Boy/Girl probability in Monty Hall

Cecil's Columns/Staff Reports
81

Okay….I set this challenge

We have 12 families. Why 12? Twelve is the lowest number that is both divisible by four (the number of possible families) and three ( the denominator for 2/3 ).

The Abbotts, the Adams, and the Andersons each have two girls.
The Bakers, the Browns, and the Byers each have a girl and a boy.
The Carpenters, the Clarks, and the Corbins each have a boy and a girl.
The Davids, the Deans, and the Drakes each have two boys.

These families would be what you would expect to find if all families are equally likely to be present in society. There are an equal number of matched pair families (A,D) to unmatched families(B,C).

The challenge is to survey all these twelve families and apply the 2/3 rule.
If the 2/3 rule works then you should get 8 out of 12 right. 8 out of 12 is, of course, 2/3.
NOTE: To prove that the 2/3 rule works, you must apply the rule to **all **twelve families.

82

[QUOTE=Uzi]
Okay, my ignorant layman’s stab at it.

We have a Girl = A and a Boy = B, thus:

AA=25%
AB=25%
BA=25%
BB=25%

This is based upon equal distribution between boys and girls.
Ssomeone comes along and tells us that one of their children is a girl. We are left with:

AA
AB
BA

As the option for BB has been removed as being impossible. Is it still true that the numbers stay the same?:

AA=25%
AB=25%
BA=25%

No, it is no longer true because the original percentages only apply if we don’t know anything else about the family other than it exists and there is an equal distribution amongst the population. Once we know the identity of one of the members it changes. So, how does it change? The girl in this case is identified first, so that we know that a boy can’t have been first. So, shouldn’t we remove the choice BA as well because it is now just as impossible as BB? We are left with:

AA=50%
AB=50%

A 50/50 chance that the next person will be a boy or girl.

Maybe if we look at the original question somewhat differently it might make more sense:

Same sex=50%
Different sex=50%

Once we know one of the children is a girl we know that there is an equal chance that her sibling will be the same sex as she is.

Okay, where did I screw up? :smiley:
[/QUOTE]

You’re right that the probabilities change but they’re still equal. I.e., we’ve gone from
AA – 25%
AB – 25%
BA – 25%
BB – 25%

to

AA – 33.3%
AB – 33.3%
BA – 33.3%

by eliminating BB. They didn’t tell you anything that makes one of the remaining three more likely than another.

You are acting on the assumption that it must still be true that there are equally likely chances of having two kids of the same gender or two kids of the opposite gender. But this is wrong. One of the two same gender possibilities was eliminated, whereas none of the different gender possibilities have been eliminated. So the information given suggets that different gender is now more likely.

As far as your argument of “*shouldn’t we remove BA as well as BB, since a girl (A) was identified first?” That would only make sense if the one who was identified was chosen randomly. The reason AB, BA, BB, and AA had equal probabilities – where these refer to birth order – is that the gender of the first born kid is random. But the gender of the first revealed is not random – if they answer “Do you have any girls?”, they will always identify a girl if either child is one.

Do you understand what I mean? If it’s AA, AB, or BA, they will identify the A – that means, they do nothing to help you distinguish those possibilities. The only info they’ve given you is that it’s not BB.

83

[QUOTE=Uzi]
This is what your options are according to your scenario:
OB YB
OB YS
OS YB
OS YS

No, two of the options can be eliminated. OB YB and one of the sister choices. If the sister is younger you have to eliminate OS YB. if the sister is older then you have to eliminate OB YS.
[/QUOTE]

No, this is wrong, because the parents don’t tell you the sister is older or younger. You claim it doesn’t matter, because either the daughter is older or younger. But it does matter – the point is “How likely is the other kid to be a boy given what you know?” You know different information in these two cases.

If I ask “Is your older child a daughter?”, the parents answer yes in 50% of the cases.

If I ask, “Do you have a daughter?”, the parents answer yes in 75% of the cases.

