Another math problem

[JKing76]

Jinsterrr, there is a difference. For the first question, the probability of the missing child being a girl is 75%. Because here there is no expanded state. There are simply four choices: BBB, BBG, BGB, or GBB. Three of the four include one girl, so the chance is 75%. The simulation also demonstrates this.

But when you say that a 3C2 yielded two boys, you have more information. BBB is guaranteed to result in two boys. BBG, BGB, or GBB are have a 2/3 chance of introducing one boy and one girl. Just as DrPizza stated, this is when you look at the extended state.

 

[SinNisTeR]

I'm in the 50% boat.. The wording on the problem clearly states: "there is always a 50% chance that a woman giving birth will have a boy and a 50% chance of her having a girl"

Whoever said previously that these events are independent is right. The probability of having a girl or boy on each subsequent birth is completely unrelated to the previous birth, just as coin flips are unrelated on each flip. The likelyhood on the otherhand that he has 100 boys is very slim...

edit: Confirmed with a math genius.. it's definitely 50% because of independent events

 

[lousydood]

I'm in the 50% boat

Sigh, you contradict yourself:

the likelyhood on the otherhand that he has 100 boys is very slim...

The likelihood of having "3 boys" vs having "2 boys and 1 girl" is also slimmer. It's 75% chance against.

 

[BBock727]

The answer is DEFINATELY 50%. The first 2 children have no outcome on the 3rd.

This isn't a card game we are dealing with. There is a difference if you lined up three cards and told someone to tell you the probability of the each card either being black or red. Obviously if you use a controlled deck of cards, the probability for the 2nd and 3rd card will change based upon the cards before it.

Here the question clearly states that there is always a 50% chance of boy and girl.

 

[SinNisTeR]

Originally posted by: lousydood

I'm in the 50% boat

Sigh, you contradict yourself:

the likelyhood on the otherhand that he has 100 boys is very slim...

The likelihood of having "3 boys" vs having "2 boys and 1 girl" is also slimmer. It's 75% chance against.

Sigh, you're dilusional.. Likelyhood is something completely different than outcome.. What's the likelyhood that you'll flip a coin 100 time and get all heads?

 

[f95toli]

Originally posted by: BBock727
The answer is DEFINATELY 50%. The first 2 children have no outcome on the 3rd.

This isn't a card game we are dealing with. There is a difference if you lined up three cards and told someone to tell you the probability of the each card either being black or red. Obviously if you use a controlled deck of cards, the probability for the 2nd and 3rd card will change based upon the cards before it.

Here the question clearly states that there is always a 50% chance of boy and girl.

The point you are missing is that we are not talking about the probability of a SINGLE event (boy OR girl), but of the probability of an ENSEMBLE of event (with 3 elements).
Hence, the probability of e.g. the YOUNGEST kid in the family being a girl(single event) is of course 50%, but that does not change the fact that the probability of running into family with 3 boys (ensemble with 3 events) is lower(25%) than meeting one with 2 boys and one girl (75%).

I am pretty sure it is possible to find statistics to confirm this, if you pick 100 families with 3 kids where 2 are boys at random, you should find that in 75 families the third kid is a girl (but again, the probability of the youngest, oldest or middle kid being a girl is of course 50%).

 

[SinNisTeR]

Originally posted by: f95toli

Originally posted by: BBock727
The answer is DEFINATELY 50%. The first 2 children have no outcome on the 3rd.

This isn't a card game we are dealing with. There is a difference if you lined up three cards and told someone to tell you the probability of the each card either being black or red. Obviously if you use a controlled deck of cards, the probability for the 2nd and 3rd card will change based upon the cards before it.

Here the question clearly states that there is always a 50% chance of boy and girl.

The point you are missing is that we are not talking about the probability of a SINGLE event (boy OR girl), but of the probability of an ENSEMBLE of event (with 3 elements).
Hence, the probability of e.g. the YOUNGEST kid in the family being a girl(single event) is of course 50%, but that does not change the fact that the probability of running into family with 3 boys (ensemble with 3 events) is lower(25%) than meeting one with 2 boys and one girl (75%).

I am pretty sure it is possible to find statistics to confirm this, if you pick 100 families with 3 kids where 2 are boys at random, you should find that in 75 families the third kid is a girl (but again, the probability of the youngest, oldest or middle kid being a girl is of course 50%).

I disagree.. The probability of the third child being boy or girl is independent. Although unlikely that it'll happen. Think of the coin flip problem as this is identical. What's the probability that the third flip will result in tails when the first two were heads? The probability is still 50/50. The likelyhood that you get HHT is 1/8 since there are 8 possibilities, not considering permutations. Permutations shouldn't be considered since order in which each of the children was born is being considered.

HHH
HHT --> is one out of the eight.
HTH
HTT
THH
TTH
THT
TTT

 

[f95toli]

Originally posted by: SinNisTeR
Permutations shouldn't be considered since order in which each of the children was born is being considered.

