Boy/Girl probability in Monty Hall

Cecil's Columns/Staff Reports
121

Okay, here’s the post that shold put this question to rest. It’s long, but IMO worth reading. :wink:

Data:

Pair 1: BB
Pair 2: BG
Pair 3: GB
Pair 4: GG

Q1: A pair was selected at random. From this pair, a child was selected at random. If that child was a girl, then it was reported “At least one child from the pair is a girl.” If that child was a boy, then it was reported “At least one child from the pair is a boy.” Here is the report that was given: “At least one child from the pair is a boy.” What is the probability that the other child was a girl?

Answer: 1/2. The report “at least one child from the pair is a boy” tells us we’re dealing with the first boy in pair one, the second boy in pair one, or the boy in pair 2, or the boy in pair 3, all with equal probability. Half of these children have a boy sibling, and half a girl sibling. So the probability is 1/2.

Q2: A pair was selected at random, and examined. If pair 1 was selected, then it was reported “At least one child from the pair is a boy,” (and one of the two boys is designated as the “reported child”). If pairs 2 or 3 were selected, then a child was selected at random from the selected pair. If the selected child was a boy, then it was reported “At least one child from the pair is a boy.” If the selected child was a girl, then it was reported, “At least one child from the pair is a girl.” Finally, if pair four was selected, then it was reported “At least one child from the pair is a girl.” Here is the report that was given: “At least one child from the pair is a boy.” What is the probability that the other (i.e., the unreported or unselected child) was a girl?

Answer: 1/2 again. The report “At least one child from the pair is a boy” tells us either pair 1, 2, or 3 was initially selected, each with equal probability, and that in case pair 3 or 3 was selected, then specifically the boy child was selected, in each case with probability of 1/2. So, there is 1/4 chance that the selection of pair 1 issued forth in the report, and 1/8 chance each that the selection of pair 2 or 3 set it off. So there is a 1/4 chance that the child which is not the “reported” child from pair 1 is a boy, and there is a 1/4 chance that the “other child” (i.e., the unselected child from pair 2 or 3) is a girl. The chances the other child is a boy, then, is the same as the chances it is a girl. Hence, the probability, given the report, is 1/2.

Q3: A pair is selected from the four at random. If pair one is selected, then it is reported “at least one child is a boy,” and one of the children is designated as “the reported child.” If pairs 2, 3, or 4 are selected, then it is reported “at least one child is a girl.” The report was, in fact, “At least one child is a boy.” What is the probability that the other (“unreported”) child is a girl? In other words, what is the probability that they are not both boys, given the report?

Answer: 0/1. The report is only given when both children are boys, so if the report is given, the probability is 100% that the “other” child is a boy, or in other words, that both children are boys.

Q4: A pair is selected from the four at random, and examined. If at least one of the children is a boy, then it is reported “At least one of the children is a boy.” Otherwise, it is reported “Neither child is a boy.” The report was, in fact, “At least one of the children is a boy.” What is the probability that the other child is a girl, too, or in other words, that the pair are not both boys?

Answer: 2/3. The report means we are dealing with pair 1, 2, or 3, each with equal probability. In one of the three cases, the pair are both boys. In 2 of the three cases, the pair are not both boys. So the probability is 2/3 that the “other” child is a girl, or in other words, that the pair are not both boys.

Given Gardner et al’s way of questioning how the information was obtained, I listed the above four versions of the question in order to illustrate the fact that there are, not just two, but a plethora of ways it might have been determined what report was to be given about the children. Have I, then, proved G. et al’s point? Am I showing how badly ambiguous the question is?

I don’t think so. I maintain that the method of report described in Q4 is the one plainly implied by the wording of the puzzle as originally formulated. My basis for maintaining this is the following claim: Only in Q4 does the phrase “At least one child is a boy” actually mean “at least one child is a boy.”

To explain, I will show you the following modified versions of the four questions:
Q1’: A pair was selected at random. From this pair, a child was selected at random. If that child was a girl, then it was reported “Ugh.” If that child was a boy, then it was reported “Bug.” Here is the report that was given: “Ugh.” What is the probability that the other child was a girl?

Q2’: A pair was selected at random, and examined. If pair 1 was selected, then it was reported “Erg,” (and one of the two boys is designated as the “reported child”). If pairs 2 or 3 were selected, then a child was selected at random from the selected pair. If the selected child was a boy, then it was reported “Erg.” If the selected child was a girl, then it was reported, “Grr.” Finally, if pair four was selected, then it was reported “Grr.” Here is the report that was given: “Erg.” What is the probability that the other (i.e., the unreported or unselected child) was a girl?