Clearly, the set of parents that answered yes to “Do you have a daughter?” is different. The difference is that it contains both YB, OS and OB, YS.

84

I think we all agree that there are equally likely chances of each of the following:

First kid girl, second kid boy.
First kid boy, second kid girl.
Both boys.
Both girls.

We can represent this probability distribution by bringing four mothers into a room – one for each of these possibilities.

Initially, all four moms are seated. If we say “Those of you who have a daughter, please stand up,” then three of the four will stand up.

How many of the three standing moms also have a son? The answer is two.

85

[QUOTE=wissdok]
Okay….I set this challenge

We have 12 families. Why 12? Twelve is the lowest number that is both divisible by four (the number of possible families) and three ( the denominator for 2/3 ).

The Abbotts, the Adams, and the Andersons each have two girls.
The Bakers, the Browns, and the Byers each have a girl and a boy.
The Carpenters, the Clarks, and the Corbins each have a boy and a girl.
The Davids, the Deans, and the Drakes each have two boys.

These families would be what you would expect to find if all families are equally likely to be present in society. There are an equal number of matched pair families (A,D) to unmatched families(B,C).

The challenge is to survey all these twelve families and apply the 2/3 rule.
If the 2/3 rule works then you should get 8 out of 12 right. 8 out of 12 is, of course, 2/3.
NOTE: To prove that the 2/3 rule works, you must apply the rule to **all **twelve families.
[/QUOTE]

What do you mean “apply the rule?” What’s the “2/3 rule?”

Show me how to “apply the rule” and why “applying the rule” is relevant to the puzzle.

-FrL-

86

Really they didn’t say BB wasn’t possible, what they said was that either the first event(child) or second event(child) wasn’t a boy(or B). From this we conclude BB isn’t possible, but at the same time either AB or BA isn’t possible either. The fact we know what one child is doesn’t decrease the chance that it comes from a matching child family. For the family with two of those children it is now a gain. The 2/3 rule could only work if we restrict ourselves to either boys or girls and choose families that have only that sex of child we have chosen. The opposing matching child family would always need to be disregarded. Hence, if we already restrict ourselves to 3 of the 4 families, then the family isn’t random. And as we have already chosen the sex of the child, the child isn’t random either.
In the puzzle given, we are presented with a family with two children. We are then told that the family has a girl. These statements were separate, and wasn’t implied that the family was chosen because of the girl. It appears that this family was just a random family that just so happened to have a girl. If our source knew of a boy he could have just as well told us about him. There was no restriction whatsoever on this family or child. For the families that have both a boy and a girl, we could have been told of either.
As long as the family is randomly chosen, and the child in that family is randomly chosen, it can’t help to be anything but ½. The closer you get to having an equal number of each type family in your experiment the closer you will get to ½.

The 2/3 rule I speak of is the belief that if we are given a family that state the gender of one of their children, you are 2/3 likely to find that their other child is of the opposing gender. In my challenge all four families are equally represented at a number that can easily show if the 2/3 rule works.

87

[QUOTE=wissdok]
The 2/3 rule I speak of is the belief that if we are given a family that state the gender of one of their children, you are 2/3 likely to find that their other child is of the opposing gender. In my challenge all four families are equally represented at a number that can easily show if the 2/3 rule works.
[/QUOTE]

I’ve forgotten the wording of the original puzzle now.

Okay, where in this thread is the statement of the puzzle itself?

Does the puzzle say “the family says it has one child which is a girl?” Or does the puzzle say “the family has one child which is a girl?”

Thinking Martin-Gardner style, if the former is the case, then we might wonder whether the family says it has a girl if and only if it has a girl, or rather, whether it says it has a girl only if it has no boys, or rather, whether it says it has a girl if it has two girls or one girl and a coin was flipped and came up heads, or whatever.

I do not think this Martin-Gardner type paranoia is called for: It is a matter of course to assume that a puzzle is giving information complete for a solution to the puzzle, and since the “if and only if there is one girl” interpretation is the default one we would assume if we weren’t being paranoid, it follows that it is the correct interpretation of the puzzle.