No, thats the point. The order DOES NOT matter, hence permutations SHOULD be considered.
If we were told that the YOUNGEST (or oldest or middle) child was in the garden, the answer would have been 50%, but we are told that A child is the garden; we don't know which one.

And as I have already pointed out the answer would be 75% even if we were dealing with coins: the probability to get three heads is smaller than the probability to get two heads and one tails as long as the order is NOT taken into account (if the order DOES matter the probabilities of all outcomes are the same).

 

[SinNisTeR]

Originally posted by: f95toli

Originally posted by: SinNisTeR
Permutations shouldn't be considered since order in which each of the children was born is being considered.

No, thats the point. The order DOES NOT matter, hence permutations SHOULD be considered.
If we were told that the YOUNGEST (or oldest or middle) child was in the garden, the answer would have been 50%, but we are told that A child is the garden; we don't know which one.

And as I have already pointed out the answer would be 75% even if we were dealing with coins: the probability to get three heads is smaller than the probability to get two heads and one tails as long as the order is NOT taken into account (if the order DOES matter the probabilities of all outcomes are the same).

Actually, it's irrelevant. The problem asks what the probability is for the third child being a girl. The probability is 50%. What is the probability of having tails on the 3rd flip of a coin.. the answer is obvious.
Coin Flips

On the other hand, the probability of having two boys and one girl is a different question.

Probability of results


FTW...

 

[f95toli]

No, read the question again. The problem asks the question "What is the probability that the third kid is a girld GIVEN that we know that there are 3 kids in the family and two of those are boys". Hence, it is exactly the same problem as asking what the probablity is of having two boys and one girl.

And you are missing the point with the coins. The probability of the ONE of the flips being heads is of course 50% (all flips are independent, the outcome of one attempt does NOT affect the probabilty of the next), but the probabiliy of getting two heads and one tails (in whatever order) is LARGER than the probability of getting three heads (which can only happen in one way), simply because there are more possible series of flips that give that result.

 

[SinNisTeR]

Ahem.. "What is the probability that the child playing in the back yard is a girl?"
The third child coming out of that mother's has a 50/50 chance in being a girl..

Nowhere in that sentence can you infer that this is a conditional probability... There is no "given" anywhere. Hence, this is why it is 50%; end of discussion.

I'm not disputing what you said in your second paragraph. I absolutely concur. Lets just agree on what the question is asking, not what became of it after so many interpreted it incorrectly.

 

[xtknight]

Here's how I see it:

Originally posted by: Nathelion
For the purposes of this discussion, there is always a 50% chance that a woman giving birth will have a boy and a 50% chance of her having a girl (this is not true in reality).

Known: 50% chance of boy, 50% chance of girl

Imagine that you are going to a party. You arrive, and a man and two boys greet you. The man says "These two boys are my sons, but I have three children. The third child is in the back yard playing." You can't see the back yard from where you are standing. You have no prior knowledge of this man's family.

The other kids he has aren't relevant.

What is the probability that the child playing in the back yard is a girl?

The fact that it's his third child isn't relevant. The fact that it is a child is relevant, and we know that there is a 50% chance of a child being a girl from the known statement above.

 

[DrPizza]

If you look back, you'll see my explanation involved a sample space... I went out of my way to demonstrate that the probability, based on the way the problem was stated, was indeed 50%

Those who are saying that the 3rd child is an independent event are correct, based on how the problem was stated. It doesn't matter what the first two children are - I meet a man who is standing with two of his children that are wrapped in black construction paper. I don't know what the heck they are. What's the probability that his 3rd child is a boy? 50%. I just figured that stating it this way would still leave lingering doubts (or denial) for the 75% crowd.

 

[JKing76]

No, now you've changed the question. If you hide information about the gender of two of the children, that's a whole different problem.

 

[JKing76]

I think the biggest problem here is the use of the term "third child". Obviously a lot of people are reading that as "I've introduced my eldest two children, what's the gender of my youngest child?". The intent of the question, though, is obviously "I've introduced two of my children, what's the gender of the missing child?" So let's stop using "third" child. If you want to specify them by birth order, use "oldest, middle, youngest"; if you're talking about the introduced/unintroduced sets, say "introduced children" and "unintroduced child".

 

[Matt1970]

Originally posted by: JKing76
Jinsterrr, there is a difference. For the first question, the probability of the missing child being a girl is 75%. Because here there is no expanded state. There are simply four choices: BBB, BBG, BGB, or GBB. Three of the four include one girl, so the chance is 75%. The simulation also demonstrates this.

But when you say that a 3C2 yielded two boys, you have more information. BBB is guaranteed to result in two boys. BBG, BGB, or GBB are have a 2/3 chance of introducing one boy and one girl. Just as DrPizza stated, this is when you look at the extended state.

Since there is no mention of the order they were born or probability of somone having 2 boys and 1 girl or somone having 3 boys, BBG, BGB, or GBB are all the same so the only factors are BBB or BBG no matter where you put the G.

 

[Matt1970]

Originally posted by: f95toli

Originally posted by: Matt1970
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.

