Boy/Girl probability in Monty Hall

[tim314]

adbadqc, you still aren’t understanding the problem as stated.

Consider the choices:

RR
RB
BR
BB

What you’re saying is if one is revealed to be R, then there’s a 50-50 chance the other will be revealed to be B. However, this is only true if the one which was revealed was chosen at random. In the following, I will show that this problem (as stated) has the one which is revealed chosen systematically, which changes the odds.

First, consider the case where we reveal a card at random. Without loss of generality, we can rewrite the pairs so that the revealed card is on the left. So the possibilities are:
RR
RB
BR
BB
all equally likely.

Of the cases where the revealed card (the left one) was R, the other card is B half the time and R have the time. 50-50, like you said. But this isn’t the same as the given problem, for the reason I’m about to tell you.

In the given problem, the parents tell you whether they have at least one girl child. In other words, if they have even one girl child, they will systematically choose that child as the one to reveal. This is equivalent to systematically chosing to reveal a red card if you have one.

So, the four equally probable possibilities (with the revealed card placed on the left) are:
RR
RB
RB
BB

Note that the BR flipped to a second RB, because they never reveal a B when they have an R.

So now, there are three cases where an R was revealed, and in two of the three the other one was a B. The odds are 2/3.

The point here is that answering the question “Do you have any girl children?” is not the same as answering the request “Tell me the sex of one of your children.” In one case, you systematically reveal a girl child (if you have one), in the other you reveal one at random (presumably).

Although on further review, perhaps I’m being unfair. Cecil orignally said “You have been told they have a daughter,” which is ambiguous as to whether they systematically revealed a daughter or just happened to reveal that child by chance. His subsequent explanation makes it clear that he meant it to be revealed systematically (as it would be if you had asked them “Do you have any daughters?”) but he probably should have been more clear in the original statement.

[tim314]

So, in summary, it all depends on whether the revealed child is chosen randomly or systematically.

If you always reveal a girl if you’ve got one, then you’ll reveal a girl three times out of four, and two out of those three times the other is a boy. Answer: 2/3

If you reveal a kid at random, then you’ll reveal a girl two times out of four, and one out of those two times the other is a boy. Answer: 1/2

So it depends on how you interpret “You have been told this family has a daughter.” Cecil’s explanation makes it clear he meant it in the way that gives 2/3, but he could have been more clear. If he had said “You ask if they have any daughters, and they say ‘yes’” then the answer would be 2/3 unambiguously.

Can we all agree on this?

[Xema]

@adbadqc

Once you know you have one girl (let’s use a capital G for the “known girl”, and lower case for the unknowns), these are the following possibilities (using the column example):
25% G g
25% g G
25% b G
25% G b

The first two possibilities presented here are in fact one possibility (i.e. one family) with a probability equal to the other two: 1/3.

Focus on the fact that the problem specifices chosing a family, not a child.

[adbadqc]

I really do understand what you’re saying, but what I’m trying to say is Statistics doesn’t work that way and since it’s somewhat counter-intuitive (and I’ve had a long day) I’m not doing a very good job of explaining it.

You can’t flip a BR and make it an RB. The order-sensitive distribution model differentiates between a B showing up before an R and vice versa.

BB
BR
BR
RR

is NOT a statistically accurate distribution model for this problem; RB is not the same thing as BR. So let’s try this with the other model (where order doesn’t matter) and see where that takes us.

25% BB
50% BR
25% RR

So… you know you have an R. Therefore since 2/3rds of what’s left of the original distribution has a B, then it must be twice as likely as an R, right?

Wrong. Here’s why. YOU HAVE TO THROW AWAY HALF OF THE BRs!

???

That’s right. Why? Because I said so. No; just kidding.

Because the selection from that subset of the population results in a R only 50% of the time; therefore you can’t include the entire subset in the next stage of the problem, you have to take into account that your actual selection (if random) had a 50% chance of selecting if it came out of this cohort*, therefore only 50% can be included in the next sampling, along with 100% of the RR cohort, and 0% of the BB cohort.

