are two distinct paths to this outcome.
G/B, B/G
You are choosing only one outcome (mixed)
This inherently includes the two 'possible' routes
one outcome is excluded completely (all boys)
the remaining outcome (all girls) has only one possible route to it, G/G
therefore, of the three possible routes, two of these lead to the mixed outcome.
You have a 2/3 probability of ending up with mixed kids. That's assuming your attitude to inter-racial relationships is open. (Guffaw, ignore that last awful quip)
If you choose B/G - how can you possibly also include G/B?
Who's choosing anything? BG is one possibility. GB is another possibility. GG is a third possibilty. You don't choose anything, you just state how likely each is to happen.
As a side note. Get the book by the dude who does the puzzles in the saturday guardian, Chris Maslanka. He does these awesome puzzles which you hack away at with algebra, probability and all this other foofurah only to flick to the answer and see that he's explained it eloquently in a single line. Awesome stuff. He has books of puzzles out, recommend.
[i]If you choose B/G - how can you possibly also include G/B? [/i]
It's the lack of specifics that keep the 2 options.
Have to go into work or I'd give you analogy that helped me see why but I'd be wasting my time with you as Glupton is never wrong.
miketually, I remember reasing something about the monty hall program and big arguments ensuing about it. One person was so massively convinced that he made some gigantic spreadsheet with 10,000 trials or something, relishing the thought of smashing it into the face of the smug maths guy. Only to see the probability drop out at 66%.Then began the fiery e-mails to microsoft regarding the massive bugs in Excel....
A guy from here (BenKinetics, I think) took quite a lot of money off someone who was convinced that it made no difference whether you switched or not on the Month Hall problem, so he did it for real with three cups and some pound coins 🙂
If I look at this logically then I get 50%, if I look at it using numbers then I agree 66%....so basically I'm going to give up and revise for my exam tomorrow!....:(
Just for reference, can everyone give their answers to the various problems discussed here recently? I'll go first:
Plane On Conveyor: It takes off.
Monty Hall: Always switch doors.
Boy-girl: 2 in 3 chance it's one of each.
I see where you're coming from but I don't agree, if I have a daughter then there's no reason why my second born is more likely to be a boy
Correct. Those are independent events - as discussed.
But the case where you have a daughter and you are waiting for a second born is entirely different to knowing that one of the children is a daughter.
If the question was:
You ask a woman how many children she has and she says two.
Then for some odd reason you ask her "Is [b]the eldest one[/b] a girl?", to which she replies "Yes".
Then it would be 50:50
But it only asks [b]is either of them a girl[/b].
As I have said previously I understand how you are getting 66%, I simply dont agree that the method you are using is correct in this scenario.
You know you have one girl , that alone removes the options that start with a boy. End of thread.
take off
Switch
66%
smee, no, you know you have one girl AFTER she's had both kids so the second child could be a girl as much as the first.
You know you have one girl , that alone removes the options that start with a boy.
But the option that begins with a boy and ends with a girl (stop s****ing at the back) is still valid, if the girl you know about is the youngest daughter.
What you're saying here is only true if the woman tells you her eldest child is a daughter.
Ok, I think that the way I've been approaching this has been flawed and I see where my logic has gone tits up!
Now listen here Mr. Smee..
[b]Here is a spreadsheet (oh yes) showing a sample of 100 parents with two children:
http://spreadsheets.google.com/ccc?key=p_H5o2Sep3PNxPhB1Cjb-Dg [/b]
Now tell me what the probability is of a parent on that spreadsheet having mixed sex siblings?
GrahamS, not heard owt from him for a good old while, I reckon he's discovered he's wrong and scuttled off quietly (yes, this is a blatant lure)
But but.. I spent ages on that spreadsheet 😯
Hopefully he has realised the error of his ways, though his "End of thread" statement suggests he has instead run off crying because he's confused and scared by what the bigger boys are saying. (blatant lure number two)
(yes, this is a blatant lure)
(blatant lure number two)
Do we need to say his name three times in front of a mirror?
[i]Ok, I think that the way I've been approaching this has been flawed and I see where my logic has gone tits up! [/i]
Yup that's where I was going wrong too, until I realised as the order was never discussed it still leaves the B/G G/B option open.
Oh - I get it now.
It is still 50:50.
What...? But but.... look at the spreadsheet... 😕
The probability of Smee making someone eat his dust, attacking someone for helping his child or refusing to admit he's wrong is 100%
Analogy I was given and roughly what was used on here.
If I tossed the coins, and hid them from you (one in each hand), and
said "at least one is a head", then you would say "OK, so I know that
one is a head, doesn't matter which one - I'll assume it's the one in
his left hand. So all that I don't know is what's in his right hand -
50/50 chance. Easy.". But that's not right. It's not obviously wrong,
but it *is* wrong. You could analyse it another way:
"What's in his left hand? Could be a head or a tail - I don't know.
Suppose it's a head, then there is an equal chance that his right hand
contains a head or a tail. So - *assuming* his left hand has a head,
then there we have two equally likely outcomes. Suppose his left hand
actually has a tail. Then his right hand *must* contain a head because
we know there's at least one head. There's a third outcome. Since the
coins were tossed fairly and randomly assigned to each hand, those
three outcomes are equally likely, and we're back to our 1/3, 2/3
probability.
