Another math problem

[Blefuscu]

i agree. because of the way the question was asked, the answer is 50%.

 

[DrPizza]

Originally posted by: jman19

Originally posted by: DrPizza
Lumbus, I'll try to keep it simple.

First of all, do you understand:

I have 3 cards,
1 black both sides
1 black one side, red one side
1 red both sides

I pull a card at random and only look at one side. I see black. What's the probability that the other side is black? Answer: 2/3

It came from a previous post. Here, allow me to complete it for you:
I have 3 cards,
1 black both sides
1 black on side, red one side
1 red both sides.

HOW you get the information about a card is relevant:
A. If I tell you that at least one side of my card has black, then there's a 50% chance that the other side is black. (It's one of two cards.)
B. If you draw a card at random and only look at one side - and that side is black, then there is a 66 2/3% chance that the other side is black.

How the OP stated the problem matches with situation (B) above.

 

[Blefuscu]

Originally posted by: DrPizza

Originally posted by: jman19

Originally posted by: DrPizza
Lumbus, I'll try to keep it simple.

First of all, do you understand:

I have 3 cards,
1 black both sides
1 black one side, red one side
1 red both sides

I pull a card at random and only look at one side. I see black. What's the probability that the other side is black? Answer: 2/3

It came from a previous post. Here, allow me to complete it for you:
I have 3 cards,
1 black both sides
1 black on side, red one side
1 red both sides.

HOW you get the information about a card is relevant:
A. If I tell you that at least one side of my card has black, then there's a 50% chance that the other side is black. (It's one of two cards.)
B. If you draw a card at random and only look at one side - and that side is black, then there is a 66 2/3% chance that the other side is black.

How the OP stated the problem matches with situation (B) above.

Hardly. There is no information whatsoever given about the third child. Another way to think of it is: goto every house in the world where there are 3 children, at least 2 of whom are boys. The third kid is a girl 50% of the time.

The question is not, "what percent of three kid families have 2 boys and a girl". That is a completely different question. If you cannot tell the difference, then you need to take a statistics course.

"What is the probability that the child playing in the back yard is a girl?" is a different question than "What percent of three kid families have 2 boys and a girl?" or "What is the probability that a 3 kid family has one girl?"

If you mean to ask the question where the answer is 50%, the OP's question is worded appropriately. If you mean to ask any other question then there are much clearer ways to pose the question. From the way it is posed, the question seems to be about the child, not the family. If you read it and think the question is about the family, well then we are not arguing about statistics anymore.

 

[TheNewGuy]

It's a girl simply by the father's response...he would have said "two of three boys/sons" or something to that effect :laugh:

IMHO, the "only" thing we know is that there's a 50/50 chance of the mother having a boy or girl, but we don't know whether the father is prone to having more boys or not...

Dave

 

[Mark R]

I'm firmly in the 50% camp.

Essentially, whether it is 75% or 50% depends on whether you are more likely to have met the 2 boys (rather than, say 2 girls or a girl and a boy)

If you were in a situation where you were only going to meet boys - maybe this was some father/boy scout trip - then the probability is 75%.
If however, you just happen to notice a guy and 2 boys in the street, then you could just as easily have met 2 girls - the probability is 50%, for reasons that DrPizza, among others has shown.

The way the question is stated implies the latter.

More formally:
Let P(A) = the probability that you meet 2 Boys and that the 3rd Child is a girl
Let P(B) = the probability that you meet 2 Boys

The question is asking P(A|B), i.e. the probability that the 3rd child is a girl given that you have met 2 boys

If we meet boys or girls at random, there are 24 possible outcomes. 8 Different families of 3 children, each with 3 possibilities for the unknown child. There are 3 out of 24 ways of meeting 2 boys, where the 3rd child is a girl. Therefore P(A) = 1/8.
There are 6 out of 24 ways of meeting 2 boys, P(B) = 1/4.