That evens up the size between the BR and the RR, and we’re back to 50-50.

• cohort: a band or group of people

[Xema]

@adbadqc

BB
BR
BR
RR
is NOT a statistically accurate distribution model for this problem; RB is not the same thing as BR.

What in the original statement of the problem makes RB different from BR? Please quote the relevant words.

More specifically, what difference is there between:

1. Choose one of 3 possible families from this list: GG, BG, GB
2. Choose one of 3 possible families from this list: GG, BG, BG

[adbadqc]

OK; I think I need to stop trying to explain this and just ask folk a few questions:

1. Based on your (speaking to the audience at large here) understanding of biology and how the world works (put DOWN that statistics textbook and step away from the table), what would you expect the sex ratio of a first child to be? I assume most of you answered 50-50 male/female.

2. Based on your understanding of biology etc. etc. etc, what would you expect the sex ratio of the SECOND child in a family to be? I’m guessing (hoping) most of you said 50-50 male/female.

3. Based on your understanding of biology etc. etc. etc., would you expect the sex of the first child in a family to have an impact on the sex of the second child in a family? (If you answered “yes”, please explain the mechanics of this effect).

4. Given the above, and leaving that statistics book lying there; what would you expect the sex ratio of the other sibling in a family to be where there is one daughter? Unless you’re cheating, you said 50-50 male/female. Think about it.

So… if your understanding of biology suggests that if a two child family has one daughter, the likelihood of another daughter is 50% and a son is 50%, and your understanding of statistical manipulation (you can pick up that textbook now) suggests that a brother is twice as likely as a sister… where’s the disconnect?

You have some choices:

1. Your understanding of biology is flawed
2. Your understanding of the appropriate statistical model/manipulation is flawed.
3. Something else (possibly I’m horribly misreading the problem statement).

I’ve tried to explain the appropriate application of statistical modeling techniques to this problem, and how they adequately explain the expected result; people seem to be having problems with that. So for those of you in that camp, what is it you’re suggesting? A daughter in a two child family is twice as likely to have a brother as a sister? That knowing that one child is a daughter increases the likelihood that the other child is a son?

This is taking longer than I thought.

[wissdok]

Posted By CJJ
In your card problem, I agree that the odds are 50-50 (as long as you can guarantee the specific distribution of cards in column 1 and column 2. However suppose you ask your buddy to pick a card, tell you the color, but NOT tell you whether he picked from the first or second column. Suppose he says he picked B. The only three combinations available then are RB, BR, or BB: 2 to 1 in favor of the other card in the pair being the opposite color. Note that if you further ask him “Which column is the B card in?”, and he says “the first column”, you can eliminate RB as well and drop the odds back to 50-50.

I once tried to explain this with coins and then cards but I couldn’t get people to agree on how to transpose this into cards. As both children in a family are independent events you should have two piles with each having a red and a black. If your friend draws one card from each pile, then it doesn’t matter which pile the card is from that he reads to you.This is because the other pile had exactly one red and one black. Hence, either card he tells you about the chances of guessing right…50% / 50% In a round about way this works with Adbadqc biology statement, there isn’t 3 cards in that other pile, just two…a black and a red.

Posted By Xema
The problem is perfectly symmetrical. If we randomly select a family and are told this it includes at least one son, we can say that the chance this boy has a sister is 2/3. If we are instead told that it includes a daughter, the probability of this girl having a brother is again 2/3.

There is no contradiction or difficulty here.

These two events are not simultaneously possible. If we take just four families (representing each of the possible types) then if the given is a son and it turns out that we get a 2/3 result…then at the same time if we are given a girl for another of the families then it has to be from the family with only girls and we would automatically get that wrong using the 2/3 “rule”. This happens because to get the a 2/3 result, the two families that have a mixed set of kids, must have given us the clue of “ a boy”…leaving only the family with two girls to answer with the clue of “a girl.” The laws of probability state that the families with both a boy and a girl would randomly give us a clue, half the time boy, half the time girl. If we use a bigger sample group then we will make the 2/3 result even less likely to happen. Either way, this is not a repeatable experiment as even in a group of four families the odds of two families with mixed kids giving matching clue is only 50%. And again I point out that we can’t have a 2/3 result for boys at the same time we have a 2/3 result for girls.

