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Probability Puzzle


Gargamel

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2 hours ago, wumpus said:

It would be 50/50 if and only if you could consider them separately.

Yeah, which is what happens in sequence.

But the statement is non-sequential.

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First of all, we don't have any info about the size human population or its gender distribution in this setting.
We just can presume that they have normal chromosome model XX/XY.

***

So, we have:

  • she-friend (female)
  • "me" (not the real me, but the questioned person, so its gender is unknown), hereinafter "my-client"
  • her primary child (male)
  • her secondary child (unknown)

***

As 

  • my-client doesn't pay alimony 
  • or in another way is aware of the she-friend's children gender, 
  • and they are still friends, not foes 

so it's highly likely that my-client is not the children's father.

That means they have another biological father, and he is most probably a male.

The secondary child should also have a biofather. Either the same, or another one, that's unknown.
And he is a male, too.

***

So, we are aware of a gender of 3 or 4 persons in the world:

  • she-friend (female)
  • primary child (male)
  • primary child's biofather (male)
  • maybe another biofather of a secondary child (male).

So, according to the given data, a gender ratio is 2:1 or 3:1, unless the secondary child is a girl.
And we don't know the gender of my-client.

But there is another omitted parameter. Does my client have its own children?

If my client is a woman, then as unlikely two female friends could be not aware of four children of theirs and still be named "friends", my client has [0;1] child.
If 1, it should have its own biofather. It may or may not be one of those ones.

If my client is a male, and has children, there should be their mom in the sampling.
Most probably, a male-client would be walking with one child, rather than with a crowd.

So, we can presume that most probably my client has [0;1] children regardless of its own gender.

Also if my client is a male and walks with a child, he may or may not be its biofather (regardless of his awareness and perception of that).
So, if my client walks with a child, we should add to the sampling a female (child's mom) and [0;1] of the mom's ex-friends (by-friends?).

***

So, we definitely know there are 1 F (she-friend) + 2 M (primary child and his biofather).
Also, there is a secondary child of unknown gender: 1 U
And there can be the secondary child biofather, so + 0..1 M

If my-client is a female
  If she has a child, this adds 1  child of unknown gender and [0;1] additional males (the child's biofather), i.e. +1 F + 1 U + [0;1] M
  otherwise +1 F

If my-client is a male
  If he has a child, this adds 1  child of unknown gender, 1 female (child's mom), and [0;1] additional males (the child's biofather, if this is not my client), i.e. +1 F + 1 U + [1;2] M
  otherwise +1 M

So

if my client: with a child without a child
is female 2 F + 2..4 M + 2 U 2 F + 2..3 M + 1 U
is male 2 F + 3..4 M + 2 U 1 F + 3..4 M + 1 U

Let's substitute:

if my client: with a child without a child
is female

2 F + 2..4 M + 2 U

4 F + 2..4 M
3 F + 3..5 M
2 F + 4..6 M

2 F + 2..3 M + 1 U

3 F + 2..3 M
2 F + 3..4 M
 

is male

2 F + 3..4 M + 2 U

4 F + 3..4 M
3 F + 4..5 M
2 F + 5..6 M

1 F + 3..4 M + 1 U

2 F + 3..4 M
1 F + 4..5 M
 

As we presume a simple XX/XY model, a theoretical gender ratio should be 1:1.
So, we can reject the uneven cases as insolvent male chauvinistic insinuations.
(The general sampling is already large, so we can presume it's ratio is close to average).

Let's exclude the most uneven cases.

if my client: with a child without a child
is female

4 F + 4 M
3 F + 3 M

3 F + 3 M

is male

4 F + 4 M

 

So, 4 most probable and balanced cases are:

if my client: with a child without a child
is female

4 F + 4 M
My client is a woman and has a daughter (and the daughter's biofather).
She-friend has a son and a daughter, from different fathers (but not from my client's boyfriend)


3 F + 3 M

My client is a woman and has a child.
She-friend has a son and a secondary child, from the same father (and he is the father of my client's child).
One of the children is a girl

3 F + 3 M
My client is a childfree woman.
She-friend has a son and a daughter, from different fathers.

is male

4 F + 4 M

My client is a male and has a daughter (and the daughter's biomother).
She-friend has a son and a daughter, from different fathers (but not from my client.)