Applying Bayes' theorem yields P(A|B) = 50%.

 

[DrPizza]

Originally posted by: Blefuscu
Hardly. There is no information whatsoever given about the third child. Another way to think of it is: goto every house in the world where there are 3 children, at least 2 of whom are boys. The third kid is a girl 50% of the time.

The question is not, "what percent of three kid families have 2 boys and a girl". That is a completely different question. If you cannot tell the difference, then you need to take a statistics course.

"What is the probability that the child playing in the back yard is a girl?" is a different question than "What percent of three kid families have 2 boys and a girl?" or "What is the probability that a 3 kid family has one girl?"

If you mean to ask the question where the answer is 50%, the OP's question is worded appropriately. If you mean to ask any other question then there are much clearer ways to pose the question. From the way it is posed, the question seems to be about the child, not the family. If you read it and think the question is about the family, well then we are not arguing about statistics anymore.

You apparently haven't read the thread or the arguments inside the thread on either side. First of all, the OP intended the answer to not be 50%. Second of all, if you go to every house in the world where there are 3 children, at least 2 of whom are boys, you'll find that the third kid is *not* a girl 50% of the time. The third child will actually be a girl 2/3's of the time, and a boy 1/3 of the time.

 

[Born2bwire]

Originally posted by: DrPizza

Originally posted by: Blefuscu
Hardly. There is no information whatsoever given about the third child. Another way to think of it is: goto every house in the world where there are 3 children, at least 2 of whom are boys. The third kid is a girl 50% of the time.

The question is not, "what percent of three kid families have 2 boys and a girl". That is a completely different question. If you cannot tell the difference, then you need to take a statistics course.

"What is the probability that the child playing in the back yard is a girl?" is a different question than "What percent of three kid families have 2 boys and a girl?" or "What is the probability that a 3 kid family has one girl?"

If you mean to ask the question where the answer is 50%, the OP's question is worded appropriately. If you mean to ask any other question then there are much clearer ways to pose the question. From the way it is posed, the question seems to be about the child, not the family. If you read it and think the question is about the family, well then we are not arguing about statistics anymore.

You apparently haven't read the thread or the arguments inside the thread on either side. First of all, the OP intended the answer to not be 50%. Second of all, if you go to every house in the world where there are 3 children, at least 2 of whom are boys, you'll find that the third kid is *not* a girl 50% of the time. The third child will actually be a girl 2/3's of the time, and a boy 1/3 of the time.

/me reads custom titles.

Are we... are we still allowed to argue with you now?

 

[Fourier Transform]

With 3 children, you have 8 possibilities:
BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG

Probability (3rd child is a girl | 2 are boys) = Probability(3rd child is a girl AND 2 are boys)/Probability (First 2 are boys).

Probability (3rd child is a girl AND First two are boys) = 3/8
Probability (First 2 are boys) = 1/2

-> Probability (3rd child is a girl | 2 are boys) = (3/8) / (1/2) = 3/4 = 75%

 

[DrPizza]

Originally posted by: Born2bwire
/me reads custom titles.

Are we... are we still allowed to argue with you now?

Sure.

 

[Nathelion]

Originally posted by: Fourier Transform
With 3 children, you have 8 possibilities:
BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG

Probability (3rd child is a girl | 2 are boys) = Probability(3rd child is a girl AND 2 are boys)/Probability (First 2 are boys).

Probability (3rd child is a girl AND First two are boys) = 3/8
Probability (First 2 are boys) = 1/2

-> Probability (3rd child is a girl | 2 are boys) = (3/8) / (1/2) = 3/4 = 75%

*sigh*

 

[hypn0tik]

Originally posted by: Nathelion

Originally posted by: Fourier Transform
With 3 children, you have 8 possibilities:
BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG

Probability (3rd child is a girl | 2 are boys) = Probability(3rd child is a girl AND 2 are boys)/Probability (First 2 are boys).

