Boy Girl Paradox

  • Thread starter Howers
  • Start date
  • #26
82
0
two children. 100% chance of one boy. 50% chance of another boy, 50% chance of a girl. if one boy is a constant than that would just be like saying that it is raining outside all the time and their is a 50 50 chance i will carry an umbrella. its always raining so i will always have a 50% chance of carrying an umbrella much like their will be a 50% chance that there will be a girl. even case to case this should work therefore there is a 50% chance of a girl.
 
  • #27
194
0
Your probabilities don't add up right. There is a .25 chance of boy boy, and a .25 chance of girl girl, and a .5 chance of boy girl. There is twice the chance of getting boy girl, than boy boy. Eliminating the chance for girl girl does not change this ratio.
the probabilities do add up if you count your chances of a boy girl family or a girl boy family... theres a .25 chance of both... then a .25 chance boy boy and .25 girl girl
 
  • #28
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
685
shamrock, you are making a classical statistical error: You are assuming two events are equiprobable when in fact they are not. This is a very common mistake, and the "paradox" here hinges on people making this classical error. There is no paradox; the odds are indeed 2/3 that the other child is a girl.
 
  • #29
194
0
i think im just confusing myself here because i can see both yours and green zach's answers as being correct
 
  • #30
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
685
i think im just confusing myself here because i can see both yours and green zach's answers as being correct
Green Zach fell into the trap of assuming equiprobable events:
two children. 100% chance of one boy. 50% chance of another boy, 50% chance of a girl.
The assumption of equiprobability is just that, an assumption. Whenever one applies this assumption one should also do some sanity check regarding the validity of making this assumption. The assumption makes sense in the case of a coin toss or the roll of a single die. The assumption does not make sense and is not valid when one looks at the problem of the sum of the rolls of two dice. There are 11 outcomes, the numbers 2 to 12. However, the 11 events are not equiprobable. The probability of rolling a seven is six times that of rolling snake eyes. Similarly, the assumption simply is not valid in the problem at hand.
 
  • #31
101
0
Assumption of equal probability: the sun will rise tomorrow or it won't, therefore there is a 50% chance the sun won't rise...
 
  • #32
CompuChip
Science Advisor
Homework Helper
4,302
47
regor, that argument will work with anything.
If I throw a die, I will get 6 or I won't. Therefore, the probability of getting 6 is 50%.
I don't think I'd agree :)
 
  • #33
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
685
Exactly. Regor gave an example where the assumption of equiprobability is obviously invalid. The same applies to a the throw of a single die.
 
  • #34
194
0
Im pretty sure that is what he was implying
 
  • #35
82
0
ok i think i understand where i went wrong so tell me if my new idea makes sence. their is a 25% chance of getting girl girl, 25% chance of getting boy boy (it would be like getting lucky twice that's why its 0.25 but were all smart here) and 50% chance of getting boy girl (the most probible outcome... just like if you flip a coin 20 times, its most likeley that you will get around 10 on one side and 10 on the other.).

so now we have boy boy and girl girl at 0.25 and boy girl at 0.5. becusae the families must have at least one boy, girl girl is eliminated. now the numbers dont add up to 1.0 becsaue one of the variables was eliminated. Now the probibilities are out of three. 1/3 = 0.25 or 25%.

therefore becasue we are left with 0.25 boy boy and 0.5 girl boy, that translates into a 1/3 chance of getting boy boy and 2/3 chance of getting boy girl.

please tell me if this makes any sence lol
 
  • #36
194
0
that makes SENSE, but the easier way to think about it is that there are 4 selections, 25% chance of each... boy boy, girl boy, boy girl, and girl girl. you cant choose girl girl, so you are left with 3 choices. two of the three choices include a girl. 2/3

the way it is worded can lead you to a conclusion that it is 50% that is why i always hate probability... especially when i play my odds i always lose anyway so i tend to stay away from probability!
 