 

 

Also, the gender ratio doesn't change from generation to generation, so more even cases are more probable.

Case Total ratio Adult ratio Children ratio
My client is a woman and has a daughter (and the daughter's biofather).
She-friend has a son and a daughter, from different fathers (but not from my client's boyfriend)
4 F + 4 M 2 F + 3 M 2 F + 1 M
My client is a woman and has a child.
She-friend has a son and a secondary child, from the same father (and he is the father of my client's child).
One of the children is a girl
3 F + 3 M 2 F + 1 M 1 F + 2 M
My client is a childfree woman.
She-friend has a son and a daughter, from different fathers.
3 F + 3 M 2 F + 2 M 1 F + 1 M
My client is a male and has a daughter (and the daughter's biomother).
She-friend has a son and a daughter, from different fathers (but not from my client.)
4 F + 4 M 2 F + 3 M 2 F + 1 M

As we can see, the most balanced case is the 2nd one.
Other three cases are less balanced in sense of generations size, but all of them presume that the she-friend has a son and a daughter, so the secondary child is a girl.
So, "by default it's a girl"

***

So, unless it's default, we now can tell a whole story.

My she-client and her she-friend had a common he-friend.
After my she-client got a child from the he-friend, he has left her and married to the she-friend.
For several years the girls didn't talk to each other, so my she-client isn't aware of her former she-friend children.
But currently their he-friend tried to change his side again.
But no.
Now the girls are friends again, they will ask for alimony, and enslave this shameless animal.
As they will live aside and raise their children together, whose exactly is daughter is arbitrary.

Edited by kerbiloid
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5 hours ago, wumpus said:

It would be 50/50 if and only if you could consider them separately.  If you said the elder (or younger) was a boy then the other would be 50/50.  But since they are lumped together and all you know is that one is a boy you get 1/3.  This is basically the Monty Hall paradox.  And yes, the wording of where the givens are lends itself to mistakes.

As not an native English speaker to me it sounds just like you have two children: first is an boy, what is the chance of the second is an boy.
 

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6 hours ago, wumpus said:

It would be 50/50 if and only if you could consider them separately.  If you said the elder (or younger) was a boy then the other would be 50/50.  But since they are lumped together and all you know is that one is a boy you get 1/3.  This is basically the Monty Hall paradox.  And yes, the wording of where the givens are lends itself to mistakes.

No. Monty Hall problem has three options that are each directly dependent on the other two. In this case, the sex of each child is independent of the other.

As for identical twins approach to the math, dizygotic twins are ten times more common, which kind of dilutes the influence.

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This is an ambiguous problem, because of the way it is written.

If you said, "A woman has two children. One is a boy. What is the chance that the other one is a boy?" then the (approx) 1/3 answer (ignoring identical twins and biological preference for boys or girls) would be correct. But the reason for this is because you don't know if child 1 or child 2 is the one that is a boy.

But the way this is written, it is possible that you see one or both children. If you see that one is a boy, then that one is already determined and so the odds for the second one become 50/50.

Edited by mikegarrison
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3 hours ago, Shpaget said:

In this case, the sex of each child is independent of the other.

... if stated in sequence.

However, take it this way :

For any (fraternal) twin and brother/sister "pair", with a perfect 50/50 chance of one being male or female, there's :

25% chance of a 'double' male

50% chance of a male/female pair

25% chance of a 'double' female

These statements are non-sequential.

Notice how the pair configuration is twice as likely as the 'double' male configuration.

So, if someone states non-sequentially that one of their two children is a boy, then it's still more likely that their children is a pair rather than a 'double' male. But because there's only 3 configuration possible, the chances becomes 66/33 rather than 50/25.

The same apply to a coin flip : it is twice as likely for two flips to be a pair than a double head. But just because you get a head on this flip doesn't mean the next flip will more likely be a tail.

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17 hours ago, 5thHorseman said:

Something tells me this is going to be more involved than the Moon Black Hole and Plane on a Treadmill threads.

We'd be lucky if it were a Moon black hole. If a GEO Telecom satellite happened to become a black hole it'd wipe out life for sure!

Man, these bayesian puzzles always get me, and this one was no exception. There's something about the way they present information that's contrary to how I mentally tabulate statistics. It's kinda like a magician, where the trick is done the moment before you start really paying attention! Then once it's explained it's so easy to see how it was done. Well, good show.

