On Friday I posted this puzzle…..

Imagine there is a country with a lot of people. These people do not die, the people consists of monogamous families only, and there is no limit to the maximum amount of children each family can have. With every birth there is a 50% chance its a boy and a 50% chance it is a girl. Every family wants to have one son: they get children until they give birth to a son, then they stop having children. This means that every family eventually has one father, one mother, one son and a variable number of daughters. What percent of the children in that country are male?

If you have not tried to solve it, have a go now. For everyone else the answer is after the break.

I have a funny feeling this is going to cause some debate! So…..what did you think?

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called **PUZZLED** and is available for the **Kindle** (UK here and USA here) and on the **iBookstore** (UK here in the USA here). You can try 101 of the puzzles for free here.

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Answer??

after the break

this weeks twist…definition of ‘break’

There was a big dust-up on the internet not long ago over a version of this puzzle. Hence Richard is expecting some debate over the answer.

I think that Richard’s presentation ironed out the ambiguities, though, so it should be pretty uncontroversial that the answer is 50%.

I agree with Physicalist and tested two methods of showing that the answer is 1/2 or 50 percent. One way was to write a simple computer program and the other way was to review the mathematics of infinite series and to develop a mathematical proof. Either way, I got the same answer: 50 precent

I don’t get it.

I don’t see the solution posted. Am I missing something?

I’m sure it will cause debate since the answer seems to be missing 🙂

My answer was: 50%.

50% of first children will be female. 50% of second children will be female. 50% of third children will be female. Etc…

That’s what I think, too. The answer’s in the question.

And the mine too. 50%

50% also. Everytime there’s a round of births you get one boy and one girl

If the answer really is 50%, consider a doughnut which is eaten in rounds. In the first round, half of the doughnut is eaten and the other half is uneaten. In the second round, half of the remaining doughnut is eaten and the other half is uneaten. For every part of the doughnut that is uneaten, there is a part that is eaten.

What percentage of the doughnut is eaten? 50%. Everytime there’s a round of eating there is one part uneaten and another part eaten.

Damnit, my head hurts.

Richard

Your question is ambiguous. You do not say whether it is a jam doughnut or a ring doughnut

@Z while that may be a tasty one, I do not think that is an analogous situation

Just to explain Zs doughnut example for those who didn’t get it. After the first round he’s still eating the same doughnut, so he’s taking the 50% that are girls and changing half of them to boys (i.e. eating half of the remaining doughnut). To get a correct doughnut analogy after eating half of the first doughnut a new smaller doughnut is made and you then eat half of that. So you have smaller and smaller doughnuts but always 50% eaten and 50% not eaten.

I agree. 50%.

It’s 50%.

“With every birth there is a 50% chance its a boy and a 50% chance it is a girl.”

It does not matter how these children are distributed over the families. People stop having children after they have a boy, but you’ll know that after the birth. There is no way of influencing the result this way.

50%.

You can sum the infinite series, or realise that whatever rules parents invent for the number of children they want, 50% of all births will be boys and 50% will be girls.

I don’t think these dynamics would be stable. The population will die out as there would be more females than males, increasingly so over time.

Difficult when there are no deaths. 🙂

But the 50% answer is right (mathematically it’s a Geometric distribution problem).

That’s why the puzzle specifies that the population is immortal…

How are you getting more females? there are an equal number of males and females being born. If anything the problem is that the number of children per couple approaches 2 so you’d have the population declining.

Shall we start a debate why Richard didn’t give an answer?

Is he caught in a cell with vicious guards that will kill him if he gives the right answer and also kill him if he gives the wrong answer? And now he tries to get out by not giving an answer?

Richard did not give the answer because he wanted us to discuss .

Thats why mention of DEBATE.

My answer is 50%

Making the assumption that males and females have the same life expectancies, it will be 50%. But typically females have a slightly longer life expectancy, so there will actually be slightly more females than males.

Oops forgot that they all live forever… so 50% it is.

It can’t be 50% because 50% of parents stopped having children after their first, their family will have 2 males and 1 female, it’s still 50/50 every time but half the people will be out of the game.

So after the first round half of families will have two males and 1 female, and the other half will have 2 females and 1 male, still 50-50 overall. None are “out of the game”, that are all still counted, just half of them will have another child, of which half will be male and half female.

Lets say we start with 100 families. In the first round of births, we have 50 boys, and 50 girls. but the families with boys quit having kids. 50 families are left having babies. 50% of the next go round will be boys, so you have a total of 75 boys and 75 girls. Now 25 families are left having babies, and you have 87 boys, 87 girls and one bi-gender baby. Anyway, it is a 50 – 50 split, all the way, except for the occasional Solomon like problem…

This is the explanation I was looking for! I had come to the 50% answer also, but was having a hard time wording the reason.

Thanks – now I understand the reason behind the answer I got using 50 lots of births in Excel.

Just to add a little mystery to all this waiting for the actual answer… I got an answer that isn’t 50% 🙂

It could be the correct answer. Is it 1/2 or a half?

I think I’ve changed my mind now. I think it’s 50%. But of course, that could just be me giving in to peer pressure 😉

I got 33% are male

Me too !

It is 50%.

In an average of 100 families, there will be 100 Sons. The number of Daughters born to the same 100 families will also be 100, as the chance of having a boy or a girl is 50% for each. The average family will be Mam, Dad, 1 Son and 1 Daughter. The percentage of male children is therefore 50%.

The principle of stopping once you have a Son doesn’t affect the 50% Male/Female birth ratio.

I think it’s slightly OVER 50% male – approx 55%

perhaps rw has provided the right answer. that this problem is formerly undecidable, as the methematics say.

remember, we simple do not know how many children there will be, so how can we say that x% of N will be a certain gender when we do not know N?

well played, richard!

fomerly? So now it is solvable?

I jest 🙂 He states in the question what the percentage is

50% here too.

I wrote a computer simulation, starting with 5 males and 5 females. When the population reaches 1 billion, 50.000236% are male. Have repeated the simulation several times with very similar results. Sometimes a tiny bit more than 50%, sometimes a tiny bit less, which is exactly what I would expect with some random variation.

That sounds as though you have misread the question. We are only asked about the children of the original families, not considering those children growing up and starting families of their own.

repton and Drew have the right answer.

For those who don’t accept it, arguing that there must be an excess of daughters, because some families have lots of daughters before they have a son, here is the point you’re overlooking: The families with more (sometimes many more) daughters than sons are balanced by the 50% of families that don’t have any daughters at all.

But the simple way to the answer is to see that choosing not to have any more children doesn’t alter the ratio of children already born, which is (per the question) exactly 50/50.

I’m afraid that’s wrong. The rule that you use to stop the size of the family does matter. Suppose that they keep on having children until in their family there is one more boy than girl. Then the percentage of boys must be more than 50%.

@Andrew Richards: Your propose counter-example doesn’t work. The thing you’re missing here is that many families will never have exactly one more boy than girl (from a probabilistic perspective, it becomes increasingly unlikely the more children that they have) and so will go on having more children forever. The extra girls to balance out the families that stop with one more boy than girl will be in this group (and as always with the Law of Large Numbers, they’ll balance things by exhaustion rather than by “catching up”).

@Andrew Richards: Nope, there is still a 50-50 chance of getting a boy or a girl you will just have some families having children indefinitely because they can’t influence the result. I know you are thinking 50% of families will have a girl and then some families after that will have more girls than boys so that more than 50% of families will have more girls than boys and that is correct BUT the majority of those families will have only one girl whereas the families with more boys will be much larger. We’re counting the number of boys vs the number of girls not the number of families with more girls than boys. The number is still 50%.