See, the last part of your statement suggests the quarter has a memory of what the first 2 flips are, and it doesn't. Your last flip of the quarter will always be 50/50 heads or tales reguardless of what the first 2 flips are. That's a trick casino's will use with the roulette wheel. The will show you which hit, either red or black, for the last 10 spins or so thinking you will see a pattern and bet at it. Reguardless of what just hit on the wheel, even if it were 5 red's in a row, there is always and will always be a 50/50 chance of red or black on the next turn.

 

[JKing76]

You are wrong. It's because there is no mention of order that it does matter where you put the 'G'.

 

[DrPizza]

Originally posted by: JKing76
No, now you've changed the question. If you hide information about the gender of two of the children, that's a whole different problem.

That's exactly what I meant by people wouldn't understand the simplest explanation. Let me say it again. You meet 2 children - it doesn't matter if it's the youngest and oldest, the two oldest, or the two youngest. As soon as someone say, "but wait, there's one more child" the probability of that child being a boy is 50%. It *does not matter* what the sexes of the two children you meet are. It doesn't make any difference if the two children you meet are boys.

However, as I explained before, if you think it's important that you met 2 boys, then we can write out a sample space to demonstrate that the probability that the 3rd child being a boy is 50%.

The 50% people are correct. Of course, among that crowd, I'm guessing that some don't quite understand the logic why they're correct. The 75% crowd overthought the problem and ignored one detail (and sometimes argued that the detail wasn't important.)

 

[f95toli]

Originally posted by: Matt1970

See, the last part of your statement suggests the quarter has a memory of what the first 2 flips are, and it doesn't. Your last flip of the quarter will always be 50/50 heads or tales reguardless of what the first 2 flips are. That's a trick casino's will use with the roulette wheel. The will show you which hit, either red or black, for the last 10 spins or so thinking you will see a pattern and bet at it. Reguardless of what just hit on the wheel, even if it were 5 red's in a row, there is always and will always be a 50/50 chance of red or black on the next turn.

No, you are talking about "gambler's fallacy" which assumes that the outcome of one flip depends on the outcome of previous flips. That is nof course not truue and is not what I wrote.
What I wrote was that you already HAVE flipped the coin three times and that you got two heads, it does not matter WHICH flips resulted in heads (it could be e.g. the first and the third); i.e. the order doe NOT matter
That makes all the difference, because (again) the probabilty of getting a series with three heads (which can only a happen one way) is lower than getting a series with two heads and one tails (which, as you can see in earlier posts, can happen in three ways).

 

[Matt1970]

Ya this has been way over thought. I wish I could show a graph or something, but since we are using BBG and whatnot, I will use that. If you want to use BBG then you haben given the first 2 and the last has only 2 options, B or G, 50%. The same applies to BGB, then you are given the first and the last and you have to perdict the odds of the middle, still 50%.

 

[JKing76]

Again, there's a lot of misintepretation going on. People are arguing about three different questions.

If you read the question as: "A man with three children introduces his oldest two, both of whom are boys. What is the probabilities that the third child is a girl?" the answer is 50%.

If you read the question as "A man has three children, two of whom are boys, what is the chance that the other child is a girl?" the answer is 75%.

If you read the question as "A man with three children randomly chooses two to introduce, both of whom turn out to be boys, what is the chance that the unintroduced child is a girl?" the answer is 50%.

 

[putnampp]

Interesting thread. We have people arguing probability and genetic/natural pre-dispositions to being more likely to have more male/female children.

My answer would be 50%. I haven't seen any studies/information suggesting that child conception can be defined conditionally, as some have assumed. I don't know, I could be wrong.

I am not a big fan of probability, but I would like to know this. Seems like the basic assumption has been that all three children come from the same mother. What if they didn't? How would that effect the probability? Seems like all three children would be completely independent then, so the result would be 50%, right?

 

[lousydood]

This is a Word Problem, like you might do in High School.
This has nothing to do with genetics, family, or biology at all. That is all window-dressing for the problem.

Put that out of your head right now.

Let X be the random variable with
P(X=0) being 0.5
P(X=1) being 0.5

read that is "the probability of X equaling 0 is 0.5" etc.

Each row in the following table is a sample of 3 events. There are 2^3=8 possible permutations. Each row in the table has an equal probability of occurrence: (0.5)^3 = 0.125

X1, X2, X3
----------
0 , 0 , 0
0 , 0 , 1
0 , 1 , 0
0 , 1 , 1
1 , 0 , 0
1 , 0 , 1
1 , 1 , 0
1 , 1 , 1

Now consider the condition "two of the samples must be 0" and eliminate rows that do not satisfy it

X1, X2, X3
----------
0 , 0 , 0
0 , 0 , 1
0 , 1 , 0
1 , 0 , 0

The remaining rows show that the probability that none of X1, X2, or X3 being equal to 1 is 0.25 under the condition stated. The complement of 0.25 is 0.75.

Re-interpret the math: "Boy=0", "Girl=1" and 3 samples meaning "a sample of 3 children"

 

[totalcommand]

my 2c. This is a semantics problem not a math problem.