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1 hour ago, DAL59 said:

Very slightly more than 50%, due to the possibility of twins.  

No, as the percentage of them being twins are the same as the percentage of twins in the whole world.

2 hours ago, mikegarrison said:

What if you know that your friend has named her children Chris and Pat?

Then you will probably know their gender already.

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4 hours ago, Xd the great said:

So, 50%. Case closed.

50% chance of any of them being male or female. But if you got two pairs it's not.

Again, just like coin flipping - if you take any 2 then by math it's 50% head and tails, 25% double heads and 25% double tails. Pairs are more likely than double head AND pairs are more likely than double tails BUT pairs are as likely as doubles of any.

But this doesn't mean if you get tails in your flip now the next is more likely to be head.

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2 hours ago, YNM said:

50% chance of any of them being male or female. But if you got two pairs it's not.

Again, just like coin flipping - if you take any 2 then by math it's 50% head and tails, 25% double heads and 25% double tails. Pairs are more likely than double head AND pairs are more likely than double tails BUT pairs are as likely as doubles of any.

But this doesn't mean if you get tails in your flip now the next is more likely to be head.

But for the kids the coins have already been flipped. The sentence "One of them is a boy" or "One of them is heads" is to be understood as "There is at least one boy" or "There is at least one heads", i.e. we are computing the probability that there are two boys, given that there is at least one boy, or in mathematical notation

P(two boys | at least one boy)

So we can use the definition of conditional probability to get

P(two boys | at least one boy) = P(two boys AND at least one boy) / P(at least one boy) = P(two boys) / P(at least one boy) = (1/4) / (3/4) = 1/3

 

Edit:

If instead the problem would state "the oldest kid is a boy" or "the first flip is heads", then we would of course get with Bayes' rule

P(two boys | oldest is boy) = P(two boys AND oldest is boy) / P(oldest is boy) = P(two boys) / P(oldest is boy) = (1/4) / (1/2) = 1/2

Alternatively, we could of course use that the gender of the two kids is independent and get to the same conclusion.

Edited by Tullius
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2 minutes ago, Tullius said:

The sentence "One of them is a boy" is to be understood as "There is at least one boy"

Well, see, that's the problem. It's not necessary to understand things that way. The problem could have been written in exactly those words ("there is at least one boy"), but it wasn't. That's why these sorts of problems are slightly unfair. In real life, if this lady was your friend you would almost certainly already know the sexes of her kids. But maybe she's not such a close friend, and last time you saw her she only had one kid (a boy). Now she has two kids. In that case, what are the odds that the other one is a boy?

I know that's not exactly the way the question was posed, but it's a much more real-world scenario, so people jump to it a lot easier than jumping to the idea that you know your friend has at least one boy but you don't know which one is the boy or what the sex of the other child is.

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19 minutes ago, mikegarrison said:

But maybe she's not such a close friend, and last time you saw her she only had one kid (a boy). Now she has two kids. In that case, what are the odds that the other one is a boy?

In that case, it is of course 1/2, cf. my edit above.

The funny thing is that as soon as we have any information which kid is the boy, the probability is 1/2. That is, if one would state that "the kid with the highest grades in school is a boy" or even "Tony is a boy", then the probability that the second kid is also a boy is 1/2. If we don't have this information, it is 1/3.

19 minutes ago, mikegarrison said:

Well, see, that's the problem. It's not necessary to understand things that way. The problem could have been written in exactly those words ("there is at least one boy"), but it wasn't. That's why these sorts of problems are slightly unfair. In real life, if this lady was your friend you would almost certainly already know the sexes of her kids. But maybe she's not such a close friend, and last time you saw her she only had one kid (a boy). Now she has two kids. In that case, what are the odds that the other one is a boy?

I know that's not exactly the way the question was posed, but it's a much more real-world scenario, so people jump to it a lot easier than jumping to the idea that you know your friend has at least one boy but you don't know which one is the boy or what the sex of the other child is.

The wording was consciously chosen this way, as "there is at least one boy" would have given away that this problem is solved by a simple application of conditional probability.

As to the real-world comparison: Mathematicians rarely care about the practical application. This problem solely serves as illustration that conditional probabilities aren't necessarily intuitive.