Q3’: A pair is selected from the four at random. If pair one is selected, then it is reported “Gah!” and one of the children is designated as “the reported child.” If pairs 2, 3, or 4 are selected, then it is reported “Augh!” The report was, in fact, “Gah!” What is the probability that the other (“unreported”) child is a girl? In other words, what is the probability that they are not both boys, given the report?

Q4’: A pair is selected from the four at random, and examined. If at least one of the children is a boy, then it is reported “Huh.” Otherwise, it is reported “Mmm.” The report was, in fact, “Huh.” What is the probability that the other child is a girl, too, or in other words, that the pair are not both boys?

The question I put forward is, which of the following phrase or phrases means “At least one child is a boy”?

Ugh
Bug
Erg
Grr
Gah!
Aug!
Huh
Mmm

It is my contention that, as they have been used in the four questions above, “Huh” and only “Huh” actually means “At least one child (from the selected pair) is a boy.” (The clearest example in what follows is that of “Gah!” so if you wish, read that one first to get the best idea of what I’m doing here.")

“Huh” is the only report that is issued if and only if either both children are boys are one is a boy and one a girl. (There are other ways to formulate what follows the “if and only if” here but that is not important. What is important is that “huh” is the only report which is like this: If at least one child from the selected pair is a boy, then “huh” is issued, and if “huh” is issued, then at least one child from the selected pair is a boy.)

“Ugh” is issued if and only if a child selected at random from among the eight constituting the pairs is a boy. So “Ugh” means, not, “At least one child from a selected pair is a boy,” but rather, “The child we’ve selected from among the eight is a boy.”

“Erg” is issued if and only if either pair 1 was selected, or pair 2 or 3 was selected and the boy was randomly selected from them. So “Erg” means “Either pair one has been selected, or pair 2 has been selected and the boy selected from it, or pair 3 has been selected and the boy selected from it.”

The clearest example: “Gah!” is issued if and only if pair one was selected. So “Gah!” means “Pair one was selected,” or in other words, “Both children are boys.”

Notice that in the case of Ugh, Erg, and Gah, it is possible for there to be at least one child which is a boy in the selected pair without those respective utterances being made. This is not possible with Huh. It is on that basis that I maintain that only “Huh” actually means “At least one child from the selected pair is a boy.” The others mean something else.

Now, of course, each of these grunts simply replaced the phrase “at least one child is a boy” from one of the original 4 versions of the question listed above. What does this mean? That though the English sentence “At least one child is a boy” is being uttered in each of the four cases, it turns out that the sentence is not being used with the same meaning in each of the four cases.

The only version in which it is being used with its normal English meaning is in the case of Q4.

Since in order to get any communication done at all we should always assume prima facie that an utterance in English is intended to be used to mean what the sentence uttered actually does mean in English, it follows that prima facie we should interpret “at least one child from the pair is a boy” in the puzzle as originally phrased as meaning exactly that “at least one child from the pair is a boy.”

Since Q4 is the only version of the puzzle which satisfies that constraint, I conclude that Q4 gives the correct interpretation of the puzzle as originally phrased.

-FrL-

122

I see. That does avoid the problem of interpretation.

My inuition is that “the other child” doesn’t mean we’re now talking about a specific child–just that we mean “whichever” child didn’t, so to speak, “trigger” the report “at least one child is a boy.”

A particular boy can “trigger” my reporting “at least one child is a boy” without my report thereby referring to a specific child. My report refers just to the pair taken as a whole, though it was the gender of a particular child which triggered my report.

But anyway, I can see why you might think otherwise, so I’m glad the original puzzle avoids this problem by being phrased the way its phrased.

-FrL-

123

I wish I’d read your post before I made my long one. But I think if you and others with qualms about “other” just reword my post, where it says, for example, “the other child is a boy,” to say instead, for example, “the pair also includes a boy” or “the pair consists in two boys,” then the point of the post still goes through.

-FrL-

124

I am not trying to pickup any brownie points….but
I would like to stop debating this puzzle for a second and thank the crowd for a civil debate. I would also like to point out to any newcomers, that in particular, I have debated this before with Bytegeist and **Frylock **on several threads and I have never read a hateful or mean spirit comment from them. I only wish all members were so courteous.
Now back to the debate……

125

No, “One of the coins…” can mean either the nickel OR the dime (or even both). This is the key fact. This results in the larger sample space, which determines the different probabilities of the two situations. It’s all about sample spaces.