Anyway, that’s if the puzzle says “the family states there is one girl.” What about the other case, in which the puzzle just says “there is one girl.” Then I don’t see how your application of a “2/3 rule” is relevant. The puzzle just gives us a fact to work with: there is one girl in the pair. This tells us the family is one of that subset of all families which have at least one girl in them. It is one of the GG, GB, or BG families.

-FrL-

88

[QUOTE=wissdok]
Really they didn’t say BB wasn’t possible, what they said was that either the first event(child) or second event(child) wasn’t a boy(or B). From this we conclude BB isn’t possible, but at the same time either AB or BA isn’t possible either.
[/QUOTE]

Well, that depends on your interpretation of the problem. Cecil’s statement of the problem was somewhat ambiguous. All he says is “You have been told this family has a daughter.”

I suppose this could mean that they told you the sex of one of their kids, selected at random. (Since birth order is random, this is equivalent to telling you the sex of the first born kid. Or to telling you the sex of their second born kid.) In that case, yes, two possibilities are eliminated, and the answer is 1/2.

However, it could also mean you asked “Do you have any daughters?” and they answered “yes.” (If they just said “We have a daughter” without refering to some particular child, this in my opinion this is probably what they meant. No one would say “we don’t have a daughter” if either one of their kids was a daughter.) In this case, only one possibility was eliminated, and the answer is 2/3.

Based on Cecil’s subsequent explanation, I think it’s clear he meant the problem to be understood in the second way I just described. What I’ve been trying to do is explain why that interpretation of the problem gives the answer 2/3.

Do you agree that if the parents essentially told you “At least one of our two kids is a daughter,” then only one of the four possibilities is eliminated, and the other three are still equally likely? If so, then I don’t think we actually disagree, except perhaps about what question Cecil was trying to pose.

The key point is that answering yes to “Do you have any daughters?” is not the same as telling someone one particular child is a daughter. The yes answer for “Do you have any daughters?” encompasses AA, AB, and BA, whereas a yes answer for a particular child (whether the first or the second) encompasses only two cases.

89

As a further example of why simply saying “We have a daughter” is not the same as saying that a particular child is a daughter, consider the following:

Suppose I tell you at the start of the problem that this family has a boy and a girl (in some order). Now the only possibilities are:
AB
BA

That is, either they had a boy and then a girl, or they had a girl and then a boy.

Now, the family tells you “We have a daughter.” But you already knew that, so neither possibility is eliminated. If they said “our first child is a daughter”, that would eliminate BA. If they said “Our second child is a daughter”, that would eliminate AB. But just saying "We have a daughter can’t eliminate either one – whether they had the daughter first or second, they will still report having a daughter.

90

Can we all at least agree that if the family told you “Either our first child is a girl, or our second child is a girl, or both are girls” then the answer is 2/3?

Likewise, can we all at least agree that if they referred to a particular child, either by saying “Our oldest is a girl” or “Our youngest is a girl” or “Julia is our daughter” or whatever, then the answer is 1/2?

If we agree on this, then the question boils down to interpreting Cecil’s statement: “You have been told this family has a daughter.”

I contend that Cecil meant that the parents said “We have a daughter,” meaning “At least one of our kids is a daughter.”

I further contend that “At least one of our kids is a daughter” means the same thing as “Either our first child is a girl, or our second child is a girl, or both are girls.” (Assuming they have two kids, of course.) However, I think Cecil [or Jordan Drachman, assuming Cecil quoted the problem exactly] could have stated this more explicitly when he posed the problem, and maybe spared us all a headache.

91

[QUOTE=wissdok]
Okay….I set this challenge

We have 12 families. Why 12? Twelve is the lowest number that is both divisible by four (the number of possible families) and three ( the denominator for 2/3 ).

The Abbotts, the Adams, and the Andersons each have two girls.
The Bakers, the Browns, and the Byers each have a girl and a boy.
The Carpenters, the Clarks, and the Corbins each have a boy and a girl.
The Davids, the Deans, and the Drakes each have two boys.

