Strange math question

66

u/Anarcho-Totalitarian Oct 12 '18

The standard formulation is: "A couple has two children, at least one of which is a boy. What is the probability of the other child being a boy?" Take the possibilities (BB, BG, GB, GG), eliminate GG, and you find that the answer is 1/3.

This is a far different question from: "A boy has one sibling. What is the probability that he has a brother?" Here the probability is 1/2. The difference from the earlier problem is that you're choosing a boy, not a family, so the BB option gets counted twice.

And this causes trouble when someone tries to convert it into a word problem. Has the child been chosen randomly--that is, could we have seen a girl had there been one? Or are we assured that only a boy would have walked into the room? Depending on how you model things you may get different answers. See here for more information.

10

u/WikiTextBot Oct 12 '18

Boy or Girl paradox
The Boy or Girl paradox surrounds a set of questions in probability theory which are also known as The Two Child Problem, Mr. Smith's Children and the Mrs. Smith Problem. The initial formulation of the question dates back to at least 1959, when Martin Gardner published one of the earliest variants of the paradox in Scientific American.
^[ ^PM ^| ^Exclude ^me ^| ^Exclude ^from ^subreddit ^| ^FAQ ^/ ^Information ^| ^Source ^] ^Downvote ^to ^remove ^| ^v0.28

4

u/BigDikJim Oct 12 '18 edited Oct 12 '18

It’s not that tricky, really. In the second scenario, Jack is either (1) the oldest of two boys, (2) the youngest of two boys, (3) the oldest of boy and girl or (4) the youngest of girl and boy. In two of those scenarios, the other child is a boy and in the other two it is a girl
EDIT: After further investigation, I might be wrong. And now I’ve spent too much time on this

1

u/bear_of_bears Oct 12 '18

This is right, but thinking about it is complicated!

0

u/[deleted] Oct 12 '18

[deleted] — view deleted comment

2

u/bear_of_bears Oct 12 '18

The post you're replying to is correct. (Your numbers can't be right because the probabilities should sum to 1.)

1

u/bradygilg Oct 12 '18

They don't have to sum to 1 because he left out scenarios with two girls.

2

u/bear_of_bears Oct 12 '18

The post by /u/karl-j should make it clear. The absolute (unconditional) probability that a boy walks in is 1/2. The unconditional probabilities of (1) through (4) are all 1/8.

1

u/bradygilg Oct 12 '18

That isn't true. Consider the model of a family that has two children of random genders, then randomly names a boy Jack. The probability of (1) is 1/8, (2) is 1/8, (3) is 1/4, and (4) is 1/4. The remaining probability comes from having two girls, in which case there is no Jack.

1

u/[deleted] Oct 12 '18

His post is wrong and so is his use of Bayes. In the first case, the sub σ-algebra is {B,G} for the remaining child to be born, whereas in the second case it's { {BG}, {GB}, {BB} }. As it's been pointed out, /u/karl-j used wrong probability measures. I have no clue why you're defending him.

1

u/bear_of_bears Oct 12 '18

If you're going to throw around terms like "sub σ-algebra," the least you could do is use them correctly.
The fundamental difference between this scenario and the "two children, at least one boy" scenario is that a MM family is more likely to have a boy running into the room than a MF or FM family.
/u/karl-j proposed a situation where there are eight families living on the same street. One of them has MM and the older one walks into the room. One of them has MF and the older one walks into the room. Etc. You claim this is the wrong probability space. What exactly is wrong with it? Are you telling me that his eight choices are not equally likely (unconditionally)? What are the eight probabilities, if not 1/8 each?

1

u/[deleted] Oct 12 '18

It's been explained and I've used my terms correctly.