The answer i got was 100*ln(2)%.

Z: I got the same 🙂

Well, then you’re both wrong!

I realise that that comment was left by an advertiser, but it made me doubt my answer and go back through the maths. Same answer. No idea how to reconcile it with the very persuasive argument given by lots of others though! (that it should be 50%)

Well, then please explain how you get this result.

That was my first answer too, found by summing the infinite series of ((Probability) * (Percentage of boys in the family)).

The reason this is wrong is that each family does not contribute to the overall average uniformly, but in proportion to the total number of children they have.

The correct sum is to instead work out the expected number of boys in each family (sum of ((Probability) * (Number of boys in the family)), and similarly the expectation of girls.

The overall proportion of boys in the population of children is E[B] / (E[B] + E[G]), which comes out to be exactly 50%.

Arthur’s right. I also initially got 100*ln(2)%, but then after thinking about the Monte Carlo results I recalculated and got an overall proportion of 50%. Nice puzzle!

What happens to the 100 families, who with a 50% chance of having a boy or girl, have nothing but boys? An unlikely situation when looking at probability, but not impossible.

Or the families who, though forever trying, never have anything but girls? Just as (im)probable … ?

“What happens to the 100 families, who with a 50% chance of having a boy or girl, have nothing but boys?”

Can’t happen. The problem states that a family stops having children when they have one boy.

They could all have one boy, and no girls. That’s what I meant…

Then the answer is 100%. But by your logic, they could have an infinite number of girls, in which case the answer is 0%. Or they could have 99 families with 1 boy and 1 family with GB, so the answer is 100/101. Or they could have 98 families with 1 boy and 2 families with GB, so 100/102. And on and on…

All those answers can’t be simultaneously correct. We have to look at this as an expected percent. If you are going to ask about the case of BB…..BB (100 B), then you also have to think about the case BB…..BGB (99 B and 1 GB) and every other permutation and combination. When you look at all combinations and permutations, and also consider the likelihood that they occur, the answer is 50%.

Beautifully and clearly explained. Thank you.

Would it eventually be zero?

I got 33%

No answer, not fair, Richard!

I got a simulation running in which the percentage of women asymptotically approaches 100%. To be fair, I have taken the liberty of introducing a Maxwell’s Demon that messes up with sperm and eggs.

What have you done?? There are so many people wrong on the internet right now 😀 (not me, I’m not posting an answer)

I did it this way:

Every family will have a father, mother and son.

Then there’s a 50% chance of having 0 daughters

a 25% chance of having 1 daughter

a 12.5% chance of having 2 daughters

a 6.25% chance of having 3 daughters

etc…..

That means the expected number of daughters in each family is:

0.5*0 + 0.25*1 + 0.125*2 + 0.0625*3 ….

The sum of this infinite series is 1

http://www.wolframalpha.com/input/?i=sum+of+n*%280.5%5E%28n%2B1%29%29+from+0+to+infinity

Therefore, if there is a large number of families, we expect the ‘average’ family to be comprised of one father, one mother, one son and one daughter. Therefore the population will be 50% male, 50% female.

This is very similar reasoning to the reasoning I used to get ln(2) as the proportion of boys, except with minor differences:

Every family will have a son.

Then there’s a 50% chance of having 0 daughters => all boys

a 25% chance of having 1 daughter => 1/2 boys

a 12.5% chance of having 2 daughters => 1/3 boys

a 6.25% chance of having 3 daughters => 1/4 boys

etc.…..

So the proportion of sons in the family is:

0.5*1 + 0.25*0.5 + 0.125*0.333 + 0.0625*0.25 + …

And the sum of *this* infinite series is ln(2)

http://www.wolframalpha.com/input/?i=sum+of+%281%2Fn%29*%280.5%5E%28n%29%29+from+1+to+infinity

i.e. the proportion of sons is ln(2), and of daughters is 1 – ln(2)

@antdos : the problem is that you add up these probabilities without taking the number of childs into account. For example if you had 50 families with 1 boy and 50 families with two girls and one boy, your reasoning would give 0,5*1+0,5*1/3, which 2/3. Considering the fact that in this example there are exactly 100 boys and 100 girls, you can see that this is wrong.

That would mean that there would be more than twice as many sons as daughters overall (~2.25 times as many), which I find a bit absurd.

Considering the case where ONLY 0 or 1 daughters could be born. 50% of the time it will be 0 daughters and 50% of the time it will be 1 daughter. That means that we end up with:

50% : just a son

50% : daughter + son.

So we get a 2:1 ratio of sons to daughters. So already I’ve made the male:female ratio smaller than your final answer, and I’ve not even considered the possibility of a family having 2, 3, 4+ daughters.

I’m not exactly sure what’s wrong though!

@Yat –

Yes, that’s it! Altering my (un)logic gives:

Expected number of boys

= 0.5*1*1 + 0.25*0.5*2 + 0.125*0.333*3 + 0.0625*0.25*4 + …

= 1

Expected number of girls

= 0.5*0*1 + 0.25*0.5*2 + 0.125*0.666*3 + 0.0625*0.75*4 + …

= 1

So that’s a ratio of 1:1 = 50%. Phew!!

[I’ve got another comment awaiting moderation with wolframalpha links. This will do for now!]

Sorry, I don’t see how it can be 50%. That was my first thought, but the key thing here is that people do not die and they stop when they get boys. Some families will have more than one girl before having a boy, and although (because it is a monogamous society) only as many females will have children as there are males – ie a lot of spinsters – there will still be surplus women. I thought the puzzle was to work out the proportion of males. It has to be less than 50% surely.

Think about how the population grows as you add children one at a time.

It’s surely not hard to see that there is a 50/50 chance that each new person will be either male or female.

So as children enter the population, the ratio never changes, no matter how you try to influence it.

The puzzle is about the number of male children aka boys, not the number of males

What you’re missing is that even though there will be families with lots of girls, there will also be families with only boys. If you add up the number of boys in the single-child families it will equal the number of extra girls in the extra girl families.

You have misunderstood the question. In this question we are not considering the children growing up and having more children of their own, merely the expected distribution of all the original families having children.

Stopping having children once you get a boy only affects one statistic:

The percentage of youngest children being male in families who have at least one boy is 100%. I use the term ‘at least one boy’ to allow for multiple births involving boys (eg twins, triplets).

Simple, 50%.

i.e.

128 families produce 64 boys and 64 girls.

Those 64 families with girls produce 32 boys and 32 girls.

Those 32 families with girls produce 16 boys and 16 girls.

Those 16 families with girls produce 8 boys and 8 girls.

Those 8 families with girls produce 4 boys and 4 girls.

Those 4 families with girls produce 2 boys and 2 girls.

Those 2 families with girls produce 1 boy and 1 girl.

And the family who produce a girl……..?????

That’s the 64 families who make it to the second step. And the 32 familes who make it to the third step.

The answer is of course 50%. And it must be so if the puzzle is to have a solution and not be contradictory. The condition of monogamous families provides the result at once. The only problem is that number of people is not a continuous variable. And that is perhaps why there might be “a debate”, I think.

As several other comments have suggested, you have misunderstood the question.

The original population consists of purely monogamous families, but the question does not ask about what happens after the children have grown up and started families of their own.

You have the right answer, but for the wrong reasons.

Actually I had other reasons too 🙂 That’s why I wrote about number of people not being a continous variable. But if we assume it is, then the solution is straightforwad: with each step the proportion of boys and girls is equal (50%), only the number of birth decreases. Hence the answer.

