In a family with two children, what are the chances, if one of the children is a girl, that both children are girls?
I just dipped into a book, The Drunkard’s Walk – How Randomness Rules Our Lives, by Leonard Mlodinow, Vintage Books, 2008. On p.107 Mlodinow says the chances are 1 in 3.
It seems obvious to me that the chances are 1 in 2. Am I correct? Is this not exactly analogous to having a bowl with an infinite number of marbles, half black and half red? Without looking I draw out a black marble. The probability of the second marble I draw being black is 1/2.
I think this question confuses a lot of people because there’s a lack of intuitive context — I’ll try to supply that.
Suppose there is a birthday party to which all of the girls (and none of the boys) in a small town are invited. If you run into a mother who has dropped off a kid at this birthday party and who has two children, the chance that she has two girls is $1/3$. Why? $3/4$ of the mothers with two children will have a daughter at the birthday party, the ones with two girls ($1/4$ of the total mothers with two children) and the ones with one girl and one boy ($1/2$ of the total mothers with two children). Out of these $3/4$ of the mothers, $1/3$ have two girls.
On the other hand, if the birthday party is only for fifth-grade girls, you get a different answer. Assuming there are no siblings who are both in the fifth grade, the answer in this case is $1/2$. The child in fifth grade is a girl, but the other child has probability $1/2$ of being a girl. Situations of this kind arise in real life much more commonly than situations of the other kind, so the answer of $1/3$ is quite nonintuitive.
In a family with 2 children there are four possibilities:
1) the first child is a boy and the second child is a boy (bb)
2) the first child is a boy and the second child is a girl (bg)
3) the first child is a girl and the second child is a boy (gb)
4) the first child is a girl and the second child is a girl (gg)
Since we are given that at least one child is a girl there are three possibilities: bg, gb, or gg. Out of those three possibilities the only one with two girls is gg. Hence the probability is $\frac{1}{3}$.
Please ignore the old comments – I have radically changed my answer. It’s long, but bear with me – after reading this answer, you should feel intuitively-comfortable with almost all other probabilistic fallacies, not just this one.
The first and most important thing is to define what we mean by “probability.” Let’s define it to mean “the expected percentage of positive outcomes when repeating an observation over a large sample” (Go ahead, read that again, more slowly). Also, let’s call the method we use to chose this large sample the “model.”
This may sound trivial, but it has some important implications. For example, what is the probability you will die before age 40? According to our definition, this question has no meaning: we can’t observe you multiple times before age forty and record how many times you die. Instead, we observe other people below age 40, and record how many of them die.
So let’s say we observe all other people on earth below age 40 (our model), and find that 1/2 of them die before hitting the big four-oh. Does this mean the probability of you dying before 40 is 50%? Well, according to this model, it does! However, this is hardly a fair model. Perhaps you live in a first-world country – now when we revise our model to include only people below 40 in first-world countries, your chances become a much-less-grim 1/10^{†}. But you are also not a smoker, and don’t live in the city, and ride your bike on Sundays with your wife and crazy mother-in-law, which makes your chances 123/4567! That’s much better… however, our model still doesn’t take into account that you are also an avid skydiver 😉
^{†} I am pulling these numbers out of nowhere, they are not real statistics.
So the point is, asking for a “probability” only makes sense in the context of a certain model – a way of repeating our observation many times. Without that, asking for a probability is meaningless.
Now, back to the original question. Before we can assign a probability, we must choose a model; how are we choosing the families to sample from? I see two obvious choices, which will lead to different answers:
Do you see the difference? In the first case, every family has the same probability of being chosen. However, in the second case, the families with two girls are more likely to be chosen than families with only one girl, because every girl has an equal chance of being chosen: the two-girl families have doubled their odds by having two girls. If children were raffle tickets, they would have bought two tickets while the one-girl families bought only one.
Thus, we should expect the probabilities in these two cases to be different. Let’s calculate them more rigorously (writing BG
to mean “boy was born, then girl):
BG
, GB
, and GG
(BB
was removed from consideration, because they have no girls). Since only one of the three has two girls, our chances of having two girls are 1/3.GG
is twice as likely as BG
or GB
. Thus, the probabilities are GG: 2/4
, GB: 1/4
, and BG: 1/4
, meaning the probability of a girl-sibling is 2/4 = 1/2 (alternatively, we could have noted that there are only two equally-probable possibilities for the sibling: boy or girl).Here lies the fallacy: the model our intuition assumes is the second one, but the way the problem is worded strongly implies the first one. When we think in terms of “randomly choosing a family (over a large number of families),” our intuition meshes perfectly with the result.