Edited by Tullius
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22 minutes ago, Tullius said:

But for the kids the coins have already been flipped.

Yup. I mean, you could get a row of heads in your one trial, but if you were to ask everyone to do it, then it's more likely :

not to be a row of heads

- to be pairs than to be row of heads

- in all the set of two flips that has heads as at least one result to have a pair of heads and tails than a row of heads.

And the last one is just this question.

20 minutes ago, mikegarrison said:

It's not necessary to understand things that way.

If you were given "one of my 7 child is a boy", you have no clue which one in the row is it.

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38 minutes ago, Tullius said:

The wording was consciously chosen this way, as "there is at least one boy" would have given away that this problem is solved by a simple application of conditional probability.

And why would that have been bad?

If the idea is to teach the concept, then teach the concept. If the idea is to trick people into applying real-world knowledge into a problem that is carefully contrived to make that wrong, then it's an amusing puzzle but not necessarily a good tool for explaining the concept.

There are plenty of real-world examples that don't require tricky language to hide the problem. The classic one is like this:

A crime is committed on an island and DNA evidence is found at the scene. Other than the DNA, there is no way to know who committed the crime -- it could be any of the 10,000 people that live on that island. Later, a person has his DNA tested and it comes up as a match. The test is right 99.9% of the time. So there is only a 0.1% chance he's not the criminal, right? Wrong.

The key is that chance that he was not the criminal starts out at 9,999/10,000. If we test the DNA of all 10,000 people on the island, and there is a 0.1% chance of a false positive and a 99.9% chance of a true positive, then we should expect to get 11 matches -- 10 false positives and 1 true positive. So the chance of any one person who tests positive actually being the criminal is only 1/11.

Brandon Mayfield is something of a real world example of this. https://en.wikipedia.org/wiki/Brandon_Mayfield

Mayfield is a man from the state of Oregon in the US. He was one of 20 people in the US fingerprint records closely matched to a fingerprint left at the site of the 2004 train bombing in Madrid. He was a lawyer, had converted to Islam, and had defended some other Muslims in the US who had been charged with a terrorism plot. The FBI decided that the fingerprint match and his personal history was enough to arrest him for the bombing.

However, it turned out that Mayfield had not left North America any time in the previous decade, and Spanish authorities found another man (an Algerian named  Ouhnane Daoud) who was a match for the fingerprint and who they believed actually committed the crime. Mayfield was eventually released and was given $2M in settlement.

Edited by mikegarrison
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6 minutes ago, mikegarrison said:

And why would that have been bad?

If the idea is to teach the concept, then teach the concept. If the idea is to trick people into applying real-world knowledge into a problem that is carefully contrived to make that wrong, then it's an amusing puzzle but not necessarily a good tool for explaining the concept.

There are plenty of real-world examples that don't require tricky language to hide the problem. The classic one is like this:

A crime is committed on an island and DNA evidence is found at the scene. Other than the DNA, there is no way to know who committed the crime -- it could be any of the 10,000 people that live on that island. Later, a person has his DNA tested and it comes up as a match. The test is right 99.9% of the time. So there is only a 0.1% chance he's not the criminal, right? Wrong.

The key is that chance that he was not the criminal starts out at 9,999/10,000. If we test the DNA of all 10,000 people on the island, and there is a 0.1% chance of a false positive and a 99.9% chance of a true positive, then we should expect to get 11 matches -- 10 false positives and 1 true positive. So the chance of any one person who tests positive actually being the criminal is only 1/11.

If you only teach the concept using the standard set mathematical wording, you don't teach them something about practically using it (Very rarely in practice, you will find the precise wording "at least x of y"). You want students to stumble over the problem, when they try to intuitively give an answer to the question, so that they learn that they should only trust their mathematical computations.

Intuitively, you would say that if the mother would tell you that "one of them is a boy" or "Tony is a boy" doesn't make any difference. However, this makes all of the difference, cf. my previous post.

It is the same with your example: Unless you tripped over this problem, you will make this error time and time again.

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1 hour ago, Tullius said:

You want students to stumble over the problem, when they try to intuitively give an answer to the question, so that they learn that they should only trust their mathematical computations.

True, but also not that true.

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I guess I made my point badly.