Its 50% but not because there is a 50:50 chance of a girl or boy. The children in each family can comprise; g, gb, ggb, gggb,ggggb etc. That is more girls than boys. @nascit Green has it.

But how many of each type of family?

I meant to say b, gb, ggb etc

“the people consists of monogamous families only”

There’s your answer. The statement above says that there is, basically, a pre-existing condition where there is one male for every female.

Furthermore, by not dying you can assume that the adult non-reproducing population will be vastly larger compared to the number of children being produced in a generation.

True, I admit I’m assuming we’re talking about heterosexual families, otherwise the whole question is pointless.

This was my thought.

The question says that we start off with monogamous families only.

We are not considering what happens when the children grow up and get married themselves, just how many are born to the original set of parents.

The question can’t be answered because the probabilities don’t determine what happens in an individual instance.

For instance, toss a fair coin 1000 times and count number of heads. It is likely the number of heads will be close to 500, but you can&’t tell in advance exactly what proportion is heads. It is possible, albeit unlikely, that they were all heads.

In this case we can expect the proportion of boys to be close to 50%. The larger the population, the more likely it is that the proportion is very close to that value. This is the answer that’s missing above.

The question clearly states: ‘With every birth there is a 50% chance its a boy and a 50% chance it is a girl.’

So you do know in advance. (It’s not a real world.)

Actually, on reflection, forget that. I can see now that knowing the chances of something happening doesn’t tell you the outcome on any one occasion. Apologies.

(But since they all live for ever, and presumably continue to reproduce, doesn’t that mean that the overall chances are exactly 50/50?)

@Jeremy: not exactly. What you can say is that, as you add more and more people through generations, the proportion of males approaches 50%, but there will always be an irreducible range of variation around 50% (in statistic-speak: an error). As Andrew Richards says, the larger the population, the smaller the error, but unfortunately there is no finite number that will make the error equal to zero.

There it gets a bit tricky. The correct statement is: If the number of families is infinite then the probability that the proportion is 50% is 1. However, contrary to what they teach in schools, probability 1 doesn’t mean certain, and probability 0 doesn’t mean impossible!

The answer is 50%.

Every family has one boy. Famililies can have 0, 1, 2, …or any number of girls. The number of girls a family has is determined by how long they have to wait fir a boy.

As the probability an individual child is a boy is 1/2, then, on average,, it takes two goes to get a boy. Hence, on average, a family has one girl.

So, each family has exactly one boy, and, on average, one girl. 50/50

50% for every birth, not 50% for all families.

it would not be 50%.

1/(1+x)

Isn’t the answer within the question???? Could be a trick going on here.

but in the real world, families with very many children will suffer poverty, and perhaps some will die.

so their will be a darwkinian pressure towards to smaller families. the smaller families have more boys, so boy percentage will rise owing to the pressures of evolution.

It’s 50%, but I wasn’t totally sure so I ran a weenyscript emulation which over 10 billion iterations came out so absurdly close to 50/50 that there couldn’t possibly be any other answer.

i guess on a slight advantage for the girls…

start Family at an age of 20, end giving birth to childeren at an age of 50, if every jear one girl is born, this track ends boyless after 30 girls….

However this difference is so small that it would joust not be relevant… (and requires an additional number of prerequisites)

every flip of a coin has a 50% chance, that is 50% is alos the solution for the whole population…

It has to be noted that the reproduction coefficient is below 1, i.e. this country will end unpopulated…

Those of you writing computer simulations might do well to replace your random number generator with a completely reliable 50/50 simulator. For example, it could return boy, girl, boy, girl, alternating every time it is called. Or if that doesn’t seem random enough, have it fill a large buffer with 50% “boys” and 50% “girls”, and every time it is called, it gets the sex from a random position in that buffer, and then removes it from the buffer.

@Richard:

Why remove the selected sex from the buffer? The fact that a birth had a particular outcome doesn’t influence the chances of a future outcome. In addition, you’re substituting one pseudo-random generator for another by specifying “gets the sex from a random position in that buffer.”

If you’re concerned about the quality of randomness, try Random.org, which uses atmospheric noise as its basis for generating outputs.

I don’t see how it can’t be 50%: ‘ With every birth there is a 50% chance it’s a boy and a 50% chance it is a girl.’ Decisions on the number of children one couple will have, regardless of whether the decisions are based on the number of daughters or the temperature in Djibuti, cannot change the fact that with EVERY birth there is a 50% chance it’s a boy and a 50% chance it is a girl.

I still don’t get the point.

If their decision for stopping the size of the family had been different it could easily end up with a different proportion. Suppose the question had said: Each family continues to have children until they have twice as many boys as girls. Then the proportions would be 2:1, and the percentage of boys would be 66.7%

@Andrew Richards: Not so. Most families would never reach the 2:1 proportion.

Unfortunately that’s true. I was trying to give an example where the final result is obvious. However, a similiar result comes from use the stopping rule: stop when you have one more boy than girl. Every family will reach that point in a finite number of children. It’s clear that there will be more boys than girls, so the proportion of boys will be greater than 50%. But calculating the proportion seems difficult to me.

@andrew : Yes, every individual family will reach that point in a finite number of children, but to calculate the ratio we need the expected value of the number of children per family.

I am afraid this leads to infinitely many girls and infinitely many boys. And the difference between both values is a finite number : the number of families.

Therefore… 50%.

@Yat Yes, quite right.

@andrew Not so sure anymore : I don’t see what prevents me from applying the exact reasoning on one single family. Problem is, on one single family, the expected value of the ratio in these conditions is not 50%.

So I don’t think comparing infinite expected values of the number of girls and the number of boys is right. Would like to know why, though.

@Yat: I’ve not worked it out explicitly, but I think the expected value in a single family (or in any number of families) of (number of boys)/(number of children) would be smaller than 0.5 + d, where d is any positive number. This is because the probability of a very large family will be high enough to drag the expected value as close to 0.5 as you please.

Exactly. If you imagine there is a mystical man in the sky who controls the sex of all babies, no matter who the parents are, and he always flips a coin to decide the sex, then obviously the babies will be half boys and half girls. There’s no other way it can go down.

It eventually comes down to one family, though, doesn’t it?

What does cloud man’s coin flip say then?

The answer is 69% boys.

50% of families have a boy first. So they stop having children. (fraction of all children which are boys = .5)

Of the remainder, half have a boy next: 25% of families have a girl then a boy, then stop. (fraction of all children which are boys = .125)

12.5% of families have two girls then a boy, then stop (fraction of all children which are boys = .04167)

6.25% of families have three girls then a boy, then stop (.015625)

And so on…

If you make a spread sheet and run only about 20 repeats, each time halving the percentage of families going onto the next round (this is where the 50% comes in!), and then adding together all the “fractions of all children which are boys”, you end up with an answer which is close to 69%. (My resident maths nerd tells me this is an example of a Poisson Process http://en.wikipedia.org/wiki/Poisson_process and can be modelled using a Poisson Distribution.)

Note that the question only asks for the percentage of CHILDREN which are male, avoiding difficult questions of what happens in the next generation with an imbalance of genders!

By the way, thanks Richard for a puzzle I could get stuck into instead of the usual silly, obvious-when-you-know-it puzzles which aren’t really puzzles in my opinion. It seems everyone else was expecting something along the usual lines, but there is no obvious answer for this one, you have to work it out! More like this please!

Nope. Let’s say 2048 families. First batch, 1024 boys and 1024 girls are born (current ratio: 50/50).

The families who had boys stop, while the families who had girls now all have another batch: 512 boys and 512 girls (current ratio: 50/50).

