Straight Dope Message Board > Main Boy/Girl probability in Monty Hall
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#51
08-12-2006, 03:05 AM
 matt_mcl Charter Member Join Date: Mar 1999 Location: Montreal Posts: 20,195
Quote:
 Originally Posted by 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.
#52
08-12-2006, 07:43 AM
 Uzi Charter Member Join Date: Jul 2000 Location: Hong Kong Posts: 4,041
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?
#53
08-12-2006, 10:27 AM
 wissdok Guest Join Date: Aug 2005
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.
#54
08-12-2006, 10:42 AM
 borschevsky Guest Join Date: Sep 2001
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:

Quote:
 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.
#55
08-12-2006, 10:58 AM
 matt_mcl Charter Member Join Date: Mar 1999 Location: Montreal Posts: 20,195
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.
#56
08-12-2006, 11:15 AM
 JR Brown Guest Join Date: Mar 2006
Quote:
 Originally Posted by 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:
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:
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
#57
08-12-2006, 11:18 AM
 JR Brown Guest Join Date: Mar 2006
Aaargh. Could some kind mod please fix the first "end italics" and last "begin italics" tags in the previous post?

JRB
#58
08-12-2006, 11:26 AM
 Bytegeist Guest Join Date: Jul 2003
Quote:
 Originally Posted by 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.

Quote:
 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.

Quote:
 . . . 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.
#59
08-12-2006, 12:46 PM
 Larry Borgia Charter Member Join Date: Mar 2000 Location: Washington DC Posts: 8,207
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.
#60
08-12-2006, 01:28 PM
 wissdok Guest Join Date: Aug 2005
Quote:
 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

Quote:
 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?

Quote:
 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%.
#61
08-12-2006, 01:31 PM
 LIT123 Guest Join Date: Jul 2003
I'm lousy at these things which is bad news for the 50-50 camp since that is where I am. I might understand the 2/3 CROWD if I could grasp why they
have both a BG and a GB. Is the order of birth being addressed and if so
why not 2 BB and 2 GG which after elimination leaves _G, G_, _B, and B_.
That gets me back to 50-50. If there is some "rule of" thing that covers it
then post that phrase and I can Google to save all your time. My next questuin
is,assuming adequate O2, h20, and food, if the second child is sealed in a steel tube is it G, B, or neither?
#62
08-12-2006, 01:32 PM
 wissdok Guest Join Date: Aug 2005
Quote:
 Posted by Larry Borgia 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.
I don’t really understand why we throw out two heads. I’ll take your bet if you let me win on two heads too. And to make it worth my while….each time I win you pay me \$1.25.
I believe in the end you will owe me \$50.
#63
08-12-2006, 01:36 PM
 Larry Borgia Charter Member Join Date: Mar 2000 Location: Washington DC Posts: 8,207
Quote:
 Originally Posted by wissdok I don’t really understand why we throw out two heads. I’ll take your bet if you let me win on two heads too. And to make it worth my while….each time I win you pay me \$1.25. I believe in the end you will owe me \$50.
The rules of the game mirror the original boy girl rule. We throw out the families who have two boys, and two thirds of the families will have a boy and a girl. That was the original problem. You can make a new problem in which the answer is 50/50, but it's not the problem Cecil describes in his column.
#64
08-12-2006, 02:02 PM
 wissdok Guest Join Date: Aug 2005
Quote:
 Posted by LIT123 I'm lousy at these things which is bad news for the 50-50 camp since that is where I am. I might understand the 2/3 CROWD if I could grasp why they have both a BG and a GB. Is the order of birth being addressed and if so why not 2 BB and 2 GG which after elimination leaves _G, G_, _B, and B_. That gets me back to 50-50. If there is some "rule of" thing that covers it then post that phrase and I can Google to save all your time. My next questuin is,assuming adequate O2, h20, and food, if the second child is sealed in a steel tube is it G, B, or neither?
I am also in the 50/50 camp (as you can see from my posts) but it is generally agreed that the two children are variables that are not interchangeable. While for simplicity we can call it “age”, they are independent events that are different. In figuring this problem we assume that if we have a child, it has an equal chance to have a brother or sister.
We assume that we have an equal chance for these combinations:
Girl with a sister
Girl with a brother
Boy with a sister
Boy with a brother

In the above example girl/boy is represents event #1 and the sister/brother represents event #2.