Let’s take a look at another similar problem
In a family of two children, where the oldest child is a girl, what is the probability they are both girls?
Once again, I can see two different, plausible models for observing our random sample:
So once again, the question strongly implies the first model, though arguments for either could be made. However, when we actually calculate the probability…
BG
, where the first child was a boy. This leaves only two equally-probable possibilities, GB
and GG
. Thus, the chances are 1/2.BG
from this case, leaving GB
and GG
. However, unlike the original question, GG
is no longer twice as likely, since the younger child can no longer be the one who was randomly chosen. Thus, GB
and GG
are equally likely, and we again have a probability of 1/2 (alternatively, we could have noted that there are only two equally-probable possibilities for the sibling: boy or girl).…we find that the choice between these two models doesn’t matter, because in this case both have the same probability! Among two-child families with an older daughter, it doesn’t matter if we randomly choose the family or the older-daughter, because in both cases there is only one of each per family.
Hopefully that all made sense. For bonus points, try applying this reasoning to the Monty Hall problem. What is our model – how are we making repeated observations? Why does it clash with our intuition?
For even more bonus points, try to figure out the following question; the math isn’t too difficult, but it took me a long while to figure out why, intuitively, the answer should be correct:
In a family of two children, one of whom is a girl named Florida, what are the chances of two girls?
(If you have troubles, post a question and leave a link to it in the comments, and I’ll try to answer it there as best I can 🙂 )
I think that the reason that these puzzles are so often confusing is that they rely on the limitations of the English language rather than on any mathematical difficulties. Of course this is not unique to English, and I think you should be able to find similar similar puzzles in pretty much any natural language.
Here is an example that brings out the difficulty more clearly. First, consider the following similar puzzle:
In elementary probability class, they teach you to answer this by making a table of the four options
Robin Lindsay B B * B G G B * G G
The starred rows are the ones where Lindsay is a girl, and we compute from them that the chance that both children are girls is 1/2.
Now consider this puzzle
The elementary probability method gives the following table, and a probability of 1/3. The difference is that we gain one more row, compared to the previous puzzle.
Robin Lindsay B B * B G * G B * G G
After looking at these, you can see that the difficulty of the original puzzle comes because, in English, “one of the children” can mean several different things:
“one” can mean “a particular one”. If you read the original puzzle like this, it becomes analogous to the first puzzle I wrote, and the answer will be 1/2.
“one” can mean “at least one”. If you read the original puzzle like this, it become analogous to the second puzzle I wrote, and the answer is 1/3.
“one” can mean “exactly one”. If you read the original puzzle like this, the answer is 0.
There is a common convention in mathematics that “one” usually means “at least one”. For example, this is the sense intended in the following sentence, which is a typical example of mathematical English: “if a natural number $n$ is a multiple of a prime number $p$, and $n = ab$, then one of $a$ and $b$ is divisible by $p$.” We would not read this as saying that exactly one of $a$ and $b$ is divisible by $p$.
I don’t believe this convention is very common in non-mathematical English. If I say, “one of my children is a girl”, in normal English this means that the other is a boy. Similar discrepancies between mathematical and non-mathematical English come up with our use of the word “or” and our use of the phrase “if/then”. When we teach mathematics, we have to spend time explaining this mathematical argot to students, so they can use the same English conventions that we do.
Probability puzzles like the one you’re asking about rely on these differences of English meaning, rather than on any logical or mathematical problem. In that sense, they aren’t really puzzles, they’re just tricks.
Here’s some idea.
Suppose you sample $4n$ families with two children, where $n$ is a very large integer. Then, with a very high probability, about $n$ of them have two boys, about $n$ have two girls, and about $2n$ have one girl and one boy.
Since you ignore the families with two boys, the desired probability is given by $n/(3n) = 1/3$.
EDIT: The above solution corresponds to the following situation. You visit a large number of families. In each family, you check if there are two kids. If no, you ignore this family. If yes, you check if one of the kids is a girl. If no, you ignore this family. If yes, you check if both kids are girls.