The families who had boys stop, while the families who had girls now all have another batch: 256 boys and 256 girls (current ratio: 50/50).

The families who had boys stop, while the families who had girls now all have another batch: 128 boys and 128 girls (current ratio: 50/50).

The families who had boys stop, while the families who had girls now all have another batch: 128 boys and 128 girls (current ratio: 50/50).

The families who had boys stop, while the families who had girls now all have another batch: 64 boys and 64 girls (current ratio: 50/50).

The families who had boys stop, while the families who had girls now all have another batch: 32 boys and 32 girls (current ratio: 50/50).

And so on. You need to be able to point out an error in that logic if you want to suggest the answer isn’t 50%.

“Of the remainder, half have a boy next: 25% of families have a girl then a boy, then stop. (fraction of all children which are boys = .125)”

Those families have 2 childs. 25% of _families_ with 50%boys is not equal to 25% of _all children_ are 50% boys.

Consider a basic situation : 50% families have 1 boy, 50% families have 2 girls and 1 boy. There are obviously 100 boys and 100 girls in total. Now if I apply your reasoning on this case, here is what it would look like :

50% of families have just a boy. Fraction of all children which are boys = .5

50% of families have two girls and one boy. Fraction of all children which are boys = .166…

The sum is .6666… (2/3).

See ? This does not work.

I’m totally convinced by Nadia, but I’m also totally convinced by Richard. Therefore I propose the solution to be the mean of 50 and 69 which is 59.5%.

I love this puzzle. Well done Mr Wiseman — best one for a long time.

Your error is ‘adding together all the “fractions of all children which are boys”’.

Each family does not have equal weight towards the overall population of children, but only contribute a proportion based on the number of children in that family.

To correct this, you need to add together the fractions, each multiplied by the number of children in that family, then divide by the number of children in total. That should give you 50%.

Nadia, your math is incorrect. You can’t add .5 + .125 + .04167 + .015625… to get an answer because your fractions are incorrect.

Your first number is (at first) correct: Let’s assume all families have a child at the same time, so at the first step half the children will be boys and half will be girls. Fraction is .5.

However, in the next step, the TOTAL number of children will increase. So in Step 2, the first fraction is no longer .5, it will be 1/3. Why is that so? Let’s assume initially you have 2000 families; in step 1 there will be 2000 children, half (1000) of which are boys. In Step 2, the 1000 families with girls will have a second child, so the TOTAL number of children is 2000+1000=3000. The fraction of boys from step 1 is 1000/3000 (1/3) and the fraction of boys from step 2 is 500/3000 (1/6). In step 2 the total fraction is then 1/3 + 1/6 = 1/2.

Step 3: 500 more children, total children is 2000+1000+500=3500

fraction boys = 1000/3500 + 500/3500 + 250/3500 = 1/2

And so on…

It’s not a Poisson process (the most obvious way to see this is that it isn’t a continuous random variable, since you can’t have a fraction of a child), but it can theoretically be modeled with a Poisson distribution (of the # of daughters). However, doing so still wouldn’t be useful here: the problem with doing that in this example is that you can’t define a Poisson variable unless you already know its expected value, which is the property of the variable that you’re trying to compute.

I got 49.99999999999999…%

I do not have an answer, but consider that all the families with gb, ggb, gggb, ggggb, gggggb, etc. configurations, there are not bg, bbg, bbbg, bbbbg, bbbbbg, etc. configurations to outhweigh them and bring the balance of children born back to fifty percent.

Wouldn’t that influence the result?

Then again, only one daughter of the multiple girls families will, on averagem find a mate to procreate (thus setting the ration back to 50:50 in every new parent generation), and maybe that is what brings the ratio back to nearly 50 percent over long simulations.

Tricky.

Correction: Read that …consider that FOR all the families with …

The answer is 50 %.

However, the statistic that would be significantly influenced by this policy is overall birth rate, which would fall below levels sufficient to sustain a population.

I disagree. If you realize that 50% is the correct answer, then I’ll assume that you accept one of the above explanations for why the expected number of boys and expected number of girls per family are both equal to one. As such, the expected number of children per family is 2, so each generation will be exactly the same size as the one that preceded it.

50

Another way to present this that might make it clearer:

A man has a perfectly balanced coin. 1,000,000 people queue up in front of him. Each person, when they reach the front of the queue, is allowed to flip the coin as many times as they like until the coin comes up “tails”.

What proportion of the coin flips come up “heads”?

This is an excellent way of thinking about it. I think.

Ah but what if everyone in the queue flipped two heads then a tail. Because there is a finite number of 1,000,000 given, you would have 2,000,000 heads and only 1,000,000 tails. Statistically improbable with a perfectly balanced coin but not impossible.

Let us make a small variant, inspired by a comment on the previous post.

A man has a perfectly balanced coin. 1,000,000 people queue up in front of him. Each persion, when they reach the front of the queue is allowed to flip the coin as many times as they like until they have achieved one more tails than heads. By the time we get to the back of the queue, there will be 1,000,000 more tails than heads that have been flipped.

Does this work? Does every person who flips coins eventually reach a situation where they have achieved one more tails than heads, or is it possible that some people will flip away and never get there? Theory (1-dimensional random walk) tells us that in fact everyone will eventually get to the position of having flipped one more heads than tails, but it might take them a long time indeed to get there.

In fact there is nothing paradoxical about this result at all. Theory tells us that if you do flip a coin away, you must always eventually reach a position of having flipped 1,000,000 more tails than heads. You will also always eventually reach a position of having flipped 1,000,000 more heads than tails. We have simply devised a method to select the first mentioned end-point. But it is likely to take you a very, very long time indeed to get to either of these situations. A million might sound like a large number, but if, say, it has taken a quadrillion flips of the coin to get to that position, an excess of 1,000,000 heads or tails is actually, proportionately, a distortion of only about 1 in a billion away from 50/50 proportions. I can’t be bothered calculating exactly how flips you would expect to have to make.

Thus this example does not actually disturb our intuition that fair coin tossing, much repeated, tends to 50/50 distribution, however we attempt to select out. Eventually an excess of 1,000,000 will happen, and is not a large excess, proportionately.

If we say there are 32 families, then 16 will have girls, 16 boys.

In the 2nd pregnancy, the 16 families who had girls will have another child, 8g, 8b. (The other 16 families stopped after a single child)

8 families: 4g,4b

4 families: 2g, 2b

2 families: 1g, 1b

The remaining family (with 5 girls) is the only possibility of a difference in the number of boys and girls. If the starting number of families is larger (or we just let the society run of a longer time), the single family will be statistically insignificant.

If you add up the number of children after this, we have 31 girls, 31 boys. In other words: 50% of each gender.

I agree with this – and it is critical that we know it is a large population – so we can say it is indeed very close to 50%. Richard says as much in his opening lines.

The reason for this post is clear. Our little RW is conducting some sort of psychology test and is enjoying reading (and analysing?) these comments.

66.66%

If the chance of any given baby being a boy is 50%, how can more babies end up being boys?

I think you did it the same way I did because I got 66.66% too. However, I’ve changed my opinion: the argument about the distribution of male and female on each round of births has convinced me that it’s 50%.

Example:

16 familys with 1 boy

8 familys with 1 boy 1 girl

4 familys with 1 boy 2 girls

2 familys with 1 boy 3 girls

1 family with 1 boy 4 girls

In that small sample we have 5 boys and 10 girls, how does anyone find that hard to understand?

You forgot to multiply the number of children with the number of families. In those numbers you have 16+8+4+2+1 boys and 8*1+(4*2)+(2*3)+1*4.