The question about your friend and the two babies is fundamentally a trick. It's one of those "A plane crashes on the border between two countries; in which one do you bury the survivors?" questions. In the real world, you would be very unlikely to know what was specified, *especially* if the lady was a friend of yours.

But there are real world problems that *are* likely and illustrate the same math. There are all sorts of real issues like false positives of rare diseases or the legal issue I described where you don't have to set up a trick of the wording.

The way you are advocating teaching the concept is more likely to leave people thinking it's an esoteric math oddity than realizing it's a real thing that people have trouble with in the real world.

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2 hours ago, mikegarrison said:

In the real world, you would be very unlikely to know what was specified, *especially* if the lady was a friend of yours.

That would actually work in favor of the set here.

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2 hours ago, mikegarrison said:

I guess I made my point badly.

The question about your friend and the two babies is fundamentally a trick. It's one of those "A plane crashes on the border between two countries; in which one do you bury the survivors?" questions. In the real world, you would be very unlikely to know what was specified, *especially* if the lady was a friend of yours.

But there are real world problems that *are* likely and illustrate the same math. There are all sorts of real issues like false positives of rare diseases or the legal issue I described where you don't have to set up a trick of the wording.

The way you are advocating teaching the concept is more likely to leave people thinking it's an esoteric math oddity than realizing it's a real thing that people have trouble with in the real world.

I'm curious under what possible conditions you could realistically acquire the information such that the odds would hold for 1/3.  If each cubscout in a cubscout den had a sibling, would the odds of each sibling being a girl be 2/3? (replace "cubscouts" with "members of GetRidOfSlimygirlS" for co-ed cubscout packs).  Presumably the rarity of such events means that the built-in heuristics in our brains simply don't handle the events properly (which means that people can get excited about disease tests for even less rare diseases that aren't close to being sufficiently accurate).

I'm not sure where to put the Monty Hall problem.  Certainly it is a math oddity (that entire university math departments got wrong), but the really weird thing is that the TV show that inspired it must have run for years, and nobody realized that there was a huge advantage in following a simple strategy.  Possibly a few people did, but considering the difficulty Martin Gardner (I think his Scientific American column got the ball rolling) in convincing people that switching doors was optimal I doubt anyone believed them or cared.

To be honest, most people have issues with even basic probability issues before you get into weird bayesian interactions.  Teach those first, so at least you can understand how it works.

5 hours ago, mikegarrison said:

And why would that have been bad?

If the idea is to teach the concept, then teach the concept. If the idea is to trick people into applying real-world knowledge into a problem that is carefully contrived to make that wrong, then it's an amusing puzzle but not necessarily a good tool for explaining the concept.

There are plenty of real-world examples that don't require tricky language to hide the problem. The classic one is like this:

True, but the world rarely leaves such things in convenient little phrases that match up to traditional math problems (if it did, science would be much easier).  At least in the US, math education has been traditionally lousy at teaching word problems (although I suspect that a group who deals with the rocket equation for fun might be outliers).  I can only hope that the math bits of common core are able to fix this.  If you can't apply math to the real world (or spend your time doing "pure math" dealing with proofs), then it seems that a great deal of your education has been wasted.

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1 hour ago, wumpus said:

I'm curious under what possible conditions you could realistically acquire the information such that the odds would hold for 1/3. 

[...]

True, but the world rarely leaves such things in convenient little phrases that match up to traditional math problems (if it did, science would be much easier).  At least in the US, math education has been traditionally lousy at teaching word problems (although I suspect that a group who deals with the rocket equation for fun might be outliers).  I can only hope that the math bits of common core are able to fix this.  If you can't apply math to the real world (or spend your time doing "pure math" dealing with proofs), then it seems that a great deal of your education has been wasted.

These two parts of your answer illustrate the same thing: word problems are stupid. Textbooks start with a mathematical equation and then try to build a "word problem" around it. It's almost never anything that makes real world sense, which everybody recognizes and many people joke about.

In real life, many engineering problems are "word problems" (example: "how thick does the skin of this airplane need to be at this location of the fuselage?"), but none of them look or sound like a math textbook word problem. Instead we get "Amy has six apples and five oranges. If she can trade two apples for one tomato and five oranges for two tomatoes, how many tomatoes does she have after the trades?" Word problems are stupid because they aren't teaching you how to apply math to the real world -- they actually try to apply the real world to math.

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