These families would be what you would expect to find if all families are equally likely to be present in society. There are an equal number of matched pair families (A,D) to unmatched families(B,C).

The challenge is to survey all these twelve families and apply the 2/3 rule.
If the 2/3 rule works then you should get 8 out of 12 right. 8 out of 12 is, of course, 2/3.
NOTE: To prove that the 2/3 rule works, you must apply the rule to **all **twelve families.
[/QUOTE]

We drive through this odd neighborhood and see one family doing something with one of it’s children, who happens to be a girl. Which family could it be? It can’t be a D family, since they don’t have girls. Of the three remaining possibilities, the A’s have 2 girls and the B’s and C’s have girls and boys. Thus there is a 2/3’s chance that the observed family is a B or C family.

When you talk about surveying all the families you’re changing the game. The information we’re given at the beginning of the problem eliminates the two-boy families from the picture. So you’re only surveying 3/4 of the families, not all of them.

92

[QUOTE=tim314]
“Julia is our daughter”
[/quote]

Maybe that example just makes it more confusing – it depends how “Julia” was selected.

The key is, if you ask the family “Do you have any daughters” they will answer yes 3/4 of the time (systematically choosing to reveal the gender of the female child, if there is one), whereas if you ask the family “Is this (randomly selected) child a daughter”, then they answer yes only 1/2 the time.

93

I send a survey to all 12 families. The survey consists of two questions:

  1. “Do you have a daughter?”
  2. “Do you have a son?”

I sort the responses into two piles, based on their answer to question 1.

The Abbotts, the Adams, the Andersons, the Bakers, the Browns, the Byers, the Carpenters, the Clarks, and the Corbins all write back “Yes” to the first question. Their surveys go in the first pile.

The Davids, the Deans, and the Drakes write back “No” to the first question. Their surveys go in the second pile.

Now, let’s look at the first pile:
The Abbotts, the Adams, and the Andersons wrote “No” to the second question.
The Bakers, the Browns, and the Byers wrote “Yes” to the second question.
The Carpenters, the Clarks, and the Corbins wrote “Yes” to the second question.

So, in the first pile I have 6 yes answers (to question 2), and 3 no answers, out of 9 total answers. 6/9 is 2/3.

If you think I’m saying 2/3 of all families have a boy (8 out of 12 in this case), you’re misunderstanding what I’m saying (and what Cecil is saying).

I’m saying 2/3 of the familities that have a girl also have a boy. In other words, 2/3 of the responses in the first pile.

94

In fact, wissdok’s question is the perfect illustration of what I’m saying, except he misunderstood what 2/3 was supposed to represent. 2/3 of the families that have a girl also have a boy. Not 2/3 of the total families.

95

Among the family with two children that have a least one girl, the chance to have a boy is 2/3. But if the family and child are both randomly chosen, then the statement is ambiguous, and without more information then falls back to ½. I have tried to show that unless we are given a family that was specifically chosen because it has a girl, the 2/3 rule won’t work. The families that have both girls and boys are left to random chance on what information is given, which leaves this puzzle unrepeatable. This is why Cecil’s girl problem and Marilyn’s boy problem can’t exist in the same universe if left to random chance. Because of Marilyn’s and Cecil’s opposite problems you would need an endless supply of GB and BG, far above the numbers of GG and BB. As I tried to show with the “stock” four families, it can only work if we predetermine what we are looking for and disregard the families that give us information that is contrary. We also have to set it up for the mixed gender families to always answer with the information we need. If we are looking for girls, the mixed gender families need to say “girls,” otherwise they will be disregarded.

If the families are random along with the child chosen, then we can’t assume that the mixed gender families are going to answer the same all the time. Given to random chance 25% of the time both the GB/BG families will say boys, 25% both will say girls, and 50% of the time one will say girl and one will say boy. If we repeat this puzzle until we get an equal number of each of the four families, it would really doesn’t matter how they answered; In the end, the two GG/BB families will create a balance and make it 50/50.

