Probability Puzzle

[Gargamel]

Posting this in the S&S forum, rather than the Lounge, so we can have a intelligent conversation about probabilities vs Bayesian Probabilities.   If you know the answer, or find this trivial, just don't blurt it out, let the others figure this out.    When we've had a bit of good discussion here, I'll post up the second puzzle. 

 

So here's the Question:

 

You are walking down the street, when you run into a friend of yours.  She has her two children with her.  One of them is a boy.  What are the odds that the other child is also a boy?

[Superfluous J]

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

But I'll put my answer in a spoiler.

Spoiler

According to Huggies (hey they were the first link) the chances that the kids are Identical twins is 1 in 285, so there is 1 1/285 chance that it's a boy and we can stop. I'm ignoring triplets and higher, as you said "her two children" and not "two of her children."

According to TheBump.com (again, first link) 51% of babies are boys, and 49% are girls. This is apparently due to the male sperm being slightly smaller, and therefore slightly faster, than female sperm.

So 1/285 chances it's a boy. Of the remaining 284 chances, 51% or 144.84 will be boys, and 49% or 139.16 will be girls.

So 148.84 are boys, 139.16 are girls, so the chances are 52.4% that the other child will be a boy.

Note, I only uses 2 digits in one factor and 3 in another, so really it's more like 52/48 boy/girl.

Interestingly, having a boy and knowing nothing else about the family raises the chances of the other child being a boy by more than 1 percent, which is more than I'd think.

Now of course it's possible that she had triplets and one of them died, but that's rare (and tragic) enough to just pretend it didn't happen.

 

Edited July 18, 2018 by 5thHorseman

[YNM]

31 minutes ago, Gargamel said:

What are the odds that the other child is also a boy?

50/50.

 

[Gargamel]

41 minutes ago, YNM said:

50/50.

This is incorrect, as the wording in the puzzle is key.  It's not "what are the odds the other child is a boy", but "is also a boy".

Veritasium is a great channel.  Love his videos.   Have not watched this one (lately), but if he refutes my answer, I can direct you to a few phd statisticians who agree with it (which is where I got the puzzle from).

 

Interesting data, but we're looking for a simpler answer than that.   But yours would be more accurate then a pure 50/50 answer. 

@5thHorseman

55 minutes ago, 5thHorseman said:

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

Actually, no, a lot simpler, but alas, an argument will always happen, as the answer, when described makes complete sense, but is still somewhat counter intuitive. 

[Elthy]

Im not a native speaker, but doesnt "boy" put limits onto the age of the child?

[Gargamel]

1 minute ago, Elthy said:

Im not a native speaker, but doesnt "boy" put limits onto the age of the child?

Yes, I guess it would, but that is not relevant to the question. 

[YNM]

1 hour ago, Gargamel said:

This is incorrect, as the wording in the puzzle is key.  It's not "what are the odds the other child is a boy", but "is also a boy".

In which case, it's possible it's a form of gambler's fallacy.

However, determination of human sexes are not gambling - identical twins are more likely to be of the same gender, while fraternal twins tend to be pairs, and census (not studies) show that males are more likely (partly why there's still more males than females).

And this all sum up to not quite 50/50.

Still, these are all the "base probability". And no, it won't change.

(and by sex/gender I mean chromosomes, XX/XY and such. just in case one wonder.)

Edited July 18, 2018 by YNM

[Wjolcz]

This is the new plane on a conveyor belt thread.

[Superfluous J]

1 hour ago, Gargamel said:

the wording in the puzzle is key.  It's not "what are the odds the other child is a boy", but "is also a boy".

Ah so it's a trick like the airplane treadmill puzzle. Should have known.

At least this one is ONLY a grammar trick. And (if it is what I think it is) a massive stretch.

[MinimumSky5]

Spoiler

25% chance to be a boy.

I'm reading this as "What are the chances that the woman has two sons?". In this case, the answer is 50% X 50% = 0.5 X 0.5 = 0.25 = 25%.

More accurately, 51% X 51% = 26.01%

 

[p1t1o]

It would help if we knew whether or not the children are her biological offspring or not, they, or one of them, could be adopted, which throws a spanner in the genetics angle.

Then you could start talking about the probability of boys or girls being put up for adoption but that seems like too much detail for this question. It doesnt seem like a question that is supposed to be solved by researching census figures.