This is the same as flipping two coins, a nickel and a dime, we would record their outcomes separately. The possible outcomes are:
Nickel…………………Dime
Tails…………………..Tails

As to your second question, it is the argument of the 2/3 camp that once we have been given a girl then we can just drop the two boy family as a possibility and the remaining three families are equally likely to by her family. I can’t speak for the other 50/50 “campers” but I believe that this is a fallacy. I believe that if you remove the boy/boy family it is because we now know that either event #1 or event #2 was a girl which means we must also remove one of the GB or BG families as well.
#65
08-12-2006, 03:35 PM
 JR Brown Guest Join Date: Mar 2006
To LIT123:
Quote:
 Originally Posted by LIT123 I'm lousy at these things which is bad news for the 50-50 camp since that is where I am. I might understand the 2/3 CROWD if I could grasp why they have both a BG and a GB. Is the order of birth being addressed and if so why not 2 BB and 2 GG which after elimination leaves _G, G_, _B, and B_. That gets me back to 50-50. If there is some "rule of" thing that covers it then post that phrase and I can Google to save all your time.
The order of birth is not important; listing GB and BG is just to make it clear to all that "family with one boy and one girl" is twice as common as "family with two girls". So (BG + GB) = (BB + GG), and (BG + GB) / (BG + GB + GG) = 2/3.

In general, the class of statistics that covers these problems is called "conditional probability".

To wissdok:
Quote:
 Originally Posted by wissdok "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”.
It also works if you choose "black". All those with at least one black card move into the living room: Adam, Charlie and Dave are now in the living room. How many of them have red cards? Two: Charlie and Dave = 2/3. Note that the set {Bill, Charlie and Dave} and the set {Adam, Charlie and Dave} have the same number of people (three) and the same aggregate number of cards (six), but are different sets.

Quote:
 Originally Posted by wissdok 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?
Yes, but that is A STASTISTICALLY DIFFERENT QUESTION! It is the equivalent to the second boy/girl problem in Cecil's column ("The first child enters--a girl. The second knocks", etc).

Reading your explanations, you consistently phrase the question as if we were discussing a specific girl, and asking if she had a brother. All of your explanations are relevant to what I'm calling the second boy/girl problem, where this is indeed the question. But in the first boy/girl question, we are asking about families (as gazpacho and matt_mcl and borschevsky previously noted).

To reiterate: the problem I am solving is
"What is the probability that a pair of siblings, of which at least one is a girl, also includes a boy"
NOT
"what is the probability that a girl with one sibling has a brother"

Quote:
 Originally Posted by wissdok As to your second question, it is the argument of the 2/3 camp that once we have been given a girl then we can just drop the two boy family as a possibility and the remaining three families are equally likely to by her family. I can’t speak for the other 50/50 “campers” but I believe that this is a fallacy. I believe that if you remove the boy/boy family it is because we now know that either event #1 or event #2 was a girl which means we must also remove one of the GB or BG families as well.
OK, here's the nub. "I believe that if you remove the boy/boy family it is because we now know that either event #1 or event #2 was a girl." The key word in that sentence is EITHER. We don't know which one. As such, BOTH the GB and BG families are still possible, as well as the GG family. Only the BB family has been excluded.

Here is an article from the Journal of Statistics Education that discusses a related problem (complete with heavy-duty math):
"United States life-table data in a certain year shows that the probability of a male birth is 0.512. If two pregnant women are chosen at random from that population, what is the conditional probability that both will give birth to boys, given that we know that at least one did so?"

Why? Math, much simplified:
probability of 2 boys = square of 0.512 = 0.262
probability of 2 girls = square of (1 - 0.512) = square of 0.488 = 0.238
probability of one of each = 2 x (0.512 x 0.488) = 0.500
(check: the above sum to 1. Good)

SO:
probability of at least one boy = probability of 2 boys + probability of one of each = 0.762

NOW:
probability of two boys, GIVEN THAT there is at least one =
(probability of 2 boys) divided by (probability of at least one boy) =
0.262 / 0.762 = 0.344
QED.