Posted By Xema
I think you are introducing an unnecessary and unhelpful addition with this concept of the family “telling” something. Is the family truthful? Do all families know the sex of all their children?

Irrelevant! The family is chosen at random, and you are told (assumption: truthfully, by whom it doesn’t matter) that at least one of their children is a daughter. This allows you to conclude that they are not among the 25% of families that have two sons, and thus there is a 2/3 probability that their daughter has a brother.

I once thought there was a difference between the puzzle when you used tell, meet, were given, …etc. After I studied this problem enough I came to the conclusion that it doesn’t matter. If we are given a child, told about a child, or even meet the child…we are told all we can base our answer on. The information must be factual; there is no reason to say otherwise. There is an argument that meeting a child allows you to separate it from its sibling, but the puzzle fails long before we have to get that far into the problem.

Posted by Xema
Suppose I take 4 sheets of paper, number them 1 through 4, and write the following on them:
Sheet 1: BB
Sheet 2: GG
Sheet 3: BG
Sheet 4: GB

I proceed to take sheet #1 and burn it, taking care to stomp the ashes into the ground. I then ask a friend to randomly select one of the remaining sheets of paper by a process that makes each of the three selections equally likely.

Two questions to answer:

1. What is the probability that he selects a sheet of paper with two different letters on it?
2. How does this differ from the original problem?

Here you have kept the Sheets(families) as a random event, but what is written on them is irrelevant(letters=children). At the being of the boy/girl problem that is true too, but in the second part of the puzzle the children do matter. The children are independent events, not like the letters. While yes this problem is 2/3, there is a big difference between it and the original puzzle because of the lack of a second part.

[Xema]

@adbadqc

So… if your understanding of biology suggests that if a two child family has one daughter, the likelihood of another daughter is 50% and a son is 50%, and your understanding of statistical manipulation (you can pick up that textbook now) suggests that a brother is twice as likely as a sister… where’s the disconnect?

Where’s your acknowledgement that the essential condition of the problem is to exclude one of the four types of families? That nothing posted by those who you disagree with in any way argues aginst the premise that the population of children is half male, half female (indeed, this is stated as one of the conditions of the problem).

It’s hard for me to see why, at this stage in this thread, you would devote a bunch of space to a discussion of obvious biology instead of addressing the points that have been made. Instead, why not answer the questions in post #37?

… what is it you’re suggesting? A daughter in a two child family is twice as likely to have a brother as a sister? That knowing that one child is a daughter increases the likelihood that the other child is a son?

How about: that knowing the family isn’t BB reduces the possibilities from 4 to 3.
(Has this not already been stated enough?)

[Ivan-Osokin]

Chances that everyone in any thread covering this topic will agree or come to understand that the answer is 2/3?

1 - .9999 repeating.

[Xema]

@wissdok

At the being of the boy/girl problem that is true too, but in the second part of the puzzle the children do matter. The children are independent events, not like the letters. While yes this problem is 2/3, there is a big difference between it and the original puzzle because of the lack of a second part.

Thanks for acknowledging that the probability here is 2/3. What remains is to decide whether or not this problem is congruent to the original one.

You say it isn’t because it lacks a second part. I looked back at the original problem and found just one part: we have a family with two children and information that at least one of them is female. I say this is the same as being told that a selected sheet of paper has at least one “G” on it.

I agree if there were a “second part” to the problem - if, for example, we were to select a child at random - then the problem would not be the same as the one with sheets of paper. But I see nothing in the original problem that calls for the selection of a child. If you can find such language in the problem, please let me know.

[matt_mcl]

@wissdok

These two events are not simultaneously possible.