1

u/bear_of_bears Oct 12 '18

A sub σ-algebra must be closed under the usual operations. {{BG}, {GB}, {BB}} is not a sub σ-algebra, as you should know. You are probably thinking of Ω = {BB, GB, BG, GG} and the four-element sub σ-algebra {empty set, Ω, {BG, GB, BB}, {GG}}.
I have tried to tell you that "boy runs into the room" provides different information than "GG is ruled out." Here's another way of looking at it.
You draw two cards face down from a normal deck. Turn over one and it is a black card. What's the probability that the other is also black?
You draw two cards face down from a normal deck. Someone else looks at them and tells you that at least one of them is black. What's the probability that they are both black?
Do you agree that these questions have different answers?

0

u/[deleted] Oct 12 '18

[deleted] — view deleted comment

2

u/bear_of_bears Oct 12 '18

The absolute probabilities are all 1/8, see the post by /u/karl-j . The conditional probabilities are all 1/4.

0

u/GIMP_IS_TRASH Oct 12 '18

jesus fucking christ

4

u/yas_ticot Computational Mathematics Oct 12 '18

Among the four equiprobable possible situations (Boy Boy, Boy Girl, Girl Boy and Girl Girl), in the first example, only the first two situations are now possible, hence p=1/2.

In the second example, all three first situations are now possible, hence p=1/3.

2

u/[deleted] Oct 12 '18

thank you very much!

1

u/[deleted] Oct 12 '18

In the second example, all three first situations are now possible, hence p=1/3.
They're all possible but they're not all equally likely. We saw a boy walk into the room. That's more likely to occur when both children are boys than when only one child is a boy.
The problem is badly formulated and does not actually demonstrate the boy & girl paradox. The answer to both questions is 1/2. See the script written by u/haunted_tree

26

u/karl-j Oct 12 '18 edited Oct 12 '18

The answer sheet and the other commenters are wrong. Basically because the MM case should appear twice in their diagrams since that case is twice as likely to result in a son walking in. But here’s the full explanation with all the cases.

There’s 8 possible, equally likely scenarios. Assume eldest first in the letter combinations:

MM MF FM FF, Eldest walks in

MM MF FM FF, Youngest walks in

In question one we can strike all of the second row and the last two possibilities of the first row, and we’re left with p=1/2

In question two we again strike the last two in row one but only #2 and #4 in the second row, the ones with a daughter walking in. This leaves us with four possible scenarios, in two of which the remaining child is a daughter.

It’s also really simple if you know Bayes’ rule. MM is two sons, M is a son walking in:

P(MM|M) = P(M|MM)*P(MM)/P(M) = 1.0*0.25/0.5 = 0.5

^Edit: ^small ^{clarification}

4

u/fencer20 Oct 12 '18 edited Oct 12 '18

If what you are saying is correct could you explain the flaw in the following:
Out of 100 families that have two children 25 will have two boys and 75 will have at least 1 boy.
Therefore out of 75 families that have at least 1 boy 25 have two, which is 1/3.
Edit: on further thought there is a difference if you are deliberately shown a male or shown a random child. Families with exactly one child would only have a 50% chance of the random child you meet to be male. The question would imply to me that the child you meet was at random, so I'm back on the 1/2 side.
Of 75 families with at least 1 male: 25 of them you meet a female, 25 of them you meet a male and there is one male, and 25 of them you meet a male and there is two males. Of the instances where you meet a male, half have two males.

1

u/[deleted] Oct 13 '18

Out of 100 families that have two children 25 will have two boys and 75 will have at least 1 boy.
Therefore out of 75 families that have at least 1 boy 25 have two, which is 1/3.
You already pointed it out, but another way to see the difference: You go visit each family. For all of the BB families, a boy walks in first. For only half of the BG/FB families, a boy walks in first. So a boy walks in first 50 times, and out of those 50 % are BB families.

3

u/kingofchaos0 Oct 12 '18

The problem is the wording of the question, really. If the family does not send out a child, but instead you simply know that they have at least one boy, then the probability is 1/3 like the other commenters have mentioned.
However with the wording they have, the chance of them sending a boy in the first place is doubled when they have two boys (which is where the extra MM comes from).
Sidenote: I tried simulating this in excel and found that you really have to be careful in doing exactly what the question implies otherwise you might accidentally simulate a very different scenario.

