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    One for the mathematicians

    Lauren Laverne has just had a mathematician on her show returning for her weekly slot having had a break to have her second child. Her first child was a girl. What are the chances, she asked, that the girl now has a sister?

    She made it multiple choice giving 1/2, 1/3 or 13/27 as the possible answers.

    According to the maths the answer is 1/3. There are 4 possible sibling combinations, boy/boy, boy/girl, girl/boy, girl/girl. Knowing it can't be boy/boy leaves 3 possible combinations hence a 1 in 3 chance that the older girl now has a baby sister.

    Surely this is wrong though. The siblings in each combination aren't equal. One is older than the other. If the first child in each combination is the older child then the possible answers are reduced to girl/boy and girl/girl hence the answer to the question is 1/2.

    Who's right?

    #2
    Depends on the wording of the question, really. Is the wording really "What are the chances that the girl now has a sister?" In that case, what they are asking is, what are the chances of the new baby being a girl? That's 50%. The chances of any new child being a boy or a girl is 50%. The gender of the previous child is irrelevant.

    If the question is, "What are the chances that a mother's first two children are both girls?", then the answer is 25%.

    So I can't understand what question they are asking that results in an answer of 1/3.

    I also have no idea what the age of the siblings has to do with it.
    Last edited by anton pulisov; 09-07-2019, 09:28.

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      #3
      Unless I'm missing something I agree with you. The boy/ boy and boy/girl combination can't happen as we know the first child is a girl.

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        #4
        Is this supposed to be some variant on the Monty Hall problem maybe?

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          #5
          But what is the question being asked? Are mother, father and the first daughter trying to predict the gender of child two? That's not 1/3 chance of being a girl. It's 1/2. It's always 1/2. It's not like blackjack where the chances of drawing a 10 of Hearts again decreases because there is already a 10 of Hearts on the table.

          Let's say you have a 1 in a million chance of winning the lottery. You win the lottery. What are your chances of winning the lottery a week later? They are still the same. They haven't increased or decreased.

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            #6
            The question was definitely about the chances of the first girl having a sister in which case I think we're all in agreement that the answer to her question is 1/2. Their ages are irrelevant except in so far as it seemed the most convenient way of differentiating them.

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              #7
              In general, I would take these TV, radio and Ted Talk scientists with a grain of salt.

              A general rule of thumb is not to trust any scientist who has a Wikipedia page and is still under the age of 60 (or not yet dead).
              Last edited by anton pulisov; 09-07-2019, 09:41.

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                #8
                One can ask the question such that it becomes one about conditional probability, which can be a brain-damaging minefield. And yes, it is kind of a variant of the Monty Hall problem.

                A woman has two children. Given that one of the children is a girl, what is the probability that both children are girls?

                The answer to this question is indeed 1/3, since:
                - the probability that both children are girls is 1/4
                - the probability that one of the children is a girl is 3/4
                Hence, the probability that both children are girls _given that one of the children is a girl_, is (1/4)/(3/4) = 1/3.

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                  #9
                  This is not the answer to the question as asked on the show, since the show's question makes the mistake to stipulate that the child of which we know the gender is the _first_ child. In order to get to 1/3, we need to know that _one_ of the children is a girl, but we need to not know _which_ of the children is meant.

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                    #10
                    Ok so I've just looked up the Monty Hall problem and my intuitive answer, that it makes no difference whether or not one switches doors, is wrong.

                    Is it though, or does it depend again on how the question is phrased. In the standard version there are 3 doors, 2 hiding a goat, I hiding a car. The chance of my pick hiding a car is 1/3, the chance of one of the other doors hiding a car is 2/3. The host opens a door hiding a goat. The chances of my original pick hiding a car remains at 1/3 while the chance that the remaining unopened door hides a car rises to 2/3 hence I should always switch.

                    Does the question omit key information though. There are 3 doors, 1 hiding a car, 2 hiding goats, 1 of which the host will always open. Why does the chance that my original pick hides a car not therefore become 1/2?

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                      #11
                      Yes, the questioner has fucked this right up. The apparently paradoxical 1/3 answer only works if if the question is "there are two children, if one is a girl, what are the odds that the other is a girl too?".
                      Last edited by Rogin the Armchair fan; 09-07-2019, 10:21. Reason: Removed attempt at humour feigning outrage about categorising people by gender

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                        #12
                        That's shit, Rogin. Really, do better.

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                          #13
                          Yeah, doesn't look as sardonic as I'd hoped. Taking it down.

                          Comment


                            #14
                            Originally posted by Artificial Hipster View Post
                            Lauren Laverne has just had a mathematician on her show returning for her weekly slot having had a break to have her second child. Her first child was a girl. What are the chances, she asked, that the girl now has a sister?

                            She made it multiple choice giving 1/2, 1/3 or 13/27 as the possible answers.