EDIT: Among all the OP’s examples, only one is relevant here. Namely, “I visit this family. I know they have 2 kids. One of them, a girl, comes into the room. The probability that the 2nd kid is also a girl is 1/2, no?”. Well, in this situation the probability is indeed $1/2$, assuming the following interpretation. You visit a large number of families. In each family, you check if there are two kids. If no, you ignore this family. If yes, you wait until one of the kids comes into the room. If it is a boy, you ignore this family; otherwise you check if both kids are girls. It is readily understood that this leads to a probability of $1/2$.
One possible solution for this problem might be though using conditional probability.
Let $B_i$ and $G_i$ represent $i$th boy and $i$th girl respectively. Here the sample space could be expressed as $S =$ {$(B_1,B_2),(B_1,G_2),(G_1,B_2),(G_1,G_2)$}
Let $A$ = An event of having both girl chil.
$B$ = An event of having at-least one girl child.
$A = ${$(G_1,G_2)$} and $B =${$(B_1,G_2),(G_1,B_2),(G_1,G_2)$}
So what you are asking is $$P(A / B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} $$
It really depends on what knowledge you have to restrict the probability space. (The original question is a little ambiguous, but I agree that 1/3 is the best answer.)
The marble example and the example of the firstborn daughter are the same situation– the second trial is independent, and the odds are 1/2. In these cases you know the first trial is independent, and you can ignore it.
The original question intends you to imagine a randomly chosen 2 child family from among all families with at least 1 daughter. For example, suppose there is a social science study on 2 child families with at least 1 daughter– in this situation, about 1/3 of the families will be daughter-daughter, 1/3 will be daughter-son, and 1/3 will be son-daughter. You have to consider the full probability space of two trials (d-d,d-s,s-d,s-s) and eliminate the s-s possibility. This leaves a 1/3 chance of 2 daughters.
For other variations on this question, you have to decide the appropriate probability space before answering. (One common, confusing variation is to suppose the family has two children, one of which is a son born on a Tuesday– what are the odds of two sons?)
In your example where the first child in the room is a daughter, if we suppose that the order the children enter has nothing to do with sex, then the probability the other child is a daughter is 1/2. However, maybe the family encourages their daughters to play inside and their sons outside– then you could assume the other child is a son, since he hasn’t come in yet. Maybe the culture has a tradition that the oldest daughter in the household (if there is a daughter) offers guests drinks– then you know the family has at least one daughter, and the odds that they have another daughter is 1/3. Once you’ve chosen your assumptions, you can define the probability space and compute the answer.
This is really a continuation of BlueRaja’s answer, where I’m changing the question slightly to give another illustration of the underlying point. Consider the following two procedures:
Procedure A: I flip two coins, a nickel and a penny. After the flip, I
pick one at random, and report its value to you.
Procedure B: I flip two coins, a nickel and a penny. If exactly one lands
heads, I report that one’s value to you, i.e. I say either “the nickel
landed heads” or “the penny landed heads” as appropriate. If both land
heads, I report on one of them chosen at random. If neither lands
heads, I say “both landed tails.”
Now, if I follow Procedure A and say “the nickel landed
heads”, then the probability that the penny landed heads is 50/50. But
if I say the very same thing after following Procedure B, the
probability that the penny landed heads is 1 in 3.
Now take the following problem: I flip two coins and say to you “the
nickel landed heads”. What is the probability that the penny landed
heads? Clearly there’s no way to answer without knowing or assuming something about the procedure I’m following.
(All this having been said, I agree that the best answer to the original question is 1/3.)
In a family with two children, what are the chances, if one of the children is a girl, that both children are girls?
0%. You already said that ONE is a girl. The other must be a boy.
We can solve this from first principles (maybe they’re technically “second” principles, who knows).
Define:
We want to find: $\mathbb{P}(N_G=2 | N_G \geq 1 , N_T=2)$
$$\mathbb{P}(N_G=2 | N_G \geq 1 , N_T=2) = \frac{\mathbb{P}(N_G=2 , N_G \geq 1 , N_T=2)}{\mathbb{P}(N_G \geq 1 , N_T=2)}$$
$$=\frac{\mathbb{P}(N_G=2 , N_T=2)}{\mathbb{P}(N_G \geq 1 , N_T=2)}$$
$$=\frac{p_G^2}{p_G^2 + 2p_Gp_B}$$
If we take $p_G = p_B = \frac{1}{2}$ then we get the book’s answer of $\frac{1\over4}{\frac{1}{4}+\frac{1}{2}}$
$$=\frac{1\over4}{3\over4} = \frac{1}{3}$$
NOTE: In the above, the second factor in the denominator is derived since the order in which the children are born does matter.