(Note that this is the inverse of our problem, and that they are not setting the boy-girl probability equal).

But why get bogged down in all that? The simplest explanation of the problem was back in post #9:

Quote:
 Originally Posted by gazpacho (slightly edited for typos and clarity) If we assume that girls and boys are equally likely then the probabilities for these families are: 1/4 two boys 1/4 two girls 1/2 one of each In a population of 1000 families: 250 have two boys 250 have two girls 500 have one of each There are 750 families that have at least one child that is a girl: 250 have two girls 500 have one of each 500 out of these 750 have a boy, or 2/3 of them.
Quote:
 Originally Posted by wissdock (edited to change family ID's for clarity) *If the families are equal, then the children are not. If we are given a girl and the families are equal, then each families (GG, BG, GB) are each likely 33.3% of the time… making each girl in family GG worth half the value of the girls in families GB and BG.
So what? We aren't being given a girl, we are being given a family that includes at least one girl. The distinction between these two situations determines how the problem must be addressed; if you can't see that, then you need to brush up on your conditional probability!

JRB
#66
08-12-2006, 05:29 PM
 Bytegeist Guest Join Date: Jul 2003
Quote:
 Originally Posted by wissdok We have to conclude that the source that gave us the information about the girl, told us of all he knew.
There is no "the girl". There is no specific child we're being given information for.

Quote:
 Originally 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?
Quote:
 Originally Posted by wissdok 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.
And in the two-child problem, the boy can be in any of two places. Moreover, there is no "given girl".

My analogy is completely relevant because it's the exact same problem, only expanded to a larger number of children. To make it more general though, restate the problem this way:

Quote:
 There is a family with N children. You have been told this family has N – 1 daughters [ at least ]. What are the odds they also have a son, assuming the biological odds of having a male or female child are equal?
For N = 2, this is the exact same problem Cecil gave in his column.

In the world at large, the N-child families having exactly 1 boy and N-1 girls always outnumber the ones having all girls — outnumber them by a ratio of N to 1, in fact. So the answer to the generalized problem is N/(N+1). Which of course is 2/3 for N = 2.
#67
08-12-2006, 05:56 PM
 Frylock Guest Join Date: Jun 2001
Quote:
 Originally Posted by 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.
Damn straight.

Here's a post where I did something like this with a sample size of a hundred pairs: http://boards.straightdope.com/sdmb/...7&postcount=60

To my recollection, Wissdoc was still not convinced.

-FrL-
#68
08-12-2006, 06:25 PM
 wissdok Guest Join Date: Aug 2005
Quote:
 Posted by JRBrown So what? We aren't being given a girl, we are being given a family that includes at least one girl. The distinction between these two situations determines how the problem must be addressed; if you can't see that, then you need to brush up on your conditional probability!
Quote:
 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.)
“Many readers correctly pointed out that the answer depends on the procedure by which the information "at least one is a boy" is obtained. If from all families with two children, at least one of whom is a boy, a family is chosen at random, then the answer is 1/3. But there is another procedure that leads to exactly the same statement of the problem. From families with two children, one family is selected at random. If both children are boys, the informant says "at least one is a boy." If both are girls, he says "at least one is a girl." And if both sexes are represented, he picks a child at random and says "at least one is a ..." naming the child picked. When this procedure is followed, the probability that both children are of the same sex is clearly 1/2. (This is easy to see because the informant makes a statement in each of the four cases -- BB, BG, GB, GG -- and in half of these case both children are of the same sex.) That the best of mathematicians can overlook such ambiguities is indicated by the fact that this problem, in unanswerable form, appeared in one of the best of recent college textbooks on modern mathematics.”
Scientific American, October, 1959

Did I miss something in Cecil’s puzzle, it didn’t say anything about choosing families that had only girls. The first sentence says, “There is a family with two children.” This sentence by itself doesn’t say it limits itself to just families with girls. In the next sentence we are told of a daughter. This second sentence doesn’t say that all families that have daughters are now in a subgroup, we only know of one possible family that will always say they have girls. Without some safeguard in place those families that have both a boy and a girl, could have just as likely tell us of a boy. We don’t know how they would have answered.