Probability (3rd child is a girl AND First two are boys) = 3/8
Probability (First 2 are boys) = 1/2

-> Probability (3rd child is a girl | 2 are boys) = (3/8) / (1/2) = 3/4 = 75%

*sigh*

?

 

[Nathelion]

I actually only read the last line of that. Now that I read all of it I can't even figure out what it's supposed to say, so I suppose I'll take that sigh back. Sort of? I'm confused.

 

[Matt1970]

The way you asked the question, "What is the probability that the child playing in the back yard is a girl?" is still 50%. You are basicly flipping a quarter. The quarter has no memory of what the first two flips are. You still have the same odds as you did on the first flip.

 

[f95toli]

Originally posted by: Matt1970
The way you asked the question, "What is the probability that the child playing in the back yard is a girl?" is still 50%. You are basicly flipping a quarter. The quarter has no memory of what the first two flips are. You still have the same odds as you did on the first flip.

No, read the question again. And the same is true for quarters, if you flip a coin three times and then tells me that that you have two heads I will know that the probability of the last coin being a head is 75% since there are more ways of getting two heads and one tail than there are ways to get 3 heads.
However, if you only flip 2 coins and then asks me "What is the probability of the next flip being a head" the answer is of course 50%, since the flips are independent.

If you don't belive me you can try runing the Matlab program I posted above. It uses a random number generator to "flip a coin" many times.

The whole point of this problem is that there are more families with 2 boys and 1 girl than families with 3 boys. It should be possible to find statistics to confirm this.

 

[JKing76]

I'll admit it -- my initial answer was 50% and I thought the people saying 75% were being stupid. Then I did the math... I'll break my work into sections, and if you're still in the 50%-camp, tell me which step you disagree with:


1) What are the different combinations of children?

Can write this out like a binary sequence, with B = 0; G = 1

B B B
B B G
B G B
B G G
G B B
G B G
G G B
G G G

8 different possibilities; which is correct: 2^3 = 8

2) What is the question asking?

A man with three children introduces two of them at random. Nothing is said about if they are the eldest two, the youngest two, or the one eldest and one youngest. The question is asking: What is the probability that the unintroduced child is a girl?

2a) What is the question not asking?

The question is not asking: What is the probablity that the third child is a girl? If it has asked specifically about the third child, it would be reasonable to assume that the use of the ordinal "third" meant the question was asking about the last child born. If the question were asking about a specific "position" like first, second, or third, then the probablity of that child being female is 50%. But that is not what the question is asking.

3) How do we solve what the question is asking?

We go back to the list of all posibilities from #1. All we know is that the man has at least two sons; but again, we don't know if they're the two eldest children, or the two youngest, or the one eldest and one youngest. So all we can narrow down is to select any of the posibilities that contain at least two boys:

B B B
B B G
B G B
B G G Remove -- Not enough boys
G B B
G B G Remove -- Not enough boys
G G B Remove -- Not enough boys
G G G Remove -- Not enough boys

4) We know have to look just at the posibilities that remain:

B B B
B B G
B G B
G B B

Three of the four posibilities have the unintroduced child as a girl. Therefore the probablity that the child outside is a girl is 75%.

 

[DrPizza]

your step 2 is where your error occurs. The probability of meeting two boys is different for the 4 different families. To understand better, look at the simplified case I've posted above regarding red and black sided cards.

3 cards:
1 black both sides
1 black one side, red one side
1 red both sides

Being told that at least 1 of the two sides is black results in a 50% probability that the other side is black. Randomly being shown one side of a card, and seeing black actually results in a 2/3 chance that the other side is black. Reason: You know it's one of two cards. However, there are 3 black sides that you can possibly be looking at. 2 of those have black on the other side, 1 has red on the other side.

Get it?