Now given Cecil’s puzzle

The first sentence talks of a family, without any stated restrictions at this point, so it can be any family. The next sentence we are told of a daughter, it doesn’t say that the family of the previous sentence was chosen because of the daughter…. it just so happen this family had a daughter. There is nothing in the whole statement that implies that the source of this information restrict his search for a family to one that had daughter, nor does it put safeguards in place requiring a family that has a daughter to proclaim it. If we only meet three families (GG,BG,GB), what would stop the BG or GB from telling us they had boys which would completely throw the 2/3 thing out the window. If the puzzle itself doesn’t say that all families with girls proclaim that information, then we very well might have only have the GG family to deal with.

96

[QUOTE=wissdok]
We have 12 families. Why 12? Twelve is the lowest number that is both divisible by four (the number of possible families) and three ( the denominator for 2/3 ).

The Abbotts, the Adams, and the Andersons each have two girls.
The Bakers, the Browns, and the Byers each have a girl and a boy.
The Carpenters, the Clarks, and the Corbins each have a boy and a girl.
The Davids, the Deans, and the Drakes each have two boys.
…
The challenge is to survey all these twelve families and apply the 2/3 rule.
If the 2/3 rule works then you should get 8 out of 12 right. 8 out of 12 is, of course, 2/3.
NOTE: To prove that the 2/3 rule works, you must apply the rule to **all **twelve families.
[/QUOTE]
Right. Of all the families that have at least one daughter (All those with surnames starting with A, B, and C), 2/3 (B and C) have a son. Of all the families that have at least one son (groups B, C, and D), 2/3 (B and C) have a daughter. The groups overlap, of course–they have to–and it’s no coincidence that the overlapping groups are the mixed siblings. We can’t consider the entire group together because we are by default throwing out a group that doesn’t have sons (in the first case) or daughters (in the second case).

This is yoru error; we’re not selecting the families based upon random chance; we are in fact imposing the condition that they must have at least one daughter (or son). By trying to maintain a completely random distribution, you are wilfully ignorning the statement of the problem.

Stranger

97

Cecil’s puzzle
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.)

May I ask, where have you concluded we were given a restricted family? The first sentence where we are given a family, neither mentions a son or daughter. From this we can conclude that the family was chosen first from all available families. The second sentence where we are told of a daughter is, by its very nature, an afterthought. In neither of these sentences does it say that having a daughter was a condition for choosing this family. I will state again, from what we are given, a random family is chosen and then a random child of that family.

98

FINAL CONCLUSION:
If you know the gender of one of two children, but not the birth order, the odds of the other child is the opposite sex is 2/3.
If you know the gender and birth order of one of two children, the odds the other child is the opposite sex is 1/2.

Does anyone dispute this?

99

[QUOTE=wissdok]
Cecil’s puzzle
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.)

I will state again, from what we are given, a random family is chosen and then a random child of that family.
[/QUOTE]

So you believe that “You have been told this family has a daughter” means “A child has been randomly selected from this family and found to be female” ??

I submit that this is a strange interpretation - one that few would share. If the problem depended on the random choice of a child, most people would expect it to say so.

The mainstream interpretation of “You have been told this family has a daughter” is “The two children include at least one daughter” which is exactly equivalent to “This family does not have two sons.”

100

[QUOTE=wissdok]
Among the family with two children that have a least one girl, the chance to have a boy is 2/3. But if the family and child are both randomly chosen, then the statement is ambiguous, and without more information then falls back to ½.
[/QUOTE]

This statement is correct, and seems strongly to confirm that you understand that we are in fact arguing about what is the problem to be solved, rather than disagreeing about the probabilities.

Note that in post #22 you say that there is just one problem, not 2. It would have saved a lot of trouble if at that time you’d acknowledged that 2/3 is the correct answer to one problem - the one I’m claiming is the “mainstream” interpretation of what Cecil posed.