An example:
Cecil Adams and Marilyn vos Savant climb into a elevator together. They start talking to each other about this puzzle. In Cecil’s version, he used a girl; In Marilyn’s a boy. They decide to test the 2/3 rule. Lucky for them they have 4 other people on the elevator with them and they all have two kids apiece.
They ask the first person to tell a clue about his children. I have a least one boy. Cecil disregards the answer, Marilyn uses the 2/3 rule and predicts the other is a girl.
The man has a son and a daughter so……….Marilyn is CORRECT.
They the repeat the question to the second man. He says he has at least one son.
Again Cecil disregards this answer and Marilyn chooses opposite.
But this time the man says he has two sons…..Marilyn is INCORRECT.
They ask the third person, and she says she has at least one daughter. This time Cecil picks opposite and Marilyn passes on this question. The woman proclaims that she has a son and daughter so…….Cecil is CORRECT.
They approach the final person and she says she has a daughter too. Cecil again chooses opposite and Marilyn again passes. This woman has two daughters so….Cecil is INCORRECT.

Final score: Cecil has 1 right and 1 wrong. Marilyn has 1 right and 1 wrong. If they had answered on all questions then they both would be with 2 and 2.
#69
08-12-2006, 08:11 PM
 JR Brown Guest Join Date: Mar 2006
Quote:
 Originally Posted by wissdok “Many readers correctly pointed out that the answer depends on the procedure by which the information "at least one is a boy" is obtained. If from all families with two children, at least one of whom is a boy, a family is chosen at random, then the answer is 1/3. But there is another procedure that leads to exactly the same statement of the problem. From families with two children, one family is selected at random. If both children are boys, the informant says "at least one is a boy." If both are girls, he says "at least one is a girl." And if both sexes are represented, he picks a child at random and says "at least one is a ..." naming the child picked. When this procedure is followed, the probability that both children are of the same sex is clearly 1/2. (This is easy to see because the informant makes a statement in each of the four cases -- BB, BG, GB, GG -- and in half of these case both children are of the same sex.) That the best of mathematicians can overlook such ambiguities is indicated by the fact that this problem, in unanswerable form, appeared in one of the best of recent college textbooks on modern mathematics.” Scientific American, October, 1959
This is true, for what it's worth, and it's an example of how probabilities will change with different selection criteria that give (superficially) similar outcomes. But from my perspective Mr Gardner's readers' objections are like asking someone "Will an unsupported rock fall down?", and when they say "Yes", responding "It wouldn't if it were in geosynchronous orbit, so you're wrong!"

If I were asked about "families with two children, at least one of whom is a boy", I would never spontaneously imagine the second selection process described. Cecil's phrasing ("There is a family with two children. You have been told this family has a daughter.") implies to me that all children have been inspected and an "at least one girl" rule applied.

There are certain assumptions built in to any statement, and while those assumptions may in some cases be wrong, it is, in general, up to the questioner to mention any deviations from standard conditions. Otherwise it is always possible to say "ahh, but it could have been otherwise!"

For instance, neither Cecil's nor Gardner's questions specify a human family, so they could equally well be about a race of aliens who have three sexes (male, female, and hermaphrodite) in a 1:1:1 ratio; would you be satisfied with an answer that worked out the probabilities for that family?