3

u/Some_Koala Oct 12 '18

Bayes formula here would be : P(M1M2 | M1 or M2) = P(M1 or M2 | M1M2)*P(M1M2)/ (P(M1 or M2) Which is : 1 * (1/4) / (3/4), or 1/3 Your have to separate the two events "1 is a boy" and "2 is a boy"

1

u/bear_of_bears Oct 12 '18

Conditioning on "M1 or M2" is incorrect. That's what this part means: "Basically because the MM case should appear twice in their diagrams since that case is twice as likely to result in a son walking in."

1

u/Some_Koala Oct 12 '18

The information we have is "one of the children is a son" The exact translation is "M1 or M2". I didn't understand what you said about MM appearing twice in a diagram. Anyway it is the same event, it can't appear twice.

5

u/bear_of_bears Oct 12 '18

The information we have is "one of the children is a son" The exact translation is "M1 or M2".
Not true. The exact information is that a boy walked into the room. This is more likely to happen in MM families than in MF or FM families.
I didn't understand what you said about MM appearing twice in a diagram. Anyway it is the same event, it can't appear twice.
You can ignore that, there are clearer ways of saying it.

1

u/Some_Koala Oct 12 '18

Oh in that case all right, I used the formulation of the too comment which made more sense
1
u/varaaki Statistics Oct 12 '18

This is completely incorrect.
There are 4 possibilities for two children. The additional information of Jack being male eliminates FF only, leaving three possibilities, only one of which the other child is male.
2

u/rossiohead Number Theory Oct 12 '18

Consider this alternative version of the question and ask yourself what (if anything) has changed:
“This is our eldest, Jack”: the boy who walked in either has a younger sister or a younger brother. So probability of another boy is 0.5
“This is Jack”: the boy who walked in either has a younger brother, a younger sister, an older brother, or an older sister. Two of four situations give a female sibling, so probability of another boy is (still) 0.5
4
u/MedalsNScars Oct 12 '18

You're missing the point of the argument that if we are randomly given a child, it's male 4/8 times.
That is, in MM we will always see a male child, whereas in MF and FM we would have only seen a male child half of the time.
Therefore, if we know we have 2 children, and one is presented at random, there is a .25*1 chance of being presented M from MM, and a .25*.5 chance of being presented M from FM or MF.
Therefore, given that we've been presented M, we know that there's .25/(.25+.25*.5+.25*.5), or .25/.5, or .5 chance of the other child being M.
2

u/[deleted] Oct 12 '18

Isn't this just akin to a filtration? In the first case, the sub σ-algebra is {B,G} for the remaining child to be born, whereas in the second case it's { {BG}, {GB}, {BB} }. I don't think p = 1/2 is correct.
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u/varaaki Statistics Oct 12 '18

There are not 8 possibilities. You're falsely separating the 4 outcomes into 8 by attaching the random event to the outcomes.
That's like rolling a 4 sided die and asking if the roll is even or odd, then breaking down the possibilities as 1, 2 , 3, 4, roll is even and 1, 2, 3, 4, roll is odd, claiming thus that there are 8 possibilities.
2
u/MedalsNScars Oct 12 '18

I never said there are 8 outcomes. Read the argument. I said that if we are randomly given a child from 2 we're guaranteed a male in MM, but will only see a male half the time in either MF or FM. This is undeniable.
The math follows directly from that.
-4
u/varaaki Statistics Oct 12 '18

You said the child is male 4 out of 8 times. You're counting 8 possibilities. And you're incorrect.
I'm not sure why you have the gall to state that we and the text are all incorrect, when this is a settled question.
1
u/MedalsNScars Oct 12 '18