Thus, to relate this to the original problem, the black black card is more likely to be seen with a black side up than the black/red card. Likewise, the bbb family is more likely to be met at random with two boys showing than the bbg, bgb, or gbb families.
While we'd like to simplify this to say that there are 4 possibilities, we can't, because the event of meeting 2 boys has a different probability in 3 of those possibilities.

Likewise, while we have 2 possible cards that it could be when we see a black side, we can't conclude the probability is 50% when we are shown a black side - because the card with two black sides has a higher probability of showing a black side than the card with a black and a red side.

 

[JKing76]

I think I see. So, given that the family with three kids has at least two boys, we have our same possiblities:

B B B
B B G
B G B
G B B

But then we need to expand all the possible ways the children could be introduced. So first we update the labels so we can distinguish the different kids:

B1 B2 B3
B1 B2 G3
B1 G2 B3
G1 B2 B3

So, all the possiblities for introductions are (with unintroduced child in parens):

B1 B2 (B3)
B1 B3 (B2)
B2 B3 (B1)

B1 B2 (G3)
B1 G3 (B2)
B2 G3 (B1)

B1 G2 (B3)
B1 B3 (G2)
G2 B3 (B1)

G1 B2 (B3)
G1 B3 (B2)
B2 B3 (G1)

Then, knowing that we were introduced to two boys, we eliminate the options that include a girl in the introductions:

B1 B2 (B3)
B1 B3 (B2)
B2 B3 (B1)

B1 B2 (G3)

B1 B3 (G2)

B2 B3 (G1)

Thus, six possiblities, three where the missing child is a boy, three where it's a girl.

And the answer to the question is that there is a 50% chance that the child in the back yard is a girl.

 

[JKing76]

I guess the trick was analyzing the question. The question is not "given a family with three children, two of which are boys, what is the chance that the other child is a girl?".

The question is: Given a family with three children, if randomly selecting two of the children yields two boys, what is the chance that the other child is a girl?". Subtley different.

Computer simulation backs this up. My algorithm:

1. Initialize the families:

Families = Array of dimension [1,000,000][3]

Iterate i on the 1,000,000 dimension:
Iterate j on the 3 dimension:
Familes[ i ][j] = random(2); // 0 = boy; 1 = girl

2. Calculate probability:

Iterate i on the 1,000,000 dimension
family = families[ i ]
if (family[0] + family[1] + family[2]) <= 1 // no more than 1 girl
introduce = random(3)

if (introduce == 0) introduced1 = family[0], introduced2 = family[1], play = family[2]
if (introduce == 1) introduced1 = family[0], introduced2 = family[2], play = family[1]
if (introduce == 2) introduced1 = family[1], introduced2 = family[2], play = family[0]

if (introduced1 + introduced2 == 0): // both boys

if (play == 0), increment bbb
if (play == 1), increment bbg

After 1,000,000th iteration:

Chance girl = bbg / (bbg + bbb)

Running this shows the chance that the child in the back yard is a girl hovers around 0.5

 

[Jinsterrr]

I cannot believe this question has gotten so "controversial". The answer is clearly 50%.

Probability questions can get tricky when the conditions are introduced, sometimes falsely, as is in this case. The best way to analyze this, is to go back to the principles.

The question already states that the chance of getting a girl/boy is 50:50. There is marginal reasons, medically, why this probability might be slightly off but the question disregards that.

People in the conditional probability camp (the "75%" camp) makes the fundamental error that the probability of the third child's gender (whatever order of birth), has anything to do with the genders of the other two children. The fact that the questions tells you the genders of the two children is a red herring. Because the genders of all three children are independent, you can substitute what the question tells you with any other independent event and the mathematical outcome should not change. The question might as well be: What is the chance of an unborn baby being male given that GWB didn't find any WOMD in Iraq and Homer Simpson had diarrhoea today? The answer is 50%, because the three events are independent.

If one insists on using the conditional probability equation to solve this problem, it can be done. It is then simply 1*1/2 divided by 1 = still 50%. The numerator is the chance of all three being male given two are already definitely male, and the denominator is the chance of what is already given, that the two children are definitely male.