I feel like people are always left with a feeling of “sure, I see how you can obtain that result but why isn’t my interpretation correct?” Here’s my latest attempt to address this issue.
If you are given that the first child is a girl or that the second child is a girl, then the obvious answer of 50% is correct.
In this case we are given that one or both of the children is a girl, but not given any information about which one(s) — so a simple evaluation of the 4 possible scenarios (for a two-child family) reveals the answer of 33%.
The confusion stems from the slightly ambiguous language used. The phrase one of the children was used to mean one or both of the children. Based on that language, our brains trick us into thinking that a specific one of the children is given and therefore we can evaluate the other child independently.
Let $X$ be the event “one of the children is a girl” Let $BG$ refer to the event the family has one boy and one girl (I’m not distinguishing $BG$ and $GB$), and so forth. Everything is conditioned on the family having two children.
$$ P(GG | X) = \frac{P(GG \wedge X)}{P(X)}
= \frac{P(X|GG) P(GG)}{P(X|GG) P(GG) + P(X|BG)P(BG) + P(X|BB)P(BB)}
= \frac{1 \cdot \frac{1}{4}}{
1 \cdot \frac{1}{4} + P(X|BG) \cdot \frac{1}{2} + 0 \cdot \frac{1}{4}}
= \frac{1}{1 + 2 \cdot P(X|BG)}
$$
(note that I am assuming this isn’t taken as a “trick” question, so that $P(X|GG) = 1$ is meant, rather than $P(X|GG)=0$)
So everything boils down to the probability of “one of the children is a girl” given that the family has a boy and a girl.
Now, what does the phrase “one of the children is a girl” mean? There are two typical options
In my opinion, an English speaker would almost always interpret the phrase the first way rather than the second way.
However, there is a nasty linguistic trap: even meaning it the first way, we may often pick a girl to be the referent of “one” (e.g. so that we can speak of the “other” child). And since we’ve picked a girl, we get stuck in the second interpretation. This is exacerbated by the fact the second interpretation has a very appealing shortcut for reasoning out the solution.
You have to read carefully. Here, the probability of a girl means the probability of at least one girl. You have
$$P(\hbox{one child is a girl}) = 1 – P(\hbox{no child is a girl})
= 1 – P(\hbox{two boys}) = \frac{1}{4}. $$
Now you are calculating the conditional probabilty
$$P(\hbox{two girls}| \hbox{at least 1 girl} = {P(\hbox{two girls})\over P(\hbox{at least one girl})} = {1\over 3}.$$
You reach into a bag and pull two marbles. You look at the first and see that it is black. There is a 1/2 chance that the other is red.
You reach into a bag and pull two marbles. Your friend tells you whether or not you have 2 red marbles. He says you don’t have 2 red marbles. The chance you have 2 black marbles isn’t 1/2 or 1/4, it’s 1/3.
The assertion that at least one child is a girl is equivalent to saying that you don’t have 2 boys. It is analogous to the second case above, not the first.
There are at least three interpretations of the statement that “one child is a
girl”. Two of them lead to the 1/2 and 1/3 answers which have been discussed
at length, but I found only one brief
answer covering the third
interpretation (though I did only skim).
The meaning of the statement, and consequently the correct answer, can be
altered by considering it to answer different questions.
Suppose you’ve just had the first discussion; what do you think the other child
will be in that case? There is no chance the other is a girl.
The second question would prompt most people to search their set of children
for a girl and answer “yes” if they can find one or “no” otherwise.
Consequently, out of the 3/4 of two-child parents who can answer yes, about 1/3
of those will have another girl.
However it’s conceivable that some kind of contrarian may resolve “one of your
children” even before they have considered the rest of the question, choosing
one at random regardless of its gender, and answering the rest of the question
with respect to that random choice. This renders the situation equivalent to
the third case.
In the third case, obviously, the gender of the other child is independent of
all the information we have. So it’s a 1/2 chance.
There are many ways to explain this, but let me share with you how I do it in my class. I set $q=P(\mbox{You say that he has at least one daughter} \mid \mbox{He has $\{B,G\}$})$.
If you sample at family level, then you ask the father if he has at least a daughter or you check official records. In that case $q=1$.
If you sample at child level, by for instance meeting the father in the company of one child or otherwise happening to find out about one child, then $q=1/2$. In this case you will half the time miss this family because you happen to see the son first.