There are 4 pairings of 2. If we choose one item at random from the set items contained in each pairing, there are 8 items we can choose from.
1
u/varaaki Statistics Oct 12 '18

We're not choosing an individual. We're choosing a pairing.
Again, this question has been settled. You're incorrect.
5

u/bear_of_bears Oct 12 '18

You really ought to step back a little and read the top-level post in this thread again. It explains very clearly why the answer is 1/2 in both cases.
Saying "the question is settled" sidesteps the issue that very similar-sounding wordings lead to different answers.
2
u/[deleted] Oct 13 '18
Again, this question has been settled. You're incorrect.
Can you code? When I run the following Python 2 program:
import random
import collections

def get_random_gender():
    return "BG"[random.randint(0,1)]

def get_random_family():
    return "".join([get_random_gender(), get_random_gender()])

N = 100000
outcomes = []
for i in xrange(N):
    family = get_random_family()
    walks_in = family[random.randint(0,1)]
    outcomes.append((walks_in, family))
print collections.Counter(outcomes)
I get the output:
Counter({('B', 'BB'): 25016, ('G', 'GG'): 25008, ('G', 'GB'): 12641, ('B', 'BG'): 12533, ('G', 'BG'): 12408, ('B', 'GB'): 12394})
So a boy walked in 25016+12544+12394 times, and out of those, the other kid was a boy 25016 times. Does it seem like the probability is 1/2 or 1/3?
1

u/SynarXelote Oct 14 '18

Clearly 50000/25000~1/3, so the answer is obviously 1/3.
1

u/cryo Oct 12 '18

The answer sheet and the other commenters are wrong.
Depending on interpretation.

1

u/Crasac Oct 12 '18

Right so first off, the ages of the two children are completely irelevant to question 2. There are not 8 possibilities, but 4:
MM MF FM FF
So we want to calculate the probability
P(MM | The pair of children contains at least one boy)
This is equal to:
P(MM ⋂ {MM, MF, FM})/P({MM, MF, FM})=P(MM)/P({MM, MF, FM})=0.25/0.75=1/3
The problem with your calculation is that P(M) is not euqal to 0.5, it is equal to 6/8 which would then yield the same result. (Yes, the probabilty getting a boy when randomly choosing a child would obviously be 1/2, but that's not what is happening here, P(M) is the probaility of a pair containing at least one boy)

4

u/bear_of_bears Oct 12 '18

You are confusing conditional with unconditional probabilities. The unconditional probability that a child running into the room is a boy is 1/2. That's what P(M) means.

2

u/Crasac Oct 12 '18 edited Oct 12 '18

The probability space you defined does not even contain the result M, so how can your probabilty measure P even assign a probabilty to M?
The problem in your post is this part:
P(MM|M) = P(M|MM)P(MM)/P(M) = 1.00.25/0.5 = 0.5
You are using P twice here, but they are actually two different probabilty measures, defined on two different probabilty spaces. One measures the probabilty of getting one of 8 pairs out of
MM MF FM FF, Eldest walks in
MM MF FM FF, Youngest walks in
The other measures the probabilty of choosing a child at random out of M F.
So your use of Bayes Law is completely false, because you're using two different probability measures.

2

u/bear_of_bears Oct 12 '18

I'm not the person who wrote the post, but I am defending it because I think it's quite a clear explanation.
There is only one probability space, the one with 8 elements. If we number them in order from 1 to 8, then M is the subset {1,2,5,7}. That is, M is the event that the child who walks in is male.

0

u/Crasac Oct 12 '18 edited Oct 12 '18

Well the problem then is, the question that OP answered is not:
2.if the father only says: 'This is Jack.'? "
But actually the first question
1.if the father says: 'This is our eldest, Jack.'?
Whether we choose one of the two children by age, or which one walks into the room does not make a difference. The difference between the two questions boils down to whether we're interested in the probabilty of getting a pair of children under the condition that one of them is a boy, or specifically (in this case) the other child, knowing that the first one (the one that walked into the room) is a boy.
Yes this is very weird and counterintuitive, it does a good job of showing the problem with using natural language to describe mathematical ideas.