Im tempted to say that due to so many unknown factors, and given two possible outcomes, the probability defaults to 50/50.

Its either that or it tends towards the probability that any given developing fetus turns out genetically male (just off 50/50).

The identical twins thing seems relevant, but it can only change the probability either way, by quite a small amount.

***

You wouldnt ask the question unless the answer was surprisingly simple, or surprisingly complex. I think simple is more likely as you came out and said almost as much.

So on that basis, the answer is almost certainly 0, 0.1, 0.25, 0.5, 0.75 or 1.

0 and 1 are out, naturally, and I think 0.1, 0.25 and 0.75 are out-of-scope.

So again, 0.5 or 50/50.

***

1 minute ago, MinimumSky5 said:

  Reveal hidden contents

25% chance to be a boy.

I'm reading this as "What are the chances that the woman has two sons?". In this case, the answer is 50% X 50% = 0.5 X 0.5 = 0.25 = 25%.

More accurately, 51% X 51% = 26.01%

 

Are you sure its not 1*0.5=0.5 ? 

As we know the first child is a boy, the probability of the first child being a boy is 1.

Edited July 18, 2018 by p1t1o

[KSK]

Probability makes my brain hurt but I’ll give this a go. I’m guessing 33% (or 1 in 3 - however you want to express it) chance that the second child is a boy.

Reasoning.

Assuming that there’s an equal probability that any given child will be male or female (not quite correct as already pointed out but good enough) and that the gender of the first child has no effect on the gender of the second child, then we have three possible combinations of children (since we know that one is male):

1st child male + second child male.

1st child male + second child female.

1st child female + second child male.

Hence a 1 in three chance that both children are male and therefore that the other child is a boy.

[MinimumSky5]

Just now, KSK said:

Probability makes my brain hurt but I’ll give this a go. I’m guessing 33% (or 1 in 3 - however you want to express it) chance that the second child is a boy.

Reasoning.

Assuming that there’s an equal probability that any given child will be male or female (not quite correct as already pointed out but good enough) and that the gender of the first child has no effect on the gender of the second child, then we have three possible combinations of children (since we know that one is male):

1st child male + second child male.

1st child male + second child female.

1st child female + second child male.

Hence a 1 in three chance that both children are male and therefore that the other child is a boy.

Nooooo....! I just realised this, and was about to post it!

[Xd the great]

Bruh, 50%. Or whatever the percentage on males in the entire population. Case closed.

[KG3]

Two children are not a statistically relevant sample size for this experiment.  She would need to have at least 30 kids before getting any meaningful results.  

[YNM]

1 hour ago, KSK said:

I’m guessing 33% (or 1 in 3 - however you want to express it) chance that the second child is a boy.

Dang, I just realized it ! (while in the bathroom no less !)

1 hour ago, YNM said:

... while fraternal twins tend to be pairs ...

That actually extends to any other arrangement (like brother/sister).

There's no cause for the next one in sequence to be of a different "sign", but in general, it's more likely to have pairs.

There are 4 case possible :

Male     / Male

Male     / Female

Female / Male

Female / Female

each of which have a 25% chance.

As can be seen there's more chance of pairs than not.

So if someone gives "oh, one is a boy" then somehow it's more likely to be a pair - hence the other to be girl, by about 66%.

I suspect this have some connection with the Monty Hall Problem -thing.

30 minutes ago, KG3 said:

She would need to have at least 30 kids before getting any meaningful results.  

I was thinking 30 parents who have 2 child each.

Edited July 18, 2018 by YNM

[Bill Phil]

According to Mendel, the odds of two children in a row being sons is 1/4. It's slightly different, due to the slightly higher probability of having a son.

However... we know that one child is male. We don't know if this one was the second or first child. 

Since the two events should be independent, 1/2, or around that. Like successive coin flips, knowing previous outcomes doesn't change future outcomes. Depending on how we find out that one child is a boy. Could be 1/3 as well, since one of the four possible pairs is ruled out, the pair with two girls. 

Oh, ambiguity!

[Reactordrone]

4 hours ago, Gargamel said:

You are walking down the street, when you run into a friend of yours.  She has her two children with her.  One of them is a boy.  What are the odds that the other child is also a boy?

One of them is a boy, so the chances of the other one also being a boy is zero otherwise two of them would be boys :p

Edited July 18, 2018 by Reactordrone

[p1t1o]

20 minutes ago, Reactordrone said:

One of them is a boy, so the chances of the other one also being a boy is zero otherwise two of them would be boys :p

You're not wrong.