A related probability problem is called by M. Gardner “The Paradox of The Second Ace”:

Two people are playing a strange card came with a deck consisting of only four cards (Ace of Spades, Ace of Hearts, Jack of Clubs, and 2 of Diamonds). During one hand the deck is shuffled and one player is dealt two cards. He looks at the cards and says “I have an Ace”. The question is: what is the probability he also has the other Ace as well?
There are six combinations of 2-card hands:

  1. Ace Hearts- Ace Spades
  2. Ace Hearts - Jack Clubs
  3. Ace Hearts - 2 Diamonds
  4. Ace Spades - Jack Clubs
  5. Ace Spades - 2 Diamonds
  6. Jack Clubs - 2 Diamonds

There are 5 hands that contain an Ace (1,2,3,4,5). In only one case (1) can he have the other Ace. Therefore the probability he also has the other Ace is 1/5.

Now suppose another time the player looks at his hand and says “I have the Ace of Hearts”. The question is: what is the probability he also has the Ace of Spades?

Think about this before proceeding.

The surprising answer is 1/3!. How can naming the Ace of Hearts make a difference? There are only three hands where he can have the Ace of Hearts (1,2,3). In one case can he also have the Ace of Spades, so the probability is 1/3. The second situation has a smaller sample space therefore the one case we are looking for (#1, both Aces) in the second situation is a higher proportion of the total of number of cases we are considering.

126

Consider the two following sentences.

  1. One coin is showing heads.
  2. One and only one coin is showing heads.

Aren’t they making different assertions?

If I toss two coins and one is heads and the other tails, the assertion " One coin is showing heads" is true. If I toss two coins and both are heads the assertion " One coin is showing heads" is also true.

If one can discern the difference between the two sentences there is no question of the non-ambiguity of the first.

127

I thought I would take this opportunity to point out that the Other Child problem isn’t the only one that is stated ambiguously. If you reread Cecil’s original comments about the Monty Hall Puzzle you will find:

On another occasion 8 months later we get this from the New York Times, Sunday Edition, July 21, 1991. (After Monty Hall showed reporters that technically the 2/3 switch-door rule wasn’t necessarily right if we don’t take in the proper safeguards.)

128

I think it was always just taken as given that the host always opens a door, since, in fact, Monty Hall always did open a door.

It wasn’t mentioned when the problem was formulated, because it was simply assumed by the formulator. Any time anyone has brought up the ambiguity, it has never been in question whether the puzzler intends the puzzlee to understand that the host always opens the door. It’s just been pointed out that the puzzler may not have explicitly specified this assumption.

But you’re right: If we ask someone who’s never seen the show the following question:

“Suppose the host opens one of the other doors, to reveal a goat. Should the player switch?”

Then this person is justified in answering “There’s no reason for the player to switch.”

The right way to ask is instead

“Suppose the host always opens one of the other doors to reveal a goat. Should the player switch?”

Then the right answer is “Yes, the player should switch.”

Asking in the latter way still allows for the correct answer to be wholly counterintive.

-FrL-

-FrL-

-FrL-

129

Here’s another one.

You are shown three cards. One card is white on both sides, another is black on both sides, and the third is white on one side and black on the other. All three cards are put into a bag and the bag is shaken. You are then asked to close your eyes, pick one card out of the bag, and place the card flat on a table. You comply, the bag containing the remaining two cards is closed, and you are told to look at the card on the table. You open your eyes and see only the top side are card, and the side is white. You are asked “what is the probability the other side of the card is white?”.

The answer is further below.

Many people say 50%. They are wrong.

The probability the other side is white is 2/3. Here’s why:

Call the white/white card Card1, the white/black card Card2, and the black/black card Card3. Since you see a white side, the card can only be Card1 or Card2. These two cards together have three white card-sides (W1, W2, W3), as follows:

W1 is on Card1
W2 is on Card1
W3 is on Card2.

When you see a white card-side, 2/3 of the time it will be on Card1. Therefore the probability the other side of the card is white = 2/3. Recognizing that we asking a question about a sample space consisting of three card/sides is the key to the solution.