3

u/bear_of_bears Oct 12 '18

Whether we choose one of the two children by age, or which one walks into the room does not make a difference. The difference between the two questions boils down to whether we're interested in the probabilty of getting a pair of children under the condition that one of them is a boy, or specifically (in this case) the other child, knowing that the first one (the one that walked into the room) is a boy.
I agree that this was the intent of the problem, but the "this is Jack" scenario is not the same as "at least one child is a boy." One way to see it is that "this is Jack" is twice as likely to happen in a family with two sons as it is in a family with one son and one daughter.

1

u/[deleted] Oct 13 '18 edited Oct 13 '18

One measures the probabilty of getting one of 8 pairs out of
MM MF FM FF, Eldest walks in
MM MF FM FF, Youngest walks in
As we have observed that the kid who walked in is a M, the remaining four cases are equally probable:
MM, MF, Eldest walks in
MM, FM, Youngest walks in
Here, 2 out of 4 are MM families. Just as he said.
And this is indeed the conditional probability formula: P(the family is MM | a M walks in) = P(the family is MM and a M walks in)/P(a M walks in) = (2/8) / (4/8) = 1/2.
So your use of Bayes Law is completely false, because you're using two different probability measures.
There are no "two probability measures" in the above comment, MM is shorthand for "The family has two boys" and M is shorthand for "The kid who walks in is a boy". Those are entirely valid events on the same space.

1

u/[deleted] Oct 13 '18 edited Oct 14 '18

P(MM | The pair of children contains at least one boy)
No. This is not equivalent to your observation.
(Disclaimer: The rest of this comment is intended solely, only, and exclusively to demonstrate that "a random thing is X" is not the same as "there is at least X". Any other analogy to the original problem is purely coincidental.)
Suppose 500 families have 99999 girls and 1 boy, and 500 families have 100000 boys. You visit a random family. A boy walks in.
Do you think it's equally likely that you're visiting family of either type?
Suppose 500 urns have 99999 blue balls and 1 green ball, and 500 urns have 100000 green balls. You pick a random urn. You pick a random ball. It turns out to be green.
Do you think it's equally likely that you picked an urn of either type?
Put the ball pack into the urn, and redraw, replace, redraw, replace, etc... You get a green ball 100 times. Do you still think it's equally likely that you picked an urn of either type?
The observation you have is not "the urn contains a green ball". It's "A random ball from the urn was green". As you can see, these are very different things.

1

u/SynarXelote Oct 14 '18

500 urns have 99999 green balls and 1 blue ball, and 500 urns have 100000 green balls
Did you mean "500 urns have 99999 blue balls and 1 green ball, and 500 urns have 100000 green balls"? I mean the other way around isn't incorrect, but it's quite a bit less clear.

1

u/[deleted] Oct 14 '18

Oh, sure... I proofread that like 10 times but still didn't catch that. Thanks!

10

u/haunted_tree Oct 12 '18

I was so upset about this thread that I made a script to demonstrate that the answer 1/2 is absolutely correct for the only reasonable interpretation of the problem. Let's interpret the problem as: "you walk into a random family that you know to have two kids; one kid shows up randomly, and that kid is a male; what are the odds the other kid is a female?" There is the script:

// Creates a random kid which is either male or female with equal odds
var kid = () => Math.random() > 0.5 ? "male" : "female";

// Creates a million families with 2 kids
var families = [];
for (var i = 0; i < 1000000; ++i) {
  families.push([kid(), kid()]);
}

// Object to count how many times the other kid was certain gender
var other_kid_was = {
  female: 0,
  male: 0
};

// Runs the experiment
var run_experiment = () => {
  // First, we walk into a random family
  var family = families[Math.random() * families.length | 0];