  • #37
346
0
Here is a very famous problem:

A random two-child family with at least one boy is chosen. What is the probability that it has a girl? An equivalent and perhaps clearer way of stating the problem is "Excluding the case of two girls, what is the probability that two random children are of different gender?"
This is how I think it is worded, firstly, a family with at least a boy is chosen, we should have factored in 75/25% with this selection, but this is wrong simply because the criteria is at least one boy, therefore it is not 75/25%. In fact there isn't any such probability simply because a boy MUST AT LEAST be chosen. Probability cannot be factored in. Only after asking of what probability is there a girl.

So it is 2/3.

one simple way of thinking of this is, there is an equal chance of girls as there are boys for any slot. There are 4 combinations, if only 3 out of the 4 is used to count, then the
probability of finding one girl is 2 out of 3 while for a boy it is 3 out of a 3.
 
Last edited:
  • #38
972
0
Here is a very famous problem:

A random two-child family with at least one boy is chosen. What is the probability that it has a girl? An equivalent and perhaps clearer way of stating the problem is "Excluding the case of two girls, what is the probability that two random children are of different gender?"
Let's look at this random two child family.

Child B: Boy
Child U: Unknown gender

The question is simply this, what is the probability that "Child U" is a girl?

A good estimate may rely upon population statistics. One way is to simply compare the birth rate of male children compared to the birth rate of female children in different countries. See http://www.photius.com/rankings/population/sex_ratio_at_birth_2007_0.html

The irony in the second form of the question is that if you assume "that (the) two random children are of different gender", "the case of two girls" is automatically excluded, and therefore, stating the problem as the following is redundant, "Excluding the case of two girls, what is the probability that two random children are of different gender?"

In there, the probability is clearly 2 out of 3 (ignoring the population data above), since 1 out of 4 of the possibilities is excluded. For example, the first child could be a boy and the second a girl. The second case is a first child who is female and a second child who is male. The third case is simply two boys.

The questions are obviously not equivalent. Yet possible arrangements are exactly the same. The paradox is well known, "There are lies, damned lies, and statistics."
 
Last edited:
  • #39
HallsofIvy
Science Advisor
Homework Helper
41,833
961
I see no paradox, only a possible ambiguity in what you mean by "Excluding the case of two girls".

If you mean exactly what you say, then the "possible arrangements" are NOT exactly teh same. For the first question, the possible arrangements are BB, GB, BG, GG. For the second, BB, GB, BG because we have been instructed to "Exclude the case of two girls".

When I saw the title to this thread, I thought you would be talking about the distinction between

"Of two children, at least one of them is a girl. What is the probability that the other child is a boy?" and
"Of two children, the older is a girl. What is the probability that the other child is a boy?"

It is easy to show that the first probability is 2/3 and that the second is 1/2. That is peculiar but still not a paradox. We have different probabilities based on different information.
 
  • #40
2
0
I think there is another more-or-less equivalent problem formulated in a more interesting way.

A king of a certain distant country has decided that he wants more men in the population for military purposes. He thus decides of a new law enforcing that a couple is allowed to have only one girl. What is the boy/girl ratio resulting in the population ? (assuming we wait for several generations, after the last person born before the law has deceased, to get a stable asymptotic result for instance).

A family could have say 7 boys, the eighth kid being a girl preventing any further child in the family.
Let girls in the military and be done with it. In a while, they'll be suing the government for equal rights, anyway.
 
  • #41
101
0
regor, that argument will work with anything.
If I throw a die, I will get 6 or I won't. Therefore, the probability of getting 6 is 50%.
I don't think I'd agree :)
I think we're saying the same thing...I hope ? I'm pointing out the fallacy of that argument...unless you really believe I believe there's only a 50% chance of the sun rising, in which case I need to build up my credibility
 