In other words, 31 boys and 26 girls. You are also missing the family with 5 girls parallel to the last 1b4g family (Taking us to 31b, 31g with a single family getting another child in the next run)

I don’t, you do.

To get your 16 with 1 boy you need 32 families – write out the binary tree in full with the possible outcomes for those 32 and you get:

1-B

…

16-B

17-GB

…

24-GB

25-GGB

26-GGB

27-GGB

28-GGB

29-GGGB

30-GGGB

31-GGGGB

32-GGGGG

Boys: 16+8+4+2+1 = 31

Girls: 0+8+8+6+4+5 = 31

(And the final unresolved family (hence B being 31) that’s GGGGG will keep going)

But the split will always be 50:50

@Anataboga,

“(And the final unresolved family (hence B being 31) that’s GGGGG will keep going)

But the split will always be 50:50”

This is contradictory, when this family next has a baby, it will only remain 50:50 if they have exactly, a girl, then a boy with a ratio of 32:32.

Otherwise it will be:

A boy immediately, leaving a ratio of 32:31

More than one girl leaving a ratio of 32:(31+n)

where n are the number of girls had before a boy.

n would have to be 1 exactly.

Given the combinations of 2 births events (needed to make 1 girl before one boy), being BB,BG,GB,GG; the chance of GB is 1 in 4.

2 to 1 for boys… At 50/50 first round of kids will be 5 boys and 5 girls (if truely 50/50). The families with boys have no more children; so, only the ones with girls will have another. “IF” truely 50/50 the the 2nd kid will be a boy. They then stop having kids. SO will be 10 boys and 5 girls…

I like this idea, but Richard says:

This means that every family eventually has one father, one mother, one son and a variable number of daughters.

So it doesn’t work.

Your understanding of “truly 50/50” seems to be that, in each family, a boy will ALWAYS follow a girl. But this isn’t what “With every birth there is a 50% chance its a boy and a 50% chance it is a girl” means.

Add me to those saying that the answer is 50%.

Stripping away all the extraneous factors — the immortality, the monogamy, no limit on the number of children — the only significant sentence is: “With every birth there is a 50% chance its a boy and a 50% chance it is a girl.” Regardless of how each family decides when to stop having children, that sentence implies that the percentage of male children in the population will be 50%.

Correction: As safc4ever notes below, immortality is significant because it implies that the male:female ratio among the children remains the same as at birth.

Anyone claiming the answer is *not* 50% is basically claiming that the outcome of a coin flip is affected by how the decision of whether or not to flip it was made.

No. We’re just observing that with 10 flips, it’s not *always* 5 heads and 5 tails.

And I’m asking about 11 coin flips.

It’s an imaginary country. Therefore it doesn’t exist. Its population, therefore, is zero.

using 32 families as others have done.

16 fams have 1 boy only(!) this is the decider.

16 fams have one boy only and between 1 – X number of girls.

But the 1 boy families (no girls) will always even up the odds of having lots of girls when your numbers get big enough and the pattern settles down. That took me far too long to come to terms with!

If there is only one family there is a 50% chance it has 1 boy, 25% chance it has 1 girl, 1 boy, 12.5 % chance it has 2G 1B etc. So for one family there is no clear cut answer. With 2 families one family would have 1B 0G, but the other family could have any of a variable number of girls (but at least 1) and with the situation of more girls occurring with lessening likelihood but only 1B, again no clear cut answer. And since it’s theoretically possible for 1 family to have an infinite number of girls (as unlikely as that is) I say the question cannot be accurately answered.

Upon further review my guess is now: with an increasing number of families it will approach the limit 66.6% boys 33.3% girls.

You are basically arguing that the mathematics of probability do not exist.

If you conduct the experiment with a specific number of people, there is a chance that you can come up with any individual result. The point is that not all of these results are equally likely, so that if you extend the number of families towards infinity, the overall answer will tend towards something specific (in this case, 50% boys).

Hi Arthur- I believe in the mathematics of probability – If you flip a fair coin a certain number of times I believe you will approach a 50-50 split of heads and tails – but if you add a condition (such as stopping whenever a “head” is achieved) then I think the mathematics of probability yield a different result (2/3 to 1/3.) Either way, nice to hear from you!

@Tim: Try it with a real (fair) coin and see what happens.

The answer to the question as written (“What percent of the children in that country are male?”) is 50%: The outcome of one birth does not affect the outcome of the next, and therefore the total ratio of boys vs. girls is roughly equal across the entire population.

Now, if the question had been asked differently: “What is the average percentage of children in a SINGLE FAMILY who are male?”, the answer I get empirically seems to hover around 69%. I’d appreciate an explanation of that result, though, since the expected average number of births to get a boy is 2.

That answer comes from constructing a probability tree, then summing up each outcome, multiplied by its probability of happening (called taking the Expectation of the variable). Several people higher up the page took this approach.

The solution of this infinite series comes out to be the natural logarithm of 2 (ln(2) in common notation), which is as you say, around 69%.

There are a couple of things to consider here. Firstly, infertility and inability to have children at some point means that there must always be more boys than girls. Secondly, although the number of girls converges towards the number of boys, it never quite reaches the same number. Thirdly, presumably the next generation of (presumably) monogamous boys will not have enough monogamous girls to breed with as some families will need to produce (let’s say for the sake of argument) 20 years to produce a boy. So, the boys always win albeit by a small margin.

No, it doesn’t mean that there must always be more gurls than boys. Each birth has a 0.5 probability of being a girl, and a 0.5 probability of being a boy. Whether there is 1 family, 2, N, or infinity, the expected ratio stays at exactly 50%.

Yes, if you run any given particular trial you might end up starting wiith a run of boys, or a run of girls, but the expectation value is always precisely 50%, independent of number of families, or rules about which families stop when.

Imagine that you are playing Roulette (a perfectly fair one, without zero, betting amounts are unlimited). You betting system is the Martingale: bet 1$ to the black, if anytime you win then bet 1$ to the black (winning bets pay 1:1), if you loose bet 2$, then 4$, then 8$ etc. With no limits on the bet sizes in the long run you will win exactly 1$. No matter what betting system you choose, you cannot alter the chances.

Ok, at first i got 50%, then i thought a bit more, and came up with a very slightly different analysis.

Say, the population is 10 people, 5 male, 5 female. They have children, and they just all happen to be male. Done, and the population is 66% male. There are similar cases with 1 female child, and 2 that bias the population very slightly towards male. The rules allow a very tiny statistically insignificant bias towards males, since the initial population never dies off.

It’s not what percentage of the *population* is male, it’s what percentage of the *children* are male.

Uggh, this is simple, start with 128 families as an example, 64 have boys, 64 have girls. You now have 64 of each.

The 64 families that have girls try again, 32 of them are boys, 32 are girls. So you now have 96 boys and 96 girls.

The 32 that had girls from that group try again. 16 boys, 16 girls, leaving you with 112 of each.

The remaining 16 try again, 8 boys, 8 girls, 120 of each total since the beginning of the series.

The 8 that have yet to have a boy try again, 4 boys, 4 girls. There’s now 124 of each.

Only 4 families have yet to have a boy, they try again and get 2 get boys, 2 have another girl. Our running total is 122 boys and 122 girls.

The final 2 try yet again and 1 has a boy, the other a girl. Our running total is 123 of each.

The final remaining family without a boy tries again and has a 50% of getting a boy finally. They have a girl instead. There’s now 124 girls and 123 boys.