130

After reading comments originally written to Marilyn about the ambiguities, Frylock, I would have to say Marilyn’s (which is the same as Cecil’s) ambiguities were self-inflicted way beyond what we would expect from someone of her proclaimed intelligence. For what I have read, it was more than just opening a door. In Marilyn’s version, you open door #1, Monty opens #2 to show a goat, and you are free to switch to door #3. By having it told in a “let’s say you open door #1” fashion, it made the puzzle sound like it had little structure. What a lot of people assumed was, as the door choices would change each time, they thought what Monty showed you did too. Some thought he was free to show you the car. I believe that is at least one of the reasons some of the 1000s of PhD that wrote her, won’t apologize.

131

Wait a minute, now you’re talking about the Monty Hall problem?

In the links you gave me, people were insisting that the boy/girl problem had been stated ambiguously by this or that person.

But to my knowledge no one serious seriously challenges Marilyn Vos Savant’s answer to the Monty Hall problem. It has been taken as a matter of course, by every academic I have ever spoken with on the subject, (that comes to maybe just three, to be clear. But still, that’s three experts who think about this kind of thing more often than the rest of us) that Vos Savant’s solution is correct. I’ve never heard of anyone taking seriously the idea of criticizing her based on an ambiguity in the problem setup. This kind of criticism just isn’t interesting in this case. If I said to a philosopher or mathematician “But she formulated it ambiguously!” they would, I am sure, reply “Maybe so, but that’s beside the point: The problem we all know she was getting at is indeed an interesting one, nevertheless!”

To put it another way: You are welcome to criticize Marilyn based on the ambiguity. But by doing so, you fail to criticize the “always switch” answer to the Monty Hall problem. And that is what’s interesting in all this–the fact that, when the problem is stated correctly, the answer is that you should always switch. That’s mighty counterintuitive, and it’s what we’re all concerned to talk about when we mention the “Monty Hall Paradox.”

Regaring the boy/girl problem, I’m still waiting for someone to reply to my very long post a few posts back. I really do think it’s worth a response. :stuck_out_tongue: I’m telling you, it proves my case conclusively! :smiley:

-FrL-

132

What if the ambiguously-gendered sibling was on a treadmill?
Sorry.

133

Frylock, please read post #127…I think it is clear that Cecil, Martin Gardner, and Dr. Diaconis (Dept of Mathemathics Harvard) all felt that the Monty Hall question was ambigious. I would contend that the Monty Hall question was far more straight foward than Cecil’s or Marilyn’s version Boy/Girl question.
I would think most people understood that in the Monty Hall question you weren’t required to always pick door #1, Monty showing what was behind door #2, and you having to choice between #1 and #3. I assume most people knew you were free to choose any door, and the car could be behind any door not just #1 and #3. Everything about the doors was random chance. Now all that was really ambigious was what Monty Hall did.
On our other child puzzle we must assume that there is not random chance. All families MUST tell us that they have a daughter, never a son. How is this possible without some predetermined screening of the families by our source? And was this screening of families by our source implied in the original statement?

134

There is “the Monty Hall question” incorrectly asked, and there is “the Monty Hall question” correctly asked. Incorrectly asked, it is ambiguous. Correctly asked, it is not ambiguous, and its answer is highly counterintuitive.

When people talk about “The Monty Hall Problem,” “The Monty Hall Paradox,” or “The Monty Hall Question,” they are referring to what is asked when the question is asked correctly and unambiguously. This puzzle, with its counterintuitive answer, is what interests everyone.

It is no longer interesting to ask whether Vos Savant’s or anyone else’s formulation of it some years ago was ambiguous or not. The point is, we all know what the intended problem is, and we all find that intended problem to be an interesting one.

As to whether the boy/girl problem has been asked ambiguously, please see my long post above. I maintain that the following is not an ambiguous formulation:

“There is a family with two children. At least one is a girl. What is the probability that the family also has a boy?”

The question is unambiguous, and the answer is 2/3. See my long post above for an explanation why the question is unambiguous.

-FrL-

135

That’s quite a lot of discussion–it’s clear that ambiguity in stating the problem is at the heart of this.

Apparently, there are many ways of asking this question and it seems that “how the information was gathered” is the major bone of contention.

We now acknowledge that the Monty Hall problem can be interpreted in different ways but it turns out that one particular version is intuitively the most interesting.

I would like to restate the original problem (as I understood it) in very simple terms.

You meet an old friend in the park. They mention that they are now married with two kids. “One of them is playing over there,” they say, “come and say hello to little Debbie.” What are the chances that Debbie has a brother?