But we’re not talking about the same event, we’re talking about possibilities. Depending on whether the revealed child is a boy or a girl, you have eliminated a different set of possible families.

You are spinning it as if the same family has a 2 in 3 chance of a certain child being a boy and also a 2 in 3 chance of that child being a girl. But that isn’t what was described to you. What was described was two different cases.

Should you discover that one of the children is a boy, odds are 2 in 3 that the other child is a girl.

Should you instead discover that one of the children is a girl, that means that this family is of a different type than in the contrary case (the type of family that contains at least one girl); and the odds for that kind of a family are 2 in 3 that they also have a boy.

[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?

[wissdok]

Basic logic….two independent events have no effect on each other. The birth of a child is an independent event, so why would anyone think, all things being equal, that once I have a daughter I am 2/3 likely to have a son? Or the reverse, I have a son so now I am likely to have a daughter. In our example sample (GG, GB,BG,BB) we all seem to again that they are equally likely, but if the 2/3 rule really worked your possibilities would be:
Family #1 GG = 16.67%
Family #2 GB = 33.34%
Family #3 BG = 33.34%
Family #4 BB = 16.67%

An example, we meet 12 families at a BBQ and each have two kids apiece. Six families say they have at least one girl, six families say they have a least one boy. If we use the 2/3 rule the distribution will be:
Family #1 GG = 2 in 12 (1/3 of the 6 girl families) = 16.67%
Family #2 & Family # 3 GB/BG = 8 in 12 (2/3 of both girl and boy families) = 66.67%
Family #4 BB = 2 in 12 (1/3 of the 6 boy families) = 16.67%

We could use 100 families, 200 families, or Gazpacho’s 1000 families or anything else that all four families would have an equal chance of being represented. If the 2/3 rule was correct then the families could never be equal. We can all just look at the 1000 families, if the 2/3 rule is in effect then 2/3 of the families would be mixed with a boy and a girl.
Only 1/3 of the families would have a pair of same sex children…half these family being with boys and half being with girls. This just doesn’t add up.

The failure of this 2/3 rule is idea that if we are given a girl then it means that the two-boy family isn’t possible. While that is true, there is something more to that. The two-boy family isn’t possible BECAUSE we know that either event #1 (child #1) or event #2 (child #2) is a girl. So if we know that one child can’t be a boy then either Family #2(GB) or Family #3(BG) is also not possible.

In an earlier post we talked of cards. I put this challenge to you. Separate the reds from the blacks and then deal out the reds into 26 piles and then add the blacks to each pile. Then shuffle all 26 piles separately. Each pile now represents an independent event. Draw one card from any pile, then draw one card from any of the other 25 piles. The possible outcomes are:
Draw #1………Draw #2
Red……………Red
Red……………Black
Black………….Red
Black………….Black

If you look at one of the drawn cards, it doesn’t matter if you look at the first draw or second, nor does it matter which piles they originally came from. The result will be that you have a 50/50 chance to guess the other card, because all piles had just a red and a black card. If you repeat this experiment for all the rest of the deck you will find that the 2/3 rule doesn’t even come close. If you have a friend look at both cards and tell you either, he could temporarily cause it to appear that the 2/3 rule works but by the end of the deck (23 experiments) you will be closer to 50% then to 66.67%

** note this experiment is best when the samples dividable by 4. I would advice 24 or 28 hands.*

[borschevsky]

We’re not claiming that if you have a son that your next child is likely to be a daughter, and we’re not claiming that if you have a daughter that your next child is likely to be a son. Look at the question:

You have been told this family has a daughter. What are the odds they also have a son

It is a question about families, not about children. Find some fathers who have two children. Ask them if they have at least one daughter. 75% of them will say yes. From among those fathers who say yes, ask them if they have a son. 2/3 will say yes.

This is different than asking a random daughter if they have a brother; half would say yes to that. But that’s not what the question asks.