JRB
#70
08-12-2006, 08:29 PM
 Frylock Guest Join Date: Jun 2001
Quote:
 Originally Posted by JR Brown This is true, for what it's worth, and it's an example of how probabilities will change with different selection criteria that give (superficially) similar outcomes. But from my perspective Mr Gardner's readers' objections are like asking someone "Will an unsupported rock fall down?", and when they say "Yes", responding "It wouldn't if it were in geosynchronous orbit, so you're wrong!" If I were asked about "families with two children, at least one of whom is a boy", I would never spontaneously imagine the second selection process described. Cecil's phrasing ("There is a family with two children. You have been told this family has a daughter.") implies to me that all children have been inspected and an "at least one girl" rule applied. There are certain assumptions built in to any statement, and while those assumptions may in some cases be wrong, it is, in general, up to the questioner to mention any deviations from standard conditions. Otherwise it is always possible to say "ahh, but it could have been otherwise!" For instance, neither Cecil's nor Gardner's questions specify a human family, so they could equally well be about a race of aliens who have three sexes (male, female, and hermaphrodite) in a 1:1:1 ratio; would you be satisfied with an answer that worked out the probabilities for that family? JRB
JRB is exactly right. There are literally an infinite number of criteria by which it might have been decided that "At least one child is a boy" should be asserted. We are not told which of these criteria were adapted. But this does not mean we don't know which criterion was adapted. This is because, as the puzzle is phrased, it is not to be suspected that anything but the obvious criterion was adapted. The obvious criterion is based on the principle that, in general, an honest and unmistaken person will say "X" if and only if X is true. Similarly, since we assume the puzzle is not lying to us, and since it does not specify a criterion for decision as to when to say what about gender, we take it the obvious criterion was used: That we are told "at least one is aboy" if and only if at least one is a boy.

Based on an assumption that this was the criterion used, the answer is 2/3.

Well, that didn't turn out nearly as clear as I hoped. Trust me: I'm just saying Jr. Brown is right.

-FrL-
#71
08-12-2006, 08:55 PM
 tim314 Charter Member Join Date: Mar 2004 Posts: 3,602
Quote:
 Originally Posted by adbadqc 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
Here's where the key is how they're answering the question. If the family is reporting the sex of a random kid, then yes, the selection of that subset results in them saying "girl" half the time.

However, if they are answering the question "Do you have any girl children?" then they will say yes 100% of the time when they have at least one girl. The 2/3 number is based on this second interpretation.

Do you see the distinction I'm making? I'm almost entirely sure that this is why we're getting different answers.
#72
08-12-2006, 09:12 PM
 wissdok Guest Join Date: Aug 2005
I don't have time to write a lengthy response, so for the time being, please checkout this website I found the last time we went over this subject.

#73
08-12-2006, 09:20 PM
 tim314 Charter Member Join Date: Mar 2004 Posts: 3,602
Quote:
 Originally Posted by 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. Yes, 50-50. Even if that might not be true in reality, it's definitely supposed to be assumed for this question. 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. Yes, each child has a 50-50 chance of being either sex, and they're independent of each other. 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). No, the sex of one particular child won't affect the sex of the other. 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.
Number 4 is where we disagree. Most families with one daughter also have one son (assuming they really are independent 50-50 choices). I know this seems like it contradicts the above, but it doesn't. Let me try to explain one more time:

1) There's a 25% chance of them having two girls. (50% times 50%)
2) There's a 25% chance of them having two boys (50% times 50%)
3) There's a 50% chance of them having one of each (100% minus 25% minus 25%)

Please let me know if you disagree with any of these three.

If you accept these three points, then you can see that most families that have a girl also have a boy: 50% of the total families have one of each, whereas 25% have girls only.

The reason you think of it as contradicting your points above is that you're not drawing a distinction between saying a particular child is a girl and saying at least one of their two children is a girl. But these are distinct.

If the first child is a girl, then yes, the second child is equally likely to be a girl or boy. If the second child is a girl, then yes, the first child was equally likely to be a girl or boy.

But if I ask "Do you have any girls," you'll say yes if the first or the second one is a girl. That's a different scenario.

Again, we have these possibilities:
(A) Girl then girl: 25% chance
(B) Girl then boy: 25% chance
(C) Boy then girl: 25% chance
(D) Boy then boy: 25% chance.

If they tell you the first child is a girl, you have possibilities (A) and (B), with a 1/2 chance that the second one is a boy.

If they tell you the second child is a girl, you have possibilities (A) and (C), with a 1/2 chance that the first was a boy.

But, if all they say is "Yes, we have a daughter," then you have (A), (B), and (C), with a 2/3 chance that they had at least one boy.