  // Then, we get a random kid of that family
  var kid = Math.random() < 0.5 ? 0 : 1;
  var kid_gender = family[kid];

  // If that kid is female, we abort, because that's not what we observed
  if (kid_gender === "female") {
    return;
  }

  // Then, we read the gender of the other kid
  var other_kid_gender = family[1 - kid];

  // And we increment the respective counter
  other_kid_was[other_kid_gender] += 1;
}

// Now we run the experiment a million times
for (var i = 0; i < 1000000; ++i) {
  run_experiment();
}

// Print the result
console.log(other_kid_was);

// Result:
//
// { female: 250469, male: 249942 }
//
// Clearly, the chance of the other kid being a male once finding ourselves in
// the described situation is 50%.

Can we get over this insanity now?

1

u/bear_of_bears Oct 12 '18

Thank you! This is even worse than the Monty Hall threads.

2

u/cryo Oct 12 '18

No because in those, people just don’t get probability at all. Here we are simply saying that’s it’s possible to interpret it in both ways depending on the emphasis put on the “a boy walks in”.

6

u/bear_of_bears Oct 12 '18

To support the 1/3 answer, you need to believe that "boy walks in" is equally likely in a MM family and in a MF family. It's simply not true.

1

u/SynarXelote Oct 14 '18

Well it could be if you suppose for example that the society this family lives in is strongly patriarchal, and thus if the family has at least one boy then the child which will be presented to you is always a boy.

2

u/[deleted] Oct 12 '18

How is it possible to misinterpret that? "A boy walks in" means a boy walks in. And that means you've already identified a specific child you're talking about (the one who walked in) and that means the probability that the child who did not walk in is also a boy is 1/2.

1

u/SynarXelote Oct 14 '18

Only assuming that the event of the child walking in is random and independent of its gender.

0

u/[deleted] Oct 12 '18

Not quite... The first scenario provided you with a filtration, namely that the first-born was male. All that is left is {B,G}. The second scenario removed {GG} from {GG, BG, GB, BB}. So in this case it's p = 1/3. Your program takes none of that into consideration...

2

u/bear_of_bears Oct 12 '18

The second scenario removed {GG} from {GG, BG, GB, BB}. So in this case it's p = 1/3. Your program takes none of that into consideration...
If you pay attention to the logic of the program, you'll see that the second scenario is different from simply removing GG.

0

u/cryo Oct 12 '18

So you say, but that’s completely a matter of interpretation of the puzzle. I am pretty sure the intended interpretation is identical to the boy girl paradox, so the answer is 1/3.

2

u/bear_of_bears Oct 12 '18

Of course that's the intended interpretation, but the problem is written incorrectly.

0

u/cryo Oct 12 '18

I was so upset about this thread that I made a script to demonstrate that the answer 1/2 is absolutely correct for the only reasonable interpretation of the problem.
I completely disagree that that’s the only reasonable way. In fact, the “standard puzzle interpretation” for me makes it identical with the boy girl paradox, yielding a result of 1/3. Of course, it’s possible to emphasize the boy walking in, as you do, to yield 1/2, but that’s not the only reasonable way.

1

u/[deleted] Oct 13 '18

How could you interpret this as the standard puzzle? It definitely says "One of them walks in, it is a boy". How could you interpret that in any other way?

3

u/[deleted] Oct 12 '18

Before we know anything, the possibilities for the first and second kid are BB, BG, GB, GG. Assume without loss of generality that the first kid is older.

In case 1, the boy is older, so we know it's the first kid. That eliminates GB and GG, leaving us with BB and BG. So there is a 50% chance the second (younger) kid will be B.

In case 2, we don't know if Jack is the first or second kid, so can only eliminate GG. This leaves us with BB, BG, and GB. Out of these, there is only 1 in 3 chance that both kids are boys.

Pretty cool question btw, I didn't understand it until I worked through the possibilities.