In this case the sex of the given child is completely random, and the question seems to be simply asking about the probability of the other child having the opposite gender. A musing from the “real” world–no lab coated Poindexter checking random sample distribution tables :slight_smile:

Regardless of the many other tangents this post has exhausted, this is the question I was addressing. I still think 50/50

136

In this question, all you are asking is what is the chance that a single child is male or female. That answer is 50/50. The number of children or the gender of the one you have seen doesn’t enter into this case. This question isn’t really related to the Monty Hall question.

137

Right, the right answer to the question you ask in this post is 50/50.

And the question you asked here isn’t the interesting or puzzling question most of us are thinking of when we think of the “boy or girl problem.”

-FrL-

138

Sorry, I was busy this week so I couldn’t write an earlier reply. I read your long thread Frylock, and I would have to disagree. I think I have shown that many people believe the phrasing of the question is ambiguous. If you look back over this thread (and the other threads in the forum about this puzzle), I think you will even find most of the 2/3 camp believe the family can be random chosen. I looked through my notes and found a reference to several studies on this subject.

Ambiguities and Unstated Assumptions in Probabilistic Reasoning
(halfway down the page)
The $64,000 question is why was it presented the way it was phrased. Normally, logic puzzles are carefully designed to ensure there is exactly one correct answer. Were Cecil and Marilyn just sloppy the morning they wrote their columns…or were they trying to make the puzzle sound more amazing than it really was? In reality there isn’t much to the puzzle unless you try to trick the reader into thinking about a random family. *Among families with two children of which one is at least a girl, twice as many families will also have a boy than will have a second daughter. *This statement isn’t a very amazing or even remotely interesting but most people easily understand this statement.
The correct way to phrase this lame puzzle is:
Among the families with two children of which have at least one girl, what is the probability that the family also has a boy?

As I stated earlier, if you state the puzzle correctly it isn’t very interesting… but it is solvable. Why phrase it any other way?

139

(In all the following, just to be perfectly clear, I am pretending there is always a 50/50 chance of having a boy when a family has a child, and I am pretending that when a family has two children, the probabilities attached to each child’s gender are independent from each other. These are not interesting nitpicks as far as the problem at hand is concerned.)

You are right to give weight to what the experts you’ve cited have said. But I am convinced they are wrong*, and I know myself to have good reasons for thinking so. If you think they are not good reasons, then ideally, you should have some idea exactly which of my reasons are constituted by invalid inferences or false assumptions. If all you can say is “but experts disagree” then that’s fine, but doesn’t support interesting discussion.

For my sake, please repeat yourself again.

Say I read the following passage:

“There is a family. This family has exactly two children. This family has at least one boy. What is the probability that this family has a girl?”

Now, I’m going to list off the things I would think in interpreting this passage. I would like you to be very specific in pointing out exactly where in this thought process I go wrong, and why.

“There is a family.” Hmm, okay. I imagine before me a family.

“This family has exactly two children.” Alrighty. I envision before me two generic children.

“This family has at least one boy.” Okay. We’re talking about children in terms of their gender. I’ve been told at least one of the children I envisioned in step two is a boy. That means exactly this: that either only the child I envisioned to the left is a boy, only the child I envisioned to the right is a boy, or both are boys. These three are equally probable. I will think of them as possibilities A, B and C, respectively, each with a probability of 1/3.

“What is the probability that this family has a girl?” Well, in possibilities B and C, the family has a girl, while in possibility A, it does not. That’s 2 out of three. That gives me an answer of 2/3.

I suspect you will take issue with this line of thought at either or both of the following points:

–You may disagree that “at least one child is a boy” from the problem means “exactly this: that either only the child I envisioned to the left is a boy, only the child I envisioned to the right is a boy, or both are boys.”

–Or you may disagree that the three possibilities just outlined can safely be assigned equal probabilities based on the wording of the problem.

If the first is where you wish to disagree, my argument is this: Any english sentence of the form “at least one of X or Y is a Z” means exactly this: either X is a Z and Y isn’t, or Y is a Z and X isn’t, or X and Y are both Z’s.

If the second is where you wish to disagree, please explain exactly why. I think you want to say that since I don’t know how it was decided that I should be told “at least one child is a boy,” and since certain procedures for telling me this would lead to a different anwer than 2/3, it follows that the problem is ambiguous. If that is your objection, please explain why that line of reasoning doesn’t lead to the conclusion that all statements of all problems are always ambiguous. That is how it seems to me, since if we’re going to ask “wait, how was that report arrived at,” we might imagine any of an infinite number of stories as possible answers to that question. How do you avoid the implication, then, that all statements of all problems like this are always ambiguous.