                            According to the maths the answer is 1/3. There are 4 possible sibling combinations, boy/boy, boy/girl, girl/boy, girl/girl. Knowing it can't be boy/boy leaves 3 possible combinations hence a 1 in 3 chance that the older girl now has a baby sister.

                            Surely this is wrong though. The siblings in each combination aren't equal. One is older than the other. If the first child in each combination is the older child then the possible answers are reduced to girl/boy and girl/girl hence the answer to the question is 1/2.

                            Who's right?
                            This is a well known paradox of probability btw.

                            If you state the paradox as "I have two children, at least one of whom is a girl, what's the probability that the other is a girl", the answer to that is 1/3. That's an identical formulation to your stated problem, except in your case there's additional, irrelevant data added (that the girl is older). But the completely irrelevant data somehow changes the answer. Which doesn't make sense.

                            To pick a different example, cos older/younger is easier to get fixated on, if you meet an old friend in the street, who you know has two children but don't know their genders and he introduces you to his daughter. At that point, the probability of his other child being a girl is 1/3 using the BG/GB/GG decision matrix. If he then goes "this is my daughter Sally" you can put Sally/Not Sally at the top of your matrix and collapse the possible states into GB/GG so it becomes a half. Which doesn't make sense at all - knowing that one daughter is named Sally doesn't change the probabilities in any real sense.

                            So it's a paradox. Neither answer is correct and both of them are.

                            https://en.wikipedia.org/wiki/Boy_or_Girl_paradox
                            Last edited by Bizarre Löw Triangle; 09-07-2019, 10:56.

                            Comment


                              #15
                              Originally posted by Artificial Hipster View Post
                              Ok so I've just looked up the Monty Hall problem and my intuitive answer, that it makes no difference whether or not one switches doors, is wrong.

                              Is it though, or does it depend again on how the question is phrased. In the standard version there are 3 doors, 2 hiding a goat, I hiding a car. The chance of my pick hiding a car is 1/3, the chance of one of the other doors hiding a car is 2/3. The host opens a door hiding a goat. The chances of my original pick hiding a car remains at 1/3 while the chance that the remaining unopened door hides a car rises to 2/3 hence I should always switch.

                              Does the question omit key information though. There are 3 doors, 1 hiding a car, 2 hiding goats, 1 of which the host will always open. Why does the chance that my original pick hides a car not therefore become 1/2?
                              The way to intuit the Monty Hall problem is imagine a case with a thousand doors rather than 3. It's not a paradox, just our brains are bad at probability.
                              - You pick one out of a thousand doors - you've got a 1/1000 chance of being right and 999/1000 chance of being wrong.
                              - The host opens 998 of the other doors. Do you then switch? You still have a 1/1000 chance of being right first time so you have a 999/1000 chance of being right if you switch.

                              It's far more intuitively obvious that your chances of picking the correct door originally are very small, while the probability of the other door containing the car are very high. Even though you now have two options, they're not equally plausible.
                              Last edited by Bizarre Löw Triangle; 09-07-2019, 11:00.

                              Comment


                                #16
                                Yes, I get it now. If the host opened 998 doors at random and still there was no car I'd presumably be left with a 1/2 chance of finding it whether or not I switched, it's the fact I know he's choosing which doors to open that changes everything. Thanks.

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                                  #17
                                  if we're doing probability paradoxes, this is a fun one:
                                  https://en.wikipedia.org/wiki/Sleeping_Beauty_problem

                                  Sleeping Beauty volunteers to undergo the following experiment and is told all of the following details: On Sunday she will be put to sleep. Once or twice, during the experiment, Sleeping Beauty will be awakened, interviewed, and put back to sleep with an amnesia-inducing drug that makes her forget that awakening. A fair coin will be tossed to determine which experimental procedure to undertake:
                                  • If the coin comes up heads, Sleeping Beauty will be awakened and interviewed on Monday only.
                                  • If the coin comes up tails, she will be awakened and interviewed on Monday and Tuesday.

                                  In either case, she will be awakened on Wednesday without interview and the experiment ends.

                                  Any time Sleeping Beauty is awakened and interviewed she will not be able to tell which day it is or whether she has been awakened before. During the interview Sleeping Beauty is asked: "What is your credence now for the proposition that the coin landed heads?"
                                  Again both the thirder and halfer positions seem totally reasonable but they can't be reconciled at all. You might go down one rabbit hole and be convinced of the third position but another rabbit hole (or way of considering the problem) gets you to another answer.

                                  If the thirder position compels you imagine Sleeping Beauty is offered a bet before the experiment on the outcome of the coin toss with 3:2 odds (so unfavourable if the probability of heads is 1/3 and favourable if it's 1/2). If she makes the bet on Sunday and resolves the bet on Wednesday, if she's given the opportunity to cancel it in the middle of the experiment, should she? If the answer to that is no - she made a bet with favourable odds and she will believe she made a bet with favourable odds on Wednesday - then how can her belief that the coin was heads be 1/3.

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