The 2/3 answer is based on the assumption that the family only said "yes, we have a daughter" -- i.e., they are telling you they have at least one daughter.

This is equivalent to looking at both cards and then saying "yes" if at least one was red.

Please let me know if you don't agree that with this interpretation of the information given the answer is 2/3, and please quote the exact step in my reasoning that you think is wrong and explain why (as I have attempted to do). After all this typing, it would be really nice if we could come to some agreement.
#74
08-12-2006, 09:40 PM
 tim314 Charter Member Join Date: Mar 2004 Posts: 3,602
A Simple Three Question Quiz

If you don't agree with the 2/3 number, please take the following quiz.

True or False:

1) There are four equally likely possibilities:
- The parents had a boy, then a girl
- The parents had a girl, then a boy
- The parents had a girl, then another girl
- The parents had a boy, then another boy.

2) In three of these four cases, the parents will answer "yes" to the question "Do you have a daughter?"

3) In two out of those three cases, the parents also have a son.

So the odds are 2/3, if the parents were asked "Do you have a daughter?" rather than "Was your first child a daughter?" or "Was your second child a daughter?" The key is that "Do you have a daughter?" produces more "yes" answers than either of those other questions, and all the extra "yes" answers come from cases where they had both a girl and a boy.
#75
08-12-2006, 09:58 PM
 Frylock Guest Join Date: Jun 2001
Quote:
 Originally Posted by wissdok I don't have time to write a lengthy response, so for the time being, please checkout this website I found the last time we went over this subject. Ask Marilyn Response
Right, it looks like that link is reiterating the interpretation of the problem as ambiguous.

But I maintain the problem is not ambiguous. There is a single, plainly intended reading of the problem. And on that reading, the answer to the question is 2/3.

-FrL-
#76
08-13-2006, 03:02 AM
 Alex_Dubinsky BANNED Join Date: Jan 2003 Location: New York City Posts: 2,859
Everybody saying 2/3... I am with you. However, let's take a second to appreciate the weird train of logic that makes 1/2 seem pretty reasonable as well.

A family tells you they have a girl.

Case 1: the girl was the firstborn
Sub-case 1.1: the secondborn was a boy
Sub-case 1.2: the secondborn was a girl

Case 2: the girl was the secondborn
Sub-case 1.1: the firstborn was a boy
Sub-case 1.2: the firstborn was a girl

At every fork it seems you are making a 50/50 choice. There's a 50/50 chance the girl was firstborn or secondborn, and a 50/50 chance that the other-born was either boy or girl.

Now, which one of you, the 2/3 people, can figure out where in that logic there is a mistake. I would assert it to be a challenging thing to effectively do, and it is very clear where (at least some of) the 1/2 people are coming from.

I can't say I can explain it elegantly, but "the firstborn girl has a second born sister" and "the second born girl has a firstborn sister" are actually the same situation, because you don't know which girl the parents were actually talking about. Even if they name her Debbie, you can't separate Debbie having a younger sister and Debbie having an older sister into two cases.

The above explanation is flimsy, but the train of logic that leads to the answer "2/3" is impervious to error. Three-quarters of two-child families have at least one girl. One-half of families have one boy one girl. These are observationally proven, verified facts. The one-half with a boy and a girl is fully a subset of the three-quarters which have at least one girl. If you know you have a family which definately belongs to the superset, it's got a 2/3 chance of belonging to the subset.

Thus, now it is simply a matter of finding a good reason that the above logic is wrong.

Now, if you meet a particular girl Debbie, then the cases of her having a younger sister or an older sister become distinct. Thus, the probability of a brother is one half.
#77
08-13-2006, 03:11 AM
 Uzi Charter Member Join Date: Jul 2000 Location: Hong Kong Posts: 4,041
Quote:
 Originally Posted by tim314 If you don't agree with the 2/3 number, please take the following quiz.
#78
08-13-2006, 03:18 AM
 garygnu Guest Join Date: Feb 2006
With two children, there are four options:

1) Older brother, younger brother
2) Older brother younger sister
3) Older sister, younger brother
4) Older sister, younger sister

By knowing the sex of one child, in this case a sister, one of the four options can be completely eliminated, in this case option 1).