They try again and get a boy finally, leaving us with exactly 124 of each. If you repeated this series indefinitely to simulate a larger population, for every time the final family had yet another girl, there would be a time another final family had a boy, cancelling each other out. For every final family that has 2 more girls, the 50% stipulation says there will predictably be 2 families that have a boy on that final go, etc…

The answer is clearly 50% if you let this run indefinitely, the only way to get a number other than exactly 50% is stop the series at some arbitrary point where there is temporarily more of one than the other.

The problem is you’re assuming 50% is the only possible birth outcome. 50% is the median, but sometimes you will get more boys & more girls.

Doesn’t matter David, over time, it will hover around 50%. There will spikes in either direction when looking at short term trends but over all, the average will be 50%.

The answer to the question depends on when you take your sample. At any given moment it could be either slightly below or above 50%. The larger the population, the smaller that variation will be. But you could also choose a moment to take the sample when the ratio is exactly 50%. So the practical answer to the question is 50%, the technical answer is that you don’t know until you take the sample due to statistical fluctuations.

Does a hermaphrodite counts as 1/2 boy or 1 boy + 1 girl?

Clearly, the answer is 50%.

For every birth there is a 50% chance of a boy and another 50% chance of a girl. After the experiment terminates, all resulting children are the products of such births. Hence, it follows that the percentages of male and female children are equal and 50%.

Just to verify, I tested it on MATLAB, and the results were as follows:

Initial conditions:

Number of couples: 1,000,000

Final values:

Number of boys: Approx. 1,000,000

Number of girls: Approx. 1,000,000

Number of children: Approx. 2,000,000

post the code 🙂

Cheers,

Fellow MATLAB user

In every family, regardless of size, 50% of the parents are male.

Across all families, 50% of the children are male (though not necessarily within any given family).

So, overall, 50% of the population is male (and the average family size is four).

The question was to find what proportion of the children are male, yet you have just stated that with no working in the second line.

The algorithmic solution (java, easily portable to other PL):

*****************************************************************************

int families = 10000000;

int girls = 0;

Random random = new Random();

for (int i=0; i<families; i++){

while(true){

boolean boy = random.nextBoolean();

if(boy){

break;

}

girls++;

}

}

******************************************************************************

Result: number of families (=number of boys) = number of girls

I couldn’t work it out – so wrote a program

These were my results

Family Type : Probability

b:0.5000862

gb:0.2499698

ggb:0.1250319

gggb:0.0624308

ggggb:0.0312677

gggggb:0.0156073

ggggggb:0.0078048

gggggggb:0.0038844

ggggggggb:0.0019584

gggggggggb:0.0009913

ggggggggggb:0.0004844

gggggggggggb:0.000237

ggggggggggggb:0.0001239

gggggggggggggb:5.9E-5

ggggggggggggggb:3.38E-5

gggggggggggggggb:1.44E-5

ggggggggggggggggb:7.9E-6

gggggggggggggggggb:4.0E-6

ggggggggggggggggggb:1.8E-6

gggggggggggggggggggb:5.0E-7

ggggggggggggggggggggb:4.0E-7

gggggggggggggggggggggb:1.0E-7

ggggggggggggggggggggggb:1.0E-7

Counts

Men:10000000

Women:9996845

Ratio of Men to Women

Ratio:0.50007888744449

you picked a cut-off point at which to stop calculating, which is fine, but you then need to include the other possibility for the 23rd child (or whatever the length of that last line is). That would have had

ggggggggggggggggggggggg, with the same probability as the line above, taking men and women to the same amount.

Even among Immortals, there can be only one.

Who wants to live forever?

Actually, the answer is NOT “50%”.

The CORRECT answer is that the question is malformed: there is no answer.

IN THE LIMIT, as the total number of births and the total population tends to infinity, the number of males tends to 50%. But for any finite population (or finite number of generations), there will on average be more males than females; exactly how many more depends on that finite number.

If you don’t have the math skills to analyze an infinite series or the programming chops to knock up a simulation, think of it this way:

Imagine a generation with 10,000 couples. The first thing we can say is that when these families are done, there will be exactly 10,000 more males — everybody stops after one boy. The open question is how many girls? In any given generation with a finite number of couples, on average half the couples will have one son as the firstborn, and stop with no daughters. Clearly then, AT LEAST half of the children born in that generation will be male. Of the remaining couples, half will probably have one daughter then one son, and stop. Of the remainder of those, half will have two daughters then a son. And so on. With a big enough population and enough time, some families will produce an enormous number of daughters before having a son. But in order for those very large families to catch up to the single-child families plus the certainty of eventually producing a son, you would have to allow for infinitely large families and infinitely many iterations.

And this, boys and girls, is why you should not pose mathematical problems as word puzzles.

Actually, if after the 10,000 boys have all been born, you don’t have 10,000 girls, it means the odds of having a girl weren’t 50% after all.

Let’s be clear: who the parents of the child are doesn’t affect their odds of having one or the other sex. Whether they’ve already had a girl or several doesn’t affect the odds. Every time a baby is born it is exactly analogous to a coin flip.

Therefore, while you do have a point about how things don’t exactly follow the 50/50 rule in real life, the question can be rephrased as “a lot of coins are flipped. What percentage of them are heads?”. The immediate answer would appear to be “50% of course” but in real life it would be somewhere *around* there.

However, going on about this misses the obvious point of the puzzle. The trick is to make you think that the rules being used to decide whether or not to conceive could somehow affect the probabilities involved. The solution is just to see past that trick and see that 50% is still the relevant number.

If you’re thinking about algorithms to test it, you can actually keep simplifying it more and more until you end up with:

families = createFamilies()

while (moreChildrenNeeded(families))

{

foreach(family in familes)

{

if(family.childCount[male] = 0)

{

childSex = getRandomSex() // 50% chance of male

increment (overallSexCount[childSex])

increment (family.childCount[childSex])

}

}

}

boys = overallSexCount[male]

girls = overallSexCount[female]

ratio = boys / (boys + girls)

– which pretty clearly simplifies down to saying “how many randomly-generated sexes are ‘male'”.

Ah, my indentation went to pot.

In terms of probability, there is likely to be

very slightlymore boys than girls. Here’s an example in the small.Starting with 11 families, all reasonable (slavishly split between M/F) possibilities:

MMMMMMFFFFF(-f),MMFFF(=),MMF(-f),F(=)

MMMMMMFFFFF(-f),MMFFF(=),MFF(-m),MF(-m),M(=)

MMMMMFFFFFF(-m),MMMFFF(-m),MMF(=),F(-m),M(=)

MMMMMFFFFFF(-m),MMMFFF(-m),MMF(=),M(-f)

The parenthetical note shows if there is currently an imbalance (-f means missing a female, -m missing male, = means no imbalance). Note that 25% of the time, there will be one extra boy. An extra girl is always

eventuallybalanced out, but an extra boy may not be.Very clever… so this does seem to be slightly different than just: “What is the expected number of heads if I flip a fair coin N times.”

What of the odd occasion when every family has 3 daughters then a son — how does that balance out?

I need a picture:

Start

/ \

B G

/ \

B G

/ \

B G

/ \

B G

Start at the top and at every junction go left then right. = 50:50

Since every pregnancy has a 50:50 outcome then every step down the right leg of this tree is balanced by a B on the left leg.

Typical! My tree lost it’s formatting once it was posted!

Each fork is meant to step once more to the right!

Definitive Answer:

2nd sentence gives that nobody dies. This means that the population ratio in children, as requested, is identical to the birth ratio.

3rd sentence gives the birth ratio as 50% boys, therefore you have the answer given in the question: 50%

All other statements are irrelevant ‘waffle’ and are placed there to confuse and ignite debate. Obviously they worked!