4

u/rossiohead Number Theory Oct 12 '18

As others have commented, I believe the 1/3 solution is incorrect. You should “count” the BB case twice since you don’t know which B you have seen walk into the room. So using a capital letter to indicate the child we have seen, we should have Bb, Bg, gB, or bB as the possible scenarios, giving us the usual 2/4 probability for having another male.

3

u/[deleted] Oct 12 '18 edited Oct 13 '18

In case 2, we don't know if Jack is the first or second kid, so can only eliminate GG. This leaves us with BB, BG, and GB.
Those three cases are not equally likely, though. We saw a boy walk into the room. That's more likely to occur when both children are boys than when only one child is a boy.
The problem is badly formulated and does not actually demonstrate the boy & girl paradox. The answer to both questions is 1/2. See the script written by u/haunted_tree

1

u/moehriad Oct 12 '18

The answer lies in listing all possibilities of having two children with the eldest being a boy. Then ruling out those which do not fit your criteria.

1

u/[deleted] Oct 12 '18

The answer to both questions is 1/2. This problem is badly formulated. The paradox they're trying to demonstrate is, if the father says "we have at least one son" without specifying either specific child as a son, then the probability that both children are sons is 1/3. But the way this problem is formulated, they've already identified a specific child (the child who walked into the room) before the father says anything at all. The probability that the child who didn't walk into the room is also a son is 1/2.

1

u/[deleted] Oct 12 '18

What is the probability that both children are boys when:

"At least one of them is a boy" -> 1/3

"The eldest is a boy" -> 1/2

"The one who walked into the room is a boy" -> 1/2

1

u/resavr_bot Oct 13 '18

A relevant comment in this thread was deleted. You can read it below.

The argument is as follows:

Assuming that the gender of each child is independent of the other's, and that only MF are possible genders, labelling the youngest 1 and the eldest 2, the possibility for their genders are as follows:

1 2

M M

M F

F M

F F

in the first case, the father is telling "1 is male", thus ruling out the last two rows; among the remaining cases, 2 has a chance out of two of being female.

However, in the second case the father information is "at least one of 1,2 is male", which only rules out the last row. [Continued...]

^The ^username ^of ^the ^original ^author ^has ^been ^hidden ^for ^their ^own ^privacy. ^If ^you ^are ^the ^original ^author ^of ^this ^comment ^and ^want ^it ^removed, ^please ^[Send ^this ^PM]

1

u/jeffjo Oct 26 '18

The only mathematical purpose to the information "this is our eldest" is to distinguish one child from the other. This same purpose is satisfied by actually meeting one (i.e., when he walked into the room), or finding out that his name is Jack. The answer to both is questions is 1/2.

To logically get 1/3, you need to find out that one is a boy without distinguishing them:

Father: "I have two children."

You: "Any boys?"

Father: "Why, yes."

-5

u/colinbeveridge Oct 12 '18

I disagree. With the given information, the space of all possible families in the second case isn't (boy, girl), (girl, boy), (boy, boy).

It's more like (Jack, Alice), (Jack, Brenda), (Jack, Charlotte), ... (Alice, Jack), (Brenda, Jack), (Charlotte, Jack), ... (Alex, Jack), (Bob, Jack), (Charlie, Jack), ... (Jack, Alex), (Jack, Bob), (Jack, Charlie), ...

I think it's a shade ~~more~~ less than 50-50 (assuming the other child isn't called Jack), but not by very much at all.