  • #42
111
15
A random two-child family with at least one boy is chosen. What is the probability that it has a girl?
Sorry to necro an old thread, but - as is usual with this "famous" problem - the exact wording is what makes it seem paradoxical, and that was overlooked. Compare these possible statements of the problem:
  1. A family is selected at random from all two-child families. An observer picks one of the children at random, and tells you that child is a boy.
  2. A family is selected at random from all two-child families that have at least one boy.
  3. A two-child family is known to have at least one boy.
  4. A family is selected at random from all two-child families. An observer looks at both children, and decides to tell you that at least one is a boy.
What is the probability that the family also has a girl? Assume any random child is a boy or a girl with probability 1/2, and that siblings' genders are independent. The answers are:
  1. 1/2. The unselected child's gender is independent of the selected child's gender.
  2. 2/3. 1/4 of all two-child families have two boys, 1/2 have a boy and a girl, and 1/4 have two girls. The two-girl families are eliminated from consideration, but all of the remaining ones are possible. P(BG)/[P(2B)+P(BG)] = (1/2)/[(1/4)+(1/2)]=2/3.
  3. It is ambiguous.
  4. It is still ambiguous, but less so. The best answer, 1/2, can be defended where no other can.
The third statement of the problem is ambiguous, because neither of the first two statements is inconsistent with it. How it became known that the family had at least one boy is not described, only the fact that it is known. You can't conclude that it was an a priori constraint, like the second problem, or an a posteriori observation of one child, like the first.

The last statement is also ambiguous. It is an a posteriori observation, but both children were observed. The ambiguity boils down to how the observer decided to tell you "at least one is a boy" when there was also a girl. If he choose at random, it is equivalent to statement #1. If he was constrained a priori to mention boys whenever possible, it is equivalent to statement #2. The reason I say 1/2 is the "best" answer (and many will disagree with me), is only because no part of the problem statement suggests such a bias. And you really can't assume it. The observer could equally likely have been constrained to mention girls whenever possible, and actually mentioned a boy only because there were no girls.

The problem is famous because not everybody recognizes the ambiguities, even while misinterpreting some of the clearer versions. Both answers have been given to most version, by intelligent and educated people. Some will take the minimalist approach, you only know about one child and you can't assume that both were observed. Others will take the maximalist approach, that the implication in "at least" means both children must have been used to create the condition.

The problem is old, originating at least as long ago as 1959, in Martin Gardner's Scientific American column. His was more or less a form of statement #3 (and he actually asked for a different probability - I'm translating all answers to this one). He answered "2/3," but later retracted that and said it was ambiguous. It reappears every few years, most notably in Marilyn vos Savant's Parade Magazine column several times from 1996 on. She has used both #2 and #3, but insists the answer can only be 2/3. It also appeared in Leonard Mlodinow's book about the public's misunderstanding of probability, The Drunkard's Walk, as #4. He also claimed the answer is 2/3. But there have been several papers written on it by specialists in probability. Most acknowledge the ambiguity, but refuse to take the bold step I do, in suggesting an answer for #4.
 
  • #43
324
0
Haha, this (the original problem) reminds me of the Monty Hall Problem. The answer is indeed 2/3.
 
  • #44
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
685
Compare these possible statements of the problem:
  1. A family is selected at random from all two-child families. An observer picks one of the children at random, and tells you that child is a boy.
  2. A family is selected at random from all two-child families that have at least one boy.
  3. A two-child family is known to have at least one boy.
  4. A family is selected at random from all two-child families. An observer looks at both children, and decides to tell you that at least one is a boy.
There is no ambiguity. In cases 1 and 4, if the randomly selected family is a two family, the observer has two choices: lie about one of the children being a girl, or pick another family. The implicit assumption here is that the report of a family having at least one boy is truthful. All bets are off of course if the report is untrustworthy.
 
  • #45
CompuChip
Science Advisor
Homework Helper
4,302
47
For 4, it does seem reasonable to add that the observer will not say that there is at least one boy if there is not. If you also add the requirement that he will tell you that whenever there is at least one boy, #4 is equivalent to #2 (i.e. you know precisely that the family that has been selected in #4 belongs to the possible families in #2).
 
  • #46
111
15
The implicit assumption here is that the report of a family having at least one boy is truthful. All bets are off of course if the report is untrustworthy.
Which is why you can't treat it as a lie (any more than you should consider the possibility the family has 5 children, or that the selection wasn't done at random.)
In cases 1 and 4, if the randomly selected family is a two (did you mean to say "two-girl" here?) family, the observer has two choices: lie about one of the children being a girl, or pick another family.
The "implicit assumption" you are making here is that the observer is required to say "at least one is a boy." There is no such requirement. The observer merely states a fact.