Why are you all talking about probabilities when the question asks about odds?
Odds are 2:1 - which is not the same as a probability of 66%
Awwww! Bless look he's all confused.
Drac - show me where in the preceding 160 odd posts that someone has said that the odds are 2:1 in favour of there being one child of each gender.
2:1 is a ratio, which can also be expressed as a fraction or a percentage.
Why are you hiding behind picking tiny faults, rather than admitting being wrong?
Awwww! Bless look he's all confused.
It's the dust. It affects his thinking.
I would suggest that it is a major fault, not giving the answer in the correct format.
Just think of it like this: you go into the bank and ask them for £1000 and they give you it in $ - would you be happy?
I wouldn't care if they gave me it in £20 notes or £50 notes though, which is closer than $s and £s.
Or are you going to claim is was the use of percentages that confused you?
Why are you all talking about probabilities when the question asks about odds?
Ahhh the old [i]"try and wriggle out of it on a technicality" defence[/i]
[b][url= http://en.wikipedia.org/wiki/Odds ]From Wikipedia[/url]:[/b]
In probability theory and statistics the odds in favour of an event or a proposition are the quantity p / (1 ? p), where p is the probability of the event or proposition. The odds against the same event are (1 ? p) / p. For example, if you chose a random day of the week, then the odds that you would choose a Sunday would be 1/6, not 1/7. The odds against you choosing Sunday are 6/1. These 'odds' are actually relative probabilities.
So if we are saying that we think that the probability (expressed as a percentage) is 66.66..% then [i]p[/i] would be 0.666..
Expressed as [i]odds[/i] that would be
p / (1 ? p)
= 0.666.. / (1 - 0.666..)
= 0.666.. / 0.333..
= 2 / 1
So yes, odds of 2/1 are [u]exactly[/u] the same as a probability of 66.66..%
Fact remains - I am the first person to get the correct answer in the correct format.
1:1, 50:50 or 50% is still a perfectly valid interpretation though.
Actually I asked for the answer to be expressed in "percentage odds" in [url= http://www.singletrackworld.com/forum/topic/the-boy-girl-puzzle#post-62902 ]the sixth post[/url].
show me where in the preceding 160 odd posts that someone has said that the odds are 2:1 in favour of there being one child of each gender.
I did. [url= http://www.singletrackworld.com/forum/topic/the-boy-girl-puzzle/page/4#post-66891 ]back here[/url] where I said:
[i].."mixed sex" versus "both girls" is 50:25[/i]
Funnily enough 50:25 is the same as 2:1 or are you now going to argue about basic maths?
1:1, 50:50 or 50% is still a perfectly valid interpretation though.
Define "perfectly valid" for me. If you mean it is "perfectly valid" to come up with completely the wrong answer, which is empirically proven to be incorrect then yes, well done.
Don't feel bad Smee - it is a deliberately counter-intuitive problem - but there is only one correct answer and that is 66.66% (aka p=0.666.., aka odds of 2:1)
[i]Drac - show me where in the preceding 160 odd posts that someone has said that the odds are 2:1 in favour of there being one child of each gender. [/i]
Well Graham beat me to it but before you make wild claims I suggest your work out odds as percentages.
Well done everybody for making this thread last 175 posts so far.
And I realised my mistake on the first page! 😉
Probability is not an easy thing to explain in fairness though!
For Smee, perhaps another twist on explaining this:
If we DO NOT know the sex of any child, then the probability of there being a boy and a girl (Boy/Girl and Girl/Boy) is 2 opportunities out of 4, or 50%.
If we know the sex of the FIRST child is a girl, but not the sex of the second child, there are only 2 options... Girl/Boy and Girl/Girl. Therefore the probability of there being a boy and a girl is 1 opportunity out of 2, or 50%.
If we know the sex of the SECOND child, but not the first, then apply the same maths as above. Probability would be 1 in 2, or 50%.
Now, we only know the sex of ONE of the children, not whether it was the first or the second child. Because we don't know whether or not that the girl (that we know exists) was born first or second, we have to assume there were 3 possibilities of combinations of children being born... Boy 1st/Girl 2nd, Girl 1st/Boy 2nd, Girl 1st/Girl 2nd. As 2 of these opportunities contain a boy and a girl, in either order, and 1 doesn't, the probability of there being a boy and a girl is 2 opportunities out of 3, or 2/3, or 66.66%, or 2:1 etc.
I am now leaving the thread, never to return! 😆
Wow.
Smee, this really is incredible, so glad you came back. Here I was thinking you had walked off noble in defeat. This is much more fun!
Smee... humour me here would you by answering a simple question.
Which one of the following sets of probabilities is correct for a woman having two children, taking into account that the sum of all probabilities must equal 100%:-
Boy/Boy 25%
Boy/Girl 50%
Girl/Girl 25%
or
Boy/Boy 33.3%
Boy/Girl 33.3%
Girl/Girl 33.3%
Good one funky, here's sure to admit his mistake now