Incorrect.

The condition to stop having children is far from irrelevant. If, for example, it had been ‘stop having children once they have exactly twice as many boys as girls’, then the answer would have been different (2/3 male children).

@Arthur: Many families would never reach that stopping condition, so the answer would not be 2/3 male children.

Arthur, you have fallen into the trap which was so cleverely set by Richard.

The condition of when to stop having babies determines the make-up of each individual family and does not impact on the population as a whole.

Your decision to stop when the proportions change is an impossible condition if everyone has to fulfil it as there is still going to be a 50/50 split.

Consider a coin-toss scenario. Millions of people toss coins until they get a Heads and then stop. As the chance of each toss to turn up Heads is 50%, the number of tosses turning up heads will be the same 50% across the population, but the individual people taking part in the experiment will have a varying number of tosses before they stop. The decision of when to stop impacts on the individual, not the whole population, and IS irrelevant to the question being asked.

safc4ever, I broadly agree with your points, but allow me to elaborate.

I still believe that enforcing my suggested condition on all families would not give a 50% answer; the issue is whether or not the situation is ‘impossible’. Mathematically, I am having a hard time defining what makes the decision criterion impossible, perhaps it leads to an infinite expectation of the number of children in each family?

In summary, at any stage in the process of births we can expect a 50% split of gender as you say. With an ‘impossible’ decision criterion however, we can guarantee a specific outcome different to 50%, but also expect that birthing process to never terminate (and the distribution does not converge as it does so).

Does that mean the decision criterion is irrelevant? In my opinion not, but only just…

@Arthur: The probability that a family ever has twice as many boys as girls is (9 – 3√5)/4 ≈ 0.573; see http://www.quora.com/What-is-the-probability-that-you-flip-exactly-twice-as-many-heads-as-tails-using-a-fair-coin for a derivation . This implies that more than 2 in 5 families would never meet the condition; i.e., they would continue having children indefinitely.

How to characterise what makes a decision criterion impossible is an interesting question. I don’t have a complete answer, but any criterion that requires families to stop when their percentage of male children exceeds x% (where x > 50) will be impossible in the sense of the above paragraph.

I think you should go through the door that the guard says you shouldn’t go through.

Only if the guard is teling the truth, which you don’t know. You should actually go through the door that he says the other guard would tell you not to go through.

@John how does the other guard know what the other other guard would say. I hate/love this riddle.

BTW, although the question does ask specifically for the ratio of boys to girls among children, it does not specifically restrict this to “first generation” or “original” children–but it does not matter.

Even if some or all children started new families, those families would be subject to the same constraints. The average family size would be four; every family would have one male child plus zero or more female children such that the average number of female children per family would be one (specifically: 1/2 of the families would have zero girls, 1/4 would have one, 1/8 would have two, 1/16 would have three, 1/32 would have four, …)

In such a world, there are lots of other interesting “puzzles”… e.g.: What are the odds that a male has an uncle? an aunt? What are the odds that a female has an uncle? an aunt?

Each person has one father and one mother regardless of the number of brothers and sisters they have so the answer is 50/50.

The expected daughters per family is:

0*(1/2) + 1*(1/4) + 2*(1/8) + 3*(1/16) + 4*(1/32) + …

which can be written SUM n/(2^(n+1))

which sums to 1. The expected boys per family is obviously 1 so the ratio of boys to girls is 1:1 or 50%

I’m torn. My simulation (and my gut) says 50:50, but my logic favors girls.

In my head the only possible distributions are like the following

1 child families: 50% boys + 50% girls

2 children families: 50% G-B + 50% G-G

3: G-G-B + G-G-G

4: G-G-G-B + G-G-G-G

5: G-G-G-G-B + G-G-G-G-G

…

n: (n-1)*G-B + n*G

Of course the frequency is halfed,everytime the family gets 1 bigger, but I still think girls would be more abundant.

Put more to the point

A family with X kids _either_ has X girls or X-1 girls and 1 boy.

How could it ever be 50:50?

Correctly interpreted, your distributions lead to 50-50.

Look at the situation after all families have had 1 child. As you say, at this point we have 50% boys and 50% girls. The 50% of families who had a boy now stop. Of the 50% of families who had a girl, 50% have a boy as their second child and 50% have a girl. So again the second child is split 50-50. Then the families who had 2 girls have another child, which are again split 50-50, and so on.

So in your “2 children families: 50% G-B + 50% G-G” line above, you only count the *second* child because the first child has already been counted in the first line.

You seem to be forgetting the boys born in previous steps. For example in 3 child families you have G-G-B + G-G-G but you also need to count *-B (boys from 2 child families, but not the girls since they are counted as part of G-G-*) and B (but not G since they are counted as part of G-G-*).

Try putting some real numbers in your logic and you may see where your logic is going wrong.

For example assume you start with 1024 families:

1 child families: 512 boys + 512 girls (fraction boys = 512/1024 = 50%)

2 child families: 256 G-B + 256 G-G

The math is a little more involved here:

Num boys: 512 boys (1 child families) + 256 boys (2 child families) = 768 boys

Total children: 512+(2*512) = 1536 (Only 512 from 1 child families since the 1 G families no longer exist because they had a second child, thus 2*512)

Fraction = 768/1536 = 50%

3 child families: 128 G-G-B + 128 G-G-G

Fraction boys = 896/(512+2*256+3*256) = 50%

Or alternately: boys = 512+256+128=896

girls = 256 (GB) + 128*2 (GGB) + 128*3 (GGG) = 896

Fraction boys = 50%

And so on…

I think this conversation needs a longer “gestation” period … fitting to the range of possibilities and complexities …

Strike that! Am I dumb or what? Didn’t think it through properly. 50%, Before or after birth!

All the 50%ers are making the same basic mistake.

Let’s suppose the population starts at 4, 2 m/f couple.

1) 25% of the time, each couple has a boy and stops.

2) 50% of the time there is a boy and girl born. The boy family stops having children, the other family has another child with 50% odds, etc.

3) 25% of the time, 2 girls are born, but the families keep having children.

Population 1 and 3 don’t balance, because of the rules.

Empirically, if you draw a tree from each case and add the distributions, you get about 61% boys from a two-family population.

Your population 1 occurs 50% of the time, because the chance of the first child being a boy is 50%.

David, the original problem stated that there are lots of people and therefore lots of families, not 2 families.

But given your assumption of two families, what is your answer to the question “What percent of the children in that country are male?”?

1) 100% ?

2) and 3) 2/X? (and what value does X have?)

We are not making a mistake, your analysis is incorrect.

1) 50% of the time, the family contributes 1 boy and 0 girls to the population.

2) 25% of the time, the family contributes 1 boy and 1 girl to the population.

3) 12.5% of the time, the family contributes 1 boy and 2 girls to the population.

etc.

If you add up all of these quantities separately (using a tree diagram if you like, I did) it leads to an answer of 50%.

The answer is 50%. So much debate and equations – people are thinking too hard. If a group of people toss coins and each stops once they have tossed a head, overall what % will be heads? Obviously 50%. Same applies for this puzzle.

I think the confusion caused by this puzzle (which I suffered myself) might be caused by the mental pictures people have. Imagine each family is represent by a stack of coins, and there’s a long row of stacks. Each stack has all Tails except the top one which is Heads. Asked to imagine that picture, most people would immediately assume that there are more Tails than Heads because the Heads look like a thin surface coating on top of a “mountain of Tails”.

But the problem with that image is that it doesn’t take into account the fact that Heads and Tails are equally likely, and, this being the case, the shape of the “mountain range” is probably not that of a ‘typical’ one that most people would imagine, but is actually very flat, with only a handful of tall stacks in it.