Using Bayes’ formula, we get
$$
P(\{B,G\}) \mid \mbox{You say that he has at least one daughter})
= \frac{P(\mbox{You say that he has at least one daughter} \mid \{B,G\}) P(\{B,G\})}
{P(\mbox{You say that he has at least one daughter})}.
$$
Since
$$P(\mbox{You say that he has at least one daughter}) = P(\mbox{You say that he has at least one daughter} \mid \{B,G\}) P(\{B,G\}) + P(\mbox{You say that he has at least one daughter} \mid \{G, G\}) P(\{G, G\}),
$$
we get
$$
P(\{B,G\} \mid \mbox{You say that he has at least one daughter}) = (q \times 1/2)/(q \times 1/2 + 1 \times 1/4) = 2q/(1 + 2q),
$$
and it follows that
$$
P(\{G, G\} \mid \mbox{You say that he has at least one daughter})=1/(1 + 2q).
$$
If $q=1/2$, we get $P=1/2$, and $q=1$ gives $P=1/3$.
Ok let us consider the unknown. A couple has a girl child. The wife gets pregnant. What are the odds the second child will be a (girl) or (boy).
Now we have reduced the question to a single possibility set.
the child will have x-x chromosomes
The child will have x-y Chromosomes
The wife ONLY provides X chromosomes. The husband has the coin toss x or y.
As there is ONLY one coin being thrown (the second x or y in our fair coin toss) then the odds remain 50%. We are NOT Throwing two fair coins to find an answer in this case.
Given a fair coin, the odds are 50% of either outcome, period. We can now use this theory regardless of anything else when asking about the next occurrence (or unknown previous occurrence) as it is separate from any other occurrence and not related in any way. There are TWO options only.
You can stack all the “possibilities and probabilities” you want to, however we KNOW that given a coin toss the next toss has a 50% chance of either outcome. IT is ONLY in looking backward at “n” tosses and in the ORDER of results that any other probability exists and that does not apply to the next toss itself. Having tossed three Y’s consecutively does not increase the odds of an X on the next toss, nor reduce it.
Where the first toss is unknown (the second child is a girl) then it still makes no difference…there is an xx, xy possibility genetically as to the first toss. Therefore 50% chance. Ignore english completely…look at the basic problem, unless you care about order or statistics of results over “n” tosses etc.
The ONLY way that one ever has a statistic of
gg
gb
bg
bb
is when a fair coin is tossed n times and actually create the “expected” result given above. It is not possible to change the odds of “next throw” of a fair coin(s). Only to look backward to the odds that a specific result might be expected over n tosses in a specific order or not. We only have ONE toss involved (next or unknown previous toss) then there is only one possible answer, a 50% chance an x or y chromosome exists.
It’s 1/2 as order is not important, just the physical combination.
i.e. BG == GB. Basic stats.
Hence you only have: GG & GB combinations.
Look up permutation vs combination!!
Order would only matter if you were betting on the order they came out the womb.
“In a family with two children, what are the chances, if one of the children is a girl, that both children are girls?”
I dont see any ambiguity with this question. Clearly, the OP, didnt specify the position of the first girl. Even if the OP, asked “…if the Eldest is a girl, that both children are girls?” and we have to include the ‘order’ of the genders in determining the probability of the 2nd child the answer is still 1/2.
General basis…that there’s a 50% chance that the 2nd child is either a boy or a girl(gender is an independent event). Therefore, (ignoring reality that the ratio of boys and girls is skewed in favor of girls) its distribution within a population (multiple events) is equally, half boys and half girls, for the second child.
Proof of this is by looking at genetics. Males, carries the xy chromosomes and xx for the females.
A child can be either xx or xy.
In a family of 2 children then the distribution of chromosomes would be the following combination…
xx xx (all girls)
xy xy (all boys)
xx xy (a girl and a boy)
*note the 3rd combination, it’s not order specific, which also represent the combination xy xx
It’s given that one child is a girl therefore we can remove the combination xy xy. We end up with only 2 combination, one of the combination is xx xx (all girls) therefore the chance of that happening is 1/2.
If we have to be specific with the gender order then our combination would be…
xx xx
xy xy
xx xy
xy xx
and “…if the Eldest is a girl…”
a) one child is a girl therefore we can remove the xy xy combination
b) eldest is a girl therefore we can remove the xy xx combination (boy is older)
which leaves us with the following combination….
xx xy
xx xx
clearly 1/2
Why 1/3 is wrong…
The original assumption is that there are 4 combinations as oppose to 3. The 4 combinations only applies if the order is specified either the eldest or the youngest.