It may be taht you agree that they’re always ambiguous. That’s fine, but surely you also agree that despite ambiguities, we generally have conventions whereby we arrive at a reasonable disambiguation. Why isn’t it reasonable, then, to disambiguate the problem at hand by saying that it means to ask “Select, out of all the families which have two children, only those which have at least one boy. Now select one of those families. What is the probability (given its prior selection as a family with at least one boy) that that that family has a girl in it?” I argue that this would be a reasonable disambiguation for the following reasons:

–It is the only disambiguation according to which the report “at least one is a boy” in the original puzzle is envisioned as having been made made if and only if at least one is a boy. On all other disambiguations (see my long post for examples) the report is made under different, stricter criteria–and hence it is doubtful the report is being used to mean what the sentence constituting the report actually means in English.

–It is the only disambiguation which leads to an interesting and counterintuitive problem.

That all ended up being longer and less simple than I expected. Hopefully I’m being clear.

-FrL-

*In fact I think the experts you cite have good, practical reasons to maintain what they are maintaining even if they are strictly incorrect. In actual fact, when doing serious statistics, you’ve got to be sort of “suspicious” of the reports you make concerning your data. You’ve got to be sure you know exactly what the relationship is between the data and the way you’re reporting on the data. This is so you can be exactly sure what your reports actually mean. (A familiar theme from my longer post and from the end of the present post as well.) So it’s good to be suspicious, for practical reasons. But such suspicion is not called for in the context of a puzzle like this. The puzzle isn’t written for the rigorous statistician, but rather, for the everyday reader of everyday English. In that context, the assumption is that when we are told X, we are being told “If you had asked whether X, I would have answered ‘yes.’” (Note that interpreting the original puzzle in this post using this assumption yields the 2/3 answer.) There are situations in statistics as actually practiced where this assumption can fail us. But in everyday situations, it is the assumption which is made, and it is reasonable to expect that the assumption’s truth is being assumed by those who are making reports to us.

140

The Reason 2/3 Rule Couldn’t Work With A Random Family

Say you are at a neighborhood BBQ party and a new couple arrives. Wilma, the neighborhood busy-body, whispers to you that the couple is the Smiths and they have two kids. She believes that at least one of their children is a girl. What are the odds the other is a boy?

Without more information, you can assume that it is 50% or 1 in 2.

Why not 2 in 3? What makes the 2/3 theory works is we include all families with girls and exclude all families with any girls. While we can easily exclude answers of “ at least one boy”, the only was to include a family is to be told of a girl.

With the Girl-Girl family you will always be told “at least one girl.” (GG)
With the Boy-Boy family you will always be told “at least one boy.” (BB)
In either of the Girl-Boy or Boy-Girl families we cannot predict how what they will tell us. (GB,BG)

Now for the 2/3 theory to work (with the stated problem) we need all GG, GB, and BG to tell us that they have girls. The only way to achieve all the families giving the information we need is by us (or our source) qualifying them for this puzzle. This would include doing something like asking if one is “at least a girl.” While it would seem that by being told “ at least one is a girl” by our source is the same thing, neither in the example above or in Cecil’s version are we sure that our source knows an equal number of each type of families that have girls. All we know is how they family answered. The next several families we meet may all tell us boys. It may be that the only family that “shows” off their girls…is family that have no boys; while the families that have both, normally show off their boys. Without some safeguard requiring families with girls to tell us of them, the best we can hope for is random probability…some GB and BG families say boys and some say girls. In the end random chance will lead back to 50%/50%.

Random families cannot work; the families needed for the 2/3 answer must be qualified. To validate the answer we need to qualify not only the original family but other families as well. The result of whatever answer we give for the original family is a moot point. We will need to repeat this puzzle with all other type families to establish if it is really 2/3. The same conditions will apply with them as the original family.

Cecil’s original statement:
There is a family with two children. You have been told this family has a daughter. What are the odds they also have a son, assuming the biological odds of having a male or female child are equal? (Answer: 2/3.)

Now in Cecil’s version where does it say that the family was required to tell of a daughter if they had one? If we try to validate this puzzle, will ALL GB and BG families say they have girls? Is it possible that only the family that has two girls ever tell people of their girls? Are we to assume that families that have both a girl and boy, never speak of their boys? What are we to assume about Marilyn’s boy version, is it validate at the same time, or is it an either/or ?