Case 1 is trivial to solve by a frequentist approach: There are four possible outcomes for the random selections, each with probability 1/4: the family has two boys and you pick a boy, the family has a boy and a girl and you pick the boy, the family has a boy and a girl and you pick the girl, or the family has two girls and you pick a girl. The last two cases, where you picked a girl, are eliminated by the observation (which can't be considered to be a lie), so the answer is (1/2)/(1/2+1/2)=1/2.

Since statements 3 and 4 could result from either 1 or 2, they are ambiguous.
 
Last edited:
  • #47
101
0
I don't see 1. being any different than the original question , with the answer being 2/3. You haven't collapsed the birth order question by indicating boy, much like MH. The 1/2 probab comes from the elemental probab of a women's giving birth, but how you build the experiment from there makes all the difference
 
  • #48
111
15
I don't see 1. being any different than the original question , with the answer being 2/3. You haven't collapsed the birth order question by indicating boy, much like MH. The 1/2 probab comes from the elemental probab of a women's giving birth, but how you build the experiment from there makes all the difference
And this is why there is always a controversy about these problems, no matter how they are asked. :)

First off, birth order is only important for counting the various possibilities. It sometimes is easier if you consider a smaller number cases with unequal probability, as I did, rather than a larger number with equal probability. The penalty is that you have to keep track of the probability associated with each case individually, rather than merely counting cases.

Here's the solution for problem #1, "collapsing" the birth order: There are eight possible cases, based on birth order and which child you picked: BBO, BBY, BGO, BGY, GBO, GBY, GGO, and GGY. Since you actually picked a boy, you can eliminate the four cases where a girl would have been picked; leaving BBO, BBY, BGO, and GBY. Of these four equally-likely cases, in two of them the family also has a girl. The probability is therefore 2/4=1/2. (This solution is a varaition of the one used for another famous "paradox," called either Bertrand's Box Paradox or Gardner's Three-Card Swindle. The only difference is that those have three equally-likely sets-of-two, where we have four.)

Essentially, my problem #1 picks a child at random. Along the way it also picked a family at random, but the information you got was only about the child (just as in Bertrand's Box, you picked a coin at random, and for the There-Card Swindle, you picked one side of a card). In particular, the observer may never even have seen the "other" child.

In my problem #2, the information is about the family. In particular, whatever observer gave you the information (1) had to look at both children in the family, and (2) had pre-decided to tell you that at least one child was a boy whenever that was true. (As CompuChip pointed out, there was a "requirement that he will tell you that [at least one is a boy] whenever there is at least one boy.")

In my #3 and #4, it is not clear whether the observer saw one or two children; or whether there was a "requirement that he will tell you that [at least one is a boy] whenever there is at least one boy." In fact, I made a point of leaving such a requirement out. And that is what makes the two cases ambiguous.

=====
Let me assure you that I choose the wordings of problems #1 and #2 very carefully, so that there could be no ambiguity other than by misperception. The answers are 1/2 and 2/3, respectively. You can easily write a computer simulation to verify them, if you are so inclined.

But we have not been trained properly to see what factors are important to establishing what the probability experiment is, and what is part of the smoke and mirrors. It hasn't helped that ambiguous versions of this problem have been presented, and answered as if they always had to be similar to #2 by people who should know better. That's why I pointed ou the one person, Martin Gardner, who recognized his mistake after making it and was honest enough to admit it.

What you need to do is figure out what the information you have is "connected to," and what possibilities exist that would allow you to get that information. In my problem #1, "boy" is connected with one individual. In my problem #2, it is connected with the most liberal observation of the family. In #3 and #4, it is not clear what it is connected to.
 
Last edited:

Related Threads on Boy Girl Paradox

  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
12
Views
6K
  • Last Post
Replies
5
Views
5K
  • Last Post
Replies
1
Views
620
Replies
3
Views
2K
  • Last Post
4
Replies
85
Views
26K
  • Last Post
2
Replies
45
Views
6K
Replies
27
Views
4K
Replies
4
Views
3K
  • Last Post
Replies
10
Views
6K
Top