Using the conditional probability formula any other way either ignores the indepedence of the genders of the three children or ignores the definitive fact that two children are male.

JKing, if you must use event space to solve this question, your event space should be defined as this:

The father introduced two male children, we don't know which two, so, the event space is:
BB?
B?B
?BB

And in each case, the ? can be either B or G. Therefore, the answer is 3/6 = 50%.

Defining your event space as BBB, BBG, BGB, GBB doesn't give BBB enough probability weight because BBB is worth 3 events when the order of introduction is unknown.

 

[JKing76]

Jinsterrr, your reasoning is incorrect. You cannot simply say "Since the chance of getting a boy/girl is 50% the answer is 50%" unless the question is asking "What is the chance that the first child born is a girl", "What is the chance that the second child born is a girl?", or "What is the chance that the third child born is a girl?". So you are misinterpreting the question.

However, the people in the 75% camp are also misinterpreting the question. If the question asked "Given a family with three children, at least two of which are boys, what is the chance that the other child is a girl?" then they would be correct. That's when you just look at the possible gender orders:

B B B
B B G
B G B
G B B

In which case there is a 75% chance the other child is a girl.

But again, that's not what the question is really asking. As I said, the question really is: "Given a family with three children, if randomly selecting two of the children yields two boys, what is the chance that the other child is a girl?"

 

[Jinsterrr]

JKing, if you must use event space to solve this question, your event space should be defined as this:

The father introduced two male children, we don't know which two, so, the event space is:
BB?
B?B
?BB

And in each case, the ? can be either B or G. Therefore, the answer is 3/6 = 50%.

Defining your event space as BBB, BBG, BGB, GBB doesn't give BBB enough probability weight because BBB is worth 3 events when the order of introduction is unknown.

This question has been over complicated by analysis. Go back to fundamentals. The third child has a 50% chance of being female as per the original question.

If you must complicate the question by factoring in the order of the gender, then it is still 50% as I illustrated with the correct event space above.

 

[Jinsterrr]

JKing, if you must use event space to solve this question, your event space should be defined as this:

The father introduced two male children, we don't know which two, so, the event space is:
BB?
B?B
?BB

And in each case, the ? can be either B or G. Therefore, the answer is 3/6 = 50%.

Defining your event space as BBB, BBG, BGB, GBB doesn't give BBB enough probability weight because BBB is worth 3 events when the order of introduction is unknown.

This question has been over complicated by analysis. Go back to fundamentals. The third child has a 50% chance of being female as per the original question.

If you must complicate the question by factoring in the order of the gender, then it is still 50% as I illustrated with the correct event space above.

 

[JKing76]

Yes, I know. If you will look at my two posts before yours, I explained why the 75%-camp is wrong. 50% is correct, just not for the reasoning you first posted. The expanded state is the correct analysis.

 

[Jinsterrr]

At the risk of starting another argument in this already lengthy discussion, JKing, there is in fact no difference in saying "given a family with three children, two of which are boys, what is the chance that the other child is a girl?" or saying "Given a family with three children, if randomly selecting two of the children yields two boys, what is the chance that the other child is a girl?"

The first question implies the two children selected are boys, and selection implies randomness anyway. There is no subtle difference there. Saying "randomly selecting two of the children yields two boys" merely leads one to solve this question by the method of defining the event space as you did. One could just as easily solve the question by ignoring order of selection altogether because order doesn't matter when the events are indepedent.

Anyway, I am glad your computer program proves the 50% answer.

 

[lousydood]

Defining your event space as BBB, BBG, BGB, GBB doesn't give BBB enough probability weight because BBB is worth 3 events when the order of introduction is unknown.

This statement is incorrect, and explains why you are confused.

Please read http://mathworld.wolfram.com/BinomialCoefficient.html