In two out of those remaining three cases, a brother is present. Therefore, a sister with one and only one sibling is statistically more likely to have a brother than to have a sister by a ration of 66.67% to 33.33%.

Is there something wrong with my logic?
#79
08-13-2006, 03:19 AM
 garygnu Guest Join Date: Feb 2006
...besides my misspelling?
#80
08-13-2006, 05:57 AM
 Uzi Charter Member Join Date: Jul 2000 Location: Hong Kong Posts: 4,041
Quote:
 Originally Posted by garygnu Is there something wrong with my logic?
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.

If the sister is (O)lder you are left with:
OB YB
OB YS
OS YB
OS YS

Because the sister is older you can't have OB YS. It is impossible.

If the sister is (Y)ounger you are left with:
OB YB
OB YS
OS YB
OS YS

Because the sister is younger you can't have OS YB. Again, it is impossible.

And it doesn't matter whether you 'know' she is older or younger. She can only ever be older or younger.
#81
08-13-2006, 09:29 AM
 wissdok Guest Join Date: Aug 2005
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
08-13-2006, 10:06 AM
 tim314 Charter Member Join Date: Mar 2004 Posts: 3,602
Quote:
 Originally Posted by 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?
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 "[i]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
08-13-2006, 10:14 AM
 tim314 Charter Member Join Date: Mar 2004 Posts: 3,602
Quote:
 Originally Posted by 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.
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
08-13-2006, 10:18 AM
 tim314 Charter Member Join Date: Mar 2004 Posts: 3,602
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
08-13-2006, 10:50 AM
 Frylock Guest Join Date: Jun 2001
Quote:
 Originally Posted by 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.

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
08-13-2006, 11:22 AM
 wissdok Guest Join Date: Aug 2005
Quote:
 Posted by Tim314 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.
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.

Quote:
 Posted by Frylock 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.
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
08-13-2006, 12:00 PM
 Frylock Guest Join Date: Jun 2001
Quote:
 Originally Posted by 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.

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
08-13-2006, 12:14 PM
 tim314 Charter Member Join Date: Mar 2004 Posts: 3,602
Quote:
 Originally Posted by 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.
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
08-13-2006, 12:22 PM
 tim314 Charter Member Join Date: Mar 2004 Posts: 3,602
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
08-13-2006, 12:30 PM
 tim314 Charter Member Join Date: Mar 2004 Posts: 3,602
Can we all at least agree on this?

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
08-13-2006, 12:37 PM
 Larry Borgia Charter Member Join Date: Mar 2000 Location: Washington DC Posts: 8,207
Quote:
 Originally Posted by 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.
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
08-13-2006, 12:40 PM
 tim314 Charter Member Join Date: Mar 2004 Posts: 3,602
Quote:
 Originally Posted by tim314 "Julia is our daughter"
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
08-13-2006, 12:52 PM
 tim314 Charter Member Join Date: Mar 2004 Posts: 3,602
Quote:
 Originally Posted by wissdok 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.
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
08-13-2006, 12:57 PM
 tim314 Charter Member Join Date: Mar 2004 Posts: 3,602
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
08-13-2006, 02:00 PM
 wissdok Guest Join Date: Aug 2005
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

Quote:
 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.)
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
08-13-2006, 02:20 PM
 Stranger On A Train Guest Join Date: May 2003
Quote:
 Originally Posted by 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.
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).

Quote:
 Originally Posted by 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. 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.
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
08-13-2006, 04:08 PM
 wissdok Guest Join Date: Aug 2005
Quote:
 Posted by Stranger on a Train This is your 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 willfully ignoring the statement of the problem.
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
08-13-2006, 04:45 PM
 garygnu Guest Join Date: Feb 2006
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
08-13-2006, 05:44 PM
 Xema Guest Join Date: Mar 2002
Quote:
 Originally Posted by 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.
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
08-13-2006, 05:54 PM
 Xema Guest Join Date: Mar 2002
Quote:
 Originally Posted by 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.
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.

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