[matt_mcl]

Okay. Since I have a penny and a certain amount of free time, I just sat down to flip my silly little coin and perform an actual experiment. Here are the results of the coin tosses:

TH TH HT TT TT HT HH HT HT HT HT TH TH HT TH HH

I tossed the coin a total of 32 times. Of those, exactly 16 are tails and 16 are heads. Overall heads/tails rate: 50:50.

Considering the pairs. We will not even look at the order of the pairs. There are two TT pairs and two HH pairs, so each is 2/16 or 1/8 of the time. This is in fact less than the expected 25%. Pairs with one head and one tail make up 7/8ths of the trials.

Now, a total of 14 of the 16 trials had a head in them. Of those 14 trials, 12 also had a tail. The odds over this sample were therefore 86% over this trial that a pair containing a head also had a tail.

In other words, if you were discussing any of these pairs, and you were informed that one of the members of the pair was a head, you would be right 86% of the time - not 50% - that the other member of the pair was a tail. The order has nothing to do with it, as we can see, since we have never discussed which flip was first.

But as you can also see, if you were discussing any of these pairs the same way, and you were told that one of the pair was a tail, you would still be right 86% of the time if you were to say that the other member of the pair was a head.

[JR_Brown]

@adbadqc

25% BB
50% BR
25% RR

So… you know you have an R. Therefore since 2/3rds of what’s left of the original distribution has a B, then it must be twice as likely as an R, right?

Wrong. Here’s why. YOU HAVE TO THROW AWAY HALF OF THE BRs!

This is correct only if you chose “2-child families with a daughter” by finding all 2-child families and then asking them to pick one of their two children at random and tell you if that child is a girl. tim314 and Xema and I (and Cecil) are working on the assumption that we have selected our families by asking them *if they have any daughters at all[i/]. This makes a difference.

Consider these examples. First, the way we “2/3” people have been thinking about it:

You have four people in your kitchen, Adam, Bill, Charlie and Dave.

You give them each two cards, like so:
Adam: two black cards
Bill: two red cards
Charlie: one red and then one black
Dave: one black and then one red

You ask each person who has any red cards to move into the living room.

Question: of the people in the living room, what percentage will also have a black card?
Answer: 2/3. Bill, Charlie and Dave have at least one red card and will thus be in the living room; of these, Charlie and Dave both have a black card.
Now the way you have been thinking about it:

You have four people in your kitchen, Adam, Bill, Charlie and Dave.

You give them each two cards, like so:
Adam: two black cards
Bill: two red cards
Charlie: one red and then one black
Dave: one black and then one red

Note that so far both conditions are identical.

You ask each person to pick one card at random. If the selected card is red, they must move into the living room.

Question: of the people in the living room, what percentage will also have a black card?
Answer: On average, 1/2. Bill must go into the living room since any card he picks will be red. Charlie and Dave each have a 1/2 chance of selecting a red card, so on average one of them will go into the living room. So on average there will be one person with a black card in the living room along with Bill, who has no black cards.

So, does changing the way the people are chosen to go into the living room change the initial distribution of the cards? No. But you only discount half of the “one of each” cohort (yes, I know that word ) if you ask about a [s]specific* card.

JRB

[JR_Brown]

Aaargh. Could some kind mod please fix the first “end italics” and last “begin italics” tags in the previous post?

JRB

[Bytegeist]

@Uzi

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

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

Right. Although I’d prefer to use “G” for girl, as long as we’re abbreviating.

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 . . .

Right. Those percentages should all be 33.3% now.

The population of families allowed by the problem is a subset of the total population of two-child families, but those categories you’ve listed are still equally numerous within the restricted population.

. . . 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. . . . The girl in this case is identified first . . .

But, you don’t know the identity of one of the children. Nor are you being asked about the identity of any particular child. There is no “the girl” that’s been “identified first”. In particular, a man with two daughters can truthfully say he has at least one daughter, but without having any specific child in mind.

You are essentially told only that the “girl-count” of this family is greater than 0, and are then asked the probability that the girl-count is exactly 1.