Although "one of them is a boy" is correct no matter what the other one is.

Semantically, it works both ways.

My fave answer so far however

 

[wumpus]

4 hours ago, YNM said:

50/50.

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.

[kerbiloid]

6 hours ago, Gargamel said:

You are walking down the street, when you run into a friend of yours.

Not a native speaker, so not sure what kind and level of relations is meant as a "friend".
As the me doesn't know about my "friend's" children, does that mean we (i.e. friends) didn't speak for several years?

Also, does the question mean that the me has my real gender (male) or possesses a body of a person of an undefined gender?
So, should I guess the questioned person gender, too, or take it as given, like "me"?

Though, anyway.
As this is about Bayesian things, let's take the most general assumptions.

(Sorry, I'll continue in the next post, need to calc).

Edited July 18, 2018 by kerbiloid

[Superfluous J]

If you can see them both, then it's either 100% or 0%. If you can't, it's 50/50. If you can't see either and hear her or someone mention her son in a way that does not identify the gender of her other child, then it's 1/3. Any other situation I can think of, someone's trying to trick you so don't bet any money.

Ignoring twins and 51/49 split on purpose for ease of numbers.

[Tullius]

1 hour ago, Bill Phil said:

According to Mendel, the odds of two children in a row being sons is 1/4. It's slightly different, due to the slightly higher probability of having a son.

However... we know that one child is male. We don't know if this one was the second or first child. 

Since the two events should be independent, 1/2, or around that. Like successive coin flips, knowing previous outcomes doesn't change future outcomes. Depending on how we find out that one child is a boy. Could be 1/3 as well, since one of the four possible pairs is ruled out, the pair with two girls. 

Oh, ambiguity!

The correct answer is 1/3, since

case	oldest kid	youngest kid
1	girl	girl
2	girl	boy
3	boy	girl
4	boy	boy

As we know that "one of them is a boy", case 1 cannot happen, but 2, 3 and 4 are possible. And so we have two boys only in case 4 among the three cases 2, 3 and 4. So the probability is 1/3.

If instead we would know that "the oldest of them is a boy", we could only be in cases 3 and 4. Hence, the probability would be 1/2.

[p1t1o]

1 hour 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.

But they are seperate. The fact that there is an existing child only has an effect if they are twins. If I have a son, then I have a second child, the probability that the 2nd is a boy is exactly the same as the 1st.

Its not like a lottery draw where you have to match each number, we are given the first result. It is like asking "you  meet a woman with her 6 lottery numbers, the first 5 are 1,2,3,4 and 5, what is the probability that the 6th number is 6?" (lets say the lottery is drawn from 49 numbers and for the sake of argument, lets say each number can be drawn, multiple times)

The probability of the 6th number being 6, is not the same as the probability that the winning combination is 1,2,3,4,5 and 6. The probability of the 6th number being 6 is 1:49 - not 1:14000000 

 

The answer would be 1/3 only if we are not initially given the sex of one of them. This is only like half of Monty-Hall, we are not given any choices, only asked a probability of one outcome - we are not asked how a change in the situation might affect said probability.

 

Disclaimer: Im like 70% certain in any of this.

Edited July 18, 2018 by p1t1o

[Bill Phil]

41 minutes ago, Tullius said:

The correct answer is 1/3, since

case	oldest kid	youngest kid
1	girl	girl
2	girl	boy
3	boy	girl
4	boy	boy

As we know that "one of them is a boy", case 1 cannot happen, but 2, 3 and 4 are possible. And so we have two boys only in case 4 among the three cases 2, 3 and 4. So the probability is 1/3.

If instead we would know that "the oldest of them is a boy", we could only be in cases 3 and 4. Hence, the probability would be 1/2.

You're not wrong... But not wholly right. 

If we know the youngest is a boy, we can only be in cases 2 and 4, still 1/2 probability. But the child we know the gender of can be either the oldest or the youngest. 

If we are asking how likely the second child is the same gender as the first, it's 1/2. If we're asking how likely is the oldest child the same gender as the younger, it's also 1/2. If we're asking how likely the two children are the same gender, knowing that one child is a specific gender, then it's 1/3. 

Are we asking the likelihood of both children being the same gender while knowing the gender of one child, or are we asking the likelihood of one child having a certain gender? 

It's up to interpretation.