If the known girl is the eldest then the combination BG has to be excluded for consideration.
If the known girl is the youngest then the combination GB has to be excluded for consideration.
The 1/3 proponent assume that there is no order and represented the boy and girl combination both ways instead of one. Or, there is order but neglected to remove the combination that doesnt represent the actual position of the known girl.
Why I think the ⅓ answer is wrong…
Ok, so I understand where the ⅓ answer comes from but taking a closer look I don’t see how it can be correct. Here’s my reasoning:
What we know:
a couple has 2 children
at least one of them is a girl
What we don’t know:
the gender of the other child
whether the child we know is a girl is 1st or 2nd in birth order
with these in mind we have 3 choices:
BB (eliminated because we know one of the children is a girl)
GG (both children are girls)
GB (1st born is a girl, 2nd born is a boy)
BG (1st born is a boy, 2nd born is a girl)
so at this stage it was concluded that the probability of the other child being another girl (GG) is ⅓ since the other choices are (BG or GB) each with the probability of ⅓
However, I believe there is a fallacy with this approach which I will discuss at the end.
Right now I will look more closely at the data we have:
First of all, if one of the children is a girl we automatically assume she will either be a 1st born or a 2nd born. Since we don’t know the birth order of the known (whether the girl is first or second born) – we either have to 1) eliminate using birth order from the equation or 2) we have to use two separate calculations of birth order to come to an answer.
Choice 1) Eliminating birth order positions:
(In this situation we can actually ignore birth order completely as it is not relevant to the question. The only reason we think it’s relevant is because we assume there has to be a 1st and a 2nd due to our knowledge that one child is born before another. But the question is simply to know the odds of having another daughter, making birth order irrelevant.)
We originally had 4 choices and eliminated 1:
BB (eliminated because we know at least on child is a girl)
BG
GB
GG
If we eliminate the use of birth order positions, the choices are further narrowed:
BB (eliminated)
BG/GB (becomes one choice because order is irrelevant)
GG
So in this situation we only have 2 choices, making the probability of the other child being a girl 50% or ½
Choice 2) Using positions/birth order:
If we decide to use birth order, then we cannot just assume we don’t know the birth order and thus put it all together into one equation.
There are only 2 possibilities: the known (girl) is either a first born or a second born.
If the (known) girl is a first born we have this possibility:
GB
GG
but these two options are eliminated:
BB (because one of the children is a girl)
BG (because the firstborn is not a boy)
So again we have ½ probability that the other child is a girl
The same for when the known girl being second born. The possibilities are:
BG
GG
but these two options are eliminated:
BB (because we know one of the children is a girl)
GB (because we know the second born is a girl and not a boy)
Again, the probability of the other child being a girl is ½
Putting it all together:
The probability that the unknown (the child who’s gender we don’t know) is a 1st born or a 2nd born is 50/50 or ½a and ½b
½a = 1st born probability
½b = 2nd born probability
The probability of the the unknown child being a boy or a girl is also 50/50 (½y and ½x)
½y = probability of being a boy
½x = probability of being a girl
Therefore, the probability of the unknown child being either a 1st born or a 2nd born daughter is:
(½a x ½x) = ¼ax = ¼ probability of being a firstborn girl
(½b x ½x) = ¼bx = ¼ probability of being a secondborn girl
(¼ax + ¼bx) = ½abx = ½ is the probability of other daughter being a girl (of either birth order)
So again, the probability of the other child being a girl is ½
Fallacy of the problem yielding the ⅓ answer:
We originally had 4 choices and eliminated 1:
BB (eliminated)
BG
GB
GG
Even if we take this original problem, we cannot solve it straight (as in ⅓ odds to ⅓ odds to ⅓ odds)
Because the reality would actually be this way:
Boy1 Girl1
Girl1 Boy1
Girl1 Girl2
Girl2 Girl1
The reason is, if we allow for: BG and GB (birth order options)
then we also must allow for G1G2 and G2G1 (birth order options for the girls)
and we would end up with this situation:
¼ BG
¼ GB
¼ GG
¼ GG
the odds of having a girl being:
¼ + ¼ = ½
So the answer to all the problems, no matter how solved, is 50/50 or ½.
Please correct me if I’m wrong.