My favorite way of seeing why the answer isn’t 50% is to expand the number of children. Suppose a family has 10 children. You are told that 9 of them are daughters. What’s the probability that they also have a son?

Adopting the reasoning that some posters have been using here, you’d still conclude that the answer is 50%. However, 10-child families with 10 girls are much rarer than 10-child families with 9 girls and 1 boy. (The second category outnumbers the first by 10 to 1.) It’s much more likely this family has a boy than not. The probability then that they have a son, somewhere among the brood, given that 9 of the brood are daughters, is 10/11.

[Larry_Borgia]

I’m willing to play the following game with anyone in the 50/50 camp.

We each flip a nickel.

If we each show heads, nothing happens.
If you show a head and I show a tail, you give me a dollar.
If you show a tail and I show a head, you give me a dollar.
If we each show tails, I give you a dollar.

I’ll spot you twenty bucks and we’ll play 400 rounds.

[wissdok]

Posted By Borschevsky
It is a question about families, not about children. Find some fathers who have two children. Ask them if they have at least one daughter. 75% of them will say yes. From among those fathers who say yes, ask them if they have a son. 2/3 will say yes.

While your example is true, that isn’t the scenario we are presented. We have to conclude that the source that gave us the information about the girl, told us of all he knew. This person wasn’t being obtuse, nor was he trying to trick us. I could just as easily create this back story:

I am buying a house in a neighborhood and my real estate agent says that the family next door has two kids, at least one is a girl. What is the sex of the other child? If you assume it is 2/3 boy….what is the basis for that? The family next door could have two girls, and this girl could be either. Because of this, we are twice as likely to be told of a girl from this family than either of the families that have only one girl. It is just as likely that the real estate agent would know of a boy, which would make the family with two boys twice as likely than either of the families with just one. The families with one of each type of child are always possible but only combined together are they equal to either of the families that have two boys or two girls.

We can also use this example:
I conduct a survey in a girl’s gym class. I gather together 30 girls who have one sibling and ask if that sibling is he a boy or a girl? 2/3 boy…WRONG. Any girl should have an equal chance to have a sister as to a brother. The most likely outcome is 15 will have sisters and 15 will have brothers.
……ACROSS THE HALL ….I survey 30 boys in gym class. 15 will have brothers and 15 will have sisters. The result of my survey:
15 girls who have a sister
15 girls who have a brother
15 boys who have a sister
15 boys who have a brother

Posted by JR Brown
You have four people in your kitchen, Adam, Bill, Charlie and Dave.

You give them each two cards, like so:
Adam: two black cards
Bill: two red cards
Charlie: one red and then one black
Dave: one black and then one red

You ask each person who has any red cards to move into the living room.

Question: of the people in the living room, what percentage will also have a black card?
Answer: 2/3. Bill, Charlie and Dave have at least one red card and will thus be in the living room; of these, Charlie and Dave both have a black card.

Here you have chosen “red”. Let’s try… you go get a Mountain Dew and find a red card on the floor next to the table? What are the chances this card came from Charlie’s or Dave’s hand? Are they not combined, equal to the chance that it came from Bill’s hand?

Posted by Bytegeist
My favorite way of seeing why the answer isn’t 50% is to expand the number of children. Suppose a family has 10 children. You are told that 9 of them are daughters. What’s the probability that they also have a son?
Adopting the reasoning that some posters have been using here, you’d still conclude that the answer is 50%. However, 10-child families with 10 girls are much rarer than 10-child families with 9 girls and 1 boy. (The second category outnumbers the first by 10 to 1.) It’s much more likely this family has a boy than not. The probability then that they have a son, somewhere among the brood, given that 9 of the brood are daughters, is 10/11.

This analogy isn’t very relevant. In your question the “lost” boy could have been in any of 10 places….in our problem just the one not occupied by the given girl. I wouldn’t bet for craps 10 times in a row, but I would twice in a roll. In your example the possibilities were never equal as you pointed out, the chance for a boy is 91%.