Home Forums Chat Forum The boy-girl puzzle

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  • The boy-girl puzzle
  • Smee
    Free Member

    Oh – I get it now.

    Smee
    Free Member

    It is still 50:50.

    GrahamS
    Full Member

    What…? But but…. look at the spreadsheet… 😕

    miketually
    Free Member

    The probability of Smee making someone eat his dust, attacking someone for helping his child or refusing to admit he’s wrong is 100%

    Drac
    Full Member

    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.

    Smee
    Free Member

    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%

    Drac
    Full Member

    Awwww! Bless look he’s all confused.

    Smee
    Free Member

    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.

    miketually
    Free Member

    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?

    miketually
    Free Member

    Awwww! Bless look he’s all confused.

    It’s the dust. It affects his thinking.

    Smee
    Free Member

    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?

    miketually
    Free Member

    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?

    GrahamS
    Full Member

    Why are you all talking about probabilities when the question asks about odds?

    Ahhh the old “try and wriggle out of it on a technicality” defence

    From Wikipedia:
    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 p would be 0.666..

    Expressed as odds that would be

    p / (1 ? p)
    = 0.666.. / (1 – 0.666..)
    = 0.666.. / 0.333..
    = 2 / 1

    So yes, odds of 2/1 are exactly the same as a probability of 66.66..%

    Smee
    Free Member

    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.

    GrahamS
    Full Member

    Actually I asked for the answer to be expressed in “percentage odds” in the sixth post.

    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. back here where I said:

    ..”mixed sex” versus “both girls” is 50:25

    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)

    Drac
    Full Member

    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.

    Well Graham beat me to it but before you make wild claims I suggest your work out odds as percentages.

    mboy
    Free Member

    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! 😆

    Wiredchops
    Free Member

    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!

    funkynick
    Full Member

    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%

    Wiredchops
    Free Member

    Good one funky, here’s sure to admit his mistake now

    Smee
    Free Member

    funkynick – the first one. neither are relevant to this one though.

    Wiredchops
    Free Member

    Smee, I realise you reckon your reasoning comes from the fact you can discount the boy/girl combination of children, can you rationally justify that? I’ve not read a coherent response to this assertion yet.

    funkynick
    Full Member

    Okay then, can you please explain why neither of those sets of probabilities are relevant?

    GrahamS
    Full Member

    Smee has already admitted that the “Odds are 2:1”

    His only bone of contention was that odds of 2:1 are somehow different from a probability of 66%, which it isn’t (as shown above).

    Note that he also refuses to provide any working or answer any of the direct challenges to his “theories”.

    So I can only assume that A) he now agrees with us and/or B) he is just playing it dumb as a some mediocre form of trolling.

    Smee
    Free Member

    You have two children, you know that one or more is a girl. This means that at least one of them can’t be a boy. This removes the options of boy/boy and boy/girl.

    Why does it do this? Forget age. Lets call them kids 1 and 2. 1 is a girl – we know that – they may be the youngest, they may be the oldest it doesn’t matter – what does matter is that they are not both. That is why you can remove one of the boy/girl, girl/boy permutations.

    miketually
    Free Member

    You have two children, you know that one or more is a girl.

    Correct.

    This means that at least one of them can’t be a boy.

    Yes.

    This removes the options of boy/boy

    True.

    and boy/girl.

    Wrong.

    mboy
    Free Member

    Sorry, stepping back in.

    WRONG Smee

    1 and 2 refers to the order in which they arrive. WE DO NOT KNOW that 1 is a girl, or that 2 is a girl, merely that at least one of them is.

    Boy/Boy option is removed, Boy/Girl option is still valid as the lady said “at least one of them is a girl”. It was not stated that the first child, nor indeed the second child is the girl, merely that “one of them is”.

    Therefore Boy/Girl, Girl/Boy and Girl/Girl are all valid options. In 2 of these there is one girl and one boy, therefore a 2 out of 3 chance of there being one of each, therefore a 66.66% chance of there being a boy and a girl, given that she has already told us that “at least one of them is a girl”.

    Wiredchops
    Free Member

    Ah! That explains it, you have a fundamental misunderstanding of the question.

    You have two children, you know that one or more is a girl.

    Yup

    This removes the options of boy/boy and boy/girl.

    nope

    Smee
    Free Member

    But boy/girl has the same content as girl/boy…. i.e. one girl and one boy. So therefore they are the same.

    miketually
    Free Member

    But boy/girl has the same content as girl/boy…. i.e. one girl and one boy. So therefore they are the same.

    Also wrong.

    Wiredchops
    Free Member

    But, but, so many people have explained this subtletly so many times! And I was so proud of my version. Boo hoo hoo! 😥

    What about the little tree diagrams! And the big calculation things!

    Even the text descriptions were accurate, succintly worded and clear!

    You just don’t like us! That’s it isn’t it!

    Dear me.

    Smee
    Free Member

    Mike – are you mental? Read what you have written.

    Smee
    Free Member

    I get the idea that most of you are using I just don’t agree that it can be used here.

    miketually
    Free Member

    Mike – are you mental? Read what you have written.

    I may be. But, I understand the question and the answer, which you seemingly do not.

    miketually
    Free Member

    I get the idea that most of you are using I just don’t agree that it can be used here.

    Ah, so the world and the laws of mathematics are wrong?

    Smee
    Free Member

    Ok then Mike. Boy/Girl and Girl/Boy – remove the ages of them and what is the difference?

    Wiredchops
    Free Member

    How about this baby.

    There are two towns.

    Town A

    Town B

    Town A has two roads leading to it

    Town B had only one

    Pick a road purely at random

    Probability you’ll end in town A?

    66%!

    Think of town A as mixed kids

    Think of town B as all girls

    the two roads to mixed kids = G/B B/G

    one road to all girls = G/G

    Smee
    Free Member

    Wiredchops – you can try answering that question too.

    djglover
    Free Member

    Until now, I have resisted looking at this thread, is it 5 pages of Smee making a fool of himself?

    Wiredchops
    Free Member

    Ok, I’ll try.

    The position of the child has an inherent value with regards to order alone.

    I.e. There is only one possible way of achieving a girl girl child combo when having two children.

    There are two possibilities of achieving a mixed sex combo. Namely, G/B B/G

    If you remove the ages of the children the only difference is the order in which you have written them down. WHICH IS NOT TO SAY IT RENDERS THEM THE SAME

Viewing 40 posts - 161 through 200 (of 463 total)

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