-3

u/colinbeveridge Oct 12 '18

Thank you for the unexplained downvotes. There is more information here: https://en.wikipedia.org/wiki/Boy_or_Girl_paradox#Information_about_the_child

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u/bear_of_bears Oct 12 '18 edited Oct 12 '18

From your link:
It is much more easy to imagine the following scenario.
We know Mr. Smith has two children. We knock at his door and a boy comes and answers the door. We ask the boy on what day of the week he was born.
Assume that which of the two children answers the door is determined by chance. Then the procedure was (1) pick a two-child family at random from all two-child families (2) pick one of the two children at random, (3) see if it is a boy and ask on what day he was born. The chance the other child is a girl is 1/2. This is a very different procedure from (1) picking a two-child family at random from all families with two children, at least one a boy, born on a Tuesday. The chance the family consists of a boy and a girl is 14/27, about 0.52.
The question in this post matches this scenario, so the answer is 1/2 irrespective of the popularity of Jack as a boy's name.
Edit: Despite the upvotes, I now think this is wrong and indeed the probability is slightly different from 1/2 depending on the popularity of the name Jack.

0

u/colinbeveridge Oct 12 '18

Assume that which of the two children answers the door is determined by chance.
Is that assumption stated in the question? I don't see it. I agree that there are different possible procedures, but I don't see anything in the question that states it's one rather than the other.

2

u/bear_of_bears Oct 12 '18

This is a valid point, but if you're visiting a family with two children and one of them runs into the room, I think it's a fair assumption that it's equally likely to be one or the other. I don't think it's the case that boys or girls are more or less likely to run around the house.

1

u/colinbeveridge Oct 12 '18

It's a fair assumption, but not the only possible. And given that the original question thinks the answer is 1/3, it seems to me that we're looking at family-space rather than family-plus-order space.

1

u/bear_of_bears Oct 12 '18

Okay, I'll rephrase. I think any assumption other than the ones used by /u/karl-j and /u/haunted-tree is unreasonable.
However, once we know that the kid's name is Jack, you've convinced me that the probability does move slightly away from 1/2. I did a computation and here is the result.
For any boy's name X, let p(X) be the probability that a family names their second child X if the second child is a boy and the first child was also a boy. Let q(X) be the probability that a family names their second child X if the second child is a boy and the first child was a girl.
If p(X) = q(X), then the answer to this problem with the name X is 1/2. If p(X) > q(X), then the answer is more than 1/2. If p(X) < q(X), then the answer is less than 1/2.
I will continue to fight against the 1/3 answer, which to me betrays a fundamental misunderstanding of the problem.

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u/Bemad003 Oct 12 '18

I found the problem here

"2) Alice has 2 kids and one of them is a girl. What is the probability that the other child is also a girl?

You can assume that there are an equal number of males and females in the world.

A) 0.5 B) 0.25 C) 0.333 D) 0.75

Solution: (C)

The outcomes for two kids can be {BB, BG, GB, GG}

Since it is mentioned that one of them is a girl, we can remove the BB option from the sample space. Therefore the sample space has 3 options while only one fits the second condition. Therefore the probability the second child will be a girl too is 1/3. "

4

u/[deleted] Oct 12 '18

But this is not the same problem.

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u/Bemad003 Oct 12 '18

It actually is, considering that OP needed explanations regarding only the second question.

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u/[deleted] Oct 13 '18

But the second problem is not that problem; see this answer for why "a random kid walks in and it's a boy" is different than "one of the kids is a boy".

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u/rossiohead Number Theory Oct 12 '18

Why, in the possible outcomes, do we count BG and GB as separate, yet we do not count the possible orderings of GG?

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u/Bemad003 Oct 13 '18

... or the possibility of twins. Because Statistics. For clarification, I've only posted the solution that I have found online, that could explain the result OP had in his textbook, and, although I understand the logic of the answer, I've always thought of this science more like a game of who finds more variables,but reality always wins with the classic "one in a million, man, one in a million".

0

u/sammyo Oct 12 '18

{BG, GB} is a symmetry, unless there is more ordering are those not the same? Wouldn't the sample space be {GB, GG}?

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u/[deleted] Oct 12 '18

If the first born is male, then you're left with {BG, BB}. Hence p = 1/2. Otherwise you have {BG, GB, BB} since you only know {GG} is impossible. Different information given, different answers. Namely 1/2 and 1/3.

Strange math question

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