The above was certainly the mistake I made when I got the wrong answer, and now I understand that I was led by my visualisation to approach the maths from the wrong angle. Somehow I managed to calculate the expected number of girls per family as 0.5 (still not sure how), and of course that leads to the wrong answer of 66.66%.

Haha! And I’ve just realised that my answer makes even LESS sense in the light of the picture I painted above. I should have got the OTHER wrong answer of 33.33% 😛

Not my finest hour!

Something that occurs to me late at night is that, if the parents are immortal, then it is possible that one family will have an infinite number of daughters. Given that every family that has a boy then stops having children, the number of boys must therefore be finite. Since any finite number divided by infinity is as near as makes no odds to zero, the proportion of boys to girls must be infinitesimal.

Am I right?

No you aren’t, if the set is infinite, there would also necessarilly be an infinite number of boys. If you suppose a large set as opposed to an infinite one, the chances of one family going on an improbable streak of girls increases but It also increase the chances of an improbably long streak of families getting a boy first time out. Also, the larger the dataset, the smaller the impact of such a streak would be on a deviation from 50%.

So whether you suppose an infinite set or a finite set, the rules apply to both boys and girls. It’s 50%.

Confused…

You can’t have families with more than one boy, but you have families with more than one girl, Therefore there are more girls.

wrong, there will be more families that ONLY have a boy, cancelling out the families with multiple girls.

A slightly different approach:

In any generation if there is an imbalance of M/F the excess will not mate and will die out (monogamy was specified).

For those that do mate, the expected birth ratio is 1:1 as calculated by Martin S above.

The population ratio is self-correcting with 50% male.

The question does not ask about multiple generations; we are not considering the possibility of the children growing up and forming new families.

Why are so many people so stupid? The answer is given in the question. It says “

With every birth there is a 50% chance its a boy and a 50% chance it is a girl.” That is the answer. Everything else is irrelevant.Does this solution work?:

1st Child: 0.5 Boys, 0.5 Girls (100% chance of a 1st kid)

2nd Child: 0.25 Boys, 0.25 Girls (50% chance of a 2nd kid)

3rd: Child: 0.125 Boys, 0.125 Girls (25% chance of a 3rd kid)

….. etc.

Each subsequent child is less likely to exist, but equally likely to be a boy as a girl. So the numbers must be equal. I think this logic works regardless of the rules, unless you start killing your (immortal) children.

50% or a little less.

Half the first births are boys.

Some, (doesn’t make any difference which ones, but half the original number) have a second child. Half these are boys. A quarter of the original population have a third child; again half are boys, etc…

After a run of girls with no boy, the parents will stop having children, making the ratio a little less than 50%

A more interesting problem is what happens in subsequent generations; multiple child families have a mix of boys and girls, but single-child families have only boys. Does a boy with no sisters behave differently to a girl, who has a brother and maybe a sister or two? Also there are no “middle” children who are boys.

This is now one for the actuaries!

Your logic collapsed at the point at which you wrote

“

After a run of girls with no boy, the parents will stop having children, making the ratio a little less than 50%”even though everything you wrote beforehand is correct – which is that in every new group, “half of these are boys”. In other words, half of all of them (added up) are boys.

Incidentally, it doesn’t say that any of them will “stop having children”, but even if they did it wouldn’t make any difference to the 50/50 ratio.

Uhhh. 25%?

Oklahoma City Locksmith

I wrote my comment without seeing the others. What an amazing response, 200 comments.

I love that this problem divides people into 2 groups: those who logically or intuitively see the answer as obvious and those who have great difficulty discerning it.

Me too.

So, if the half who get it intuitively stop commenting, and the other half comment again until the penny drops then stop, what percentage of commenters will have the right answer?

Jeremy, obviously 50% will!

This has been an amusing discussion to follow.

I’d love to play poker with some of the posters. Keep drawing to that inside straight, Keep chasing. 🙂

I don’t understand why I’m getting different results with different approaches.

1) If I calculate the expected value of the “number of males in a family” and the “number of females in a family” I get both 2, so 50% males and 50% females.

2) If I calculate the expected value of X=”the ratio of males in a family” I get

E(X)=2/3*1/2+2/4*1/4+2/5*1/8+ … =~ 0.55

What’s wrong with my approach #2 ?

Something is wrong with both approaches.

You seem to be taking the parents into account in both approaches, but the question specifically asks for the proportion of children only. Thus your expectations should be 1 child of each gender.

Thus, calculating the expected value of your X (and using the word proportion, rather than ratio) is affected to 1×1/2 + 1/2×1/4 + 1/3×1/8… which sums to ln(2) = ~69%.

The reason this is still not 50% is that each family does not contribute to the population uniformly, but weighted by the number of children in that family. If you modify your summation to multiply each term by the number of children there, then divide by the total number of children, you should get the right answer (maths is harder this way, and I did not attempt it).

I don’t understand why I’m getting different results with different approaches.

1) If I calculate the expected value of the “number of males in a family” and the “number of females in a family” I get both 2, so 50% males and 50% females.

2) If I calculate the expected value of X=”the ratio of males in a family” I get

E(X)=2/3*1/2+2/4*1/4+2/5*1/8+ … =~ 0.55

What’s wrong with my approach #2 ?

Firstly, the question asks for the percentage of *children* that are male. Your approach 1 gives the expected numbers of males and females in a family (2 each) , from which we can infer the expected numbers of male and female children in a family (1 each), giving an answer of 50% male children.

Your approach 2 gives E(X) = 8*ln(2) – 5 =~ 0.545177. This is indeed the expected value of X = ”the ratio of males in a family”, but because it weights all families equally, you can’t use it to directly calculate the percentage of people in the population that are male.

I should have said: “… because it weights all families equally regardless of size…”

OK, you’ll love this.

Imagine all reproductive cycles are synchronised across society. By the time the biggest families have 10 children, and for ever more,

54.5% of society is a parent,

18.2% of society is a female child, and

27.3% of society is a male child – which is what the original question asked for.

No, the original question asked “What percent of the children in that country are male?” not “what percent of society…”

So umm.. i dont know or dont really remember right now if theres actually a mathematical solution for this

But if the 50 – 50 chance is by every birth, than it is also possible that they could keep getting daughters by each new birth :S or am i missing a whole big point here.. the only way i could be sure that one day the new birth will be a son is, if there was a % 51 chance that it will be a boy.

Even if there was a 99% chance of it being a boy, it’s still possible that every birth could be a girl. Just unlikely.

true

I came up with 50% by looking at combinatorial patterns. For example, with a set of 3 children you get 8 combinations of birth order. If you follow the rule of deleting all children after the first boy appears, you end up with 7 boys and 7 girls. The 8th family in this set has 3 girls, so they would have to keep trying, and there is a 50/50 chance of their 4th child being a boy or a girl. However, now they are a member of the set of families with 4 children, and when you look at the combinations and delete the children after the first boy, you see that those 16 families with have 15 boys and 15 girls.

For argument’s sake imagine that every family starts with no children, and then all give birth at the same time.

This first generation is 1/2 boys, 1/2 girls, but in fact only 1/3 of the whole population is children – 2/3 of the population (and of each family at this point) are parents.

So the split at generation 1 is: parents 2/3, female children 1/6, male children 1/6.

50% of the population has now ‘stopped’ growing, and is made up of 2 parents and 1 male child.

All of the remaining 50% of families go on to have another child, 1/2 of them a boy, 1/2 a girl.

So in generation 2 the population of society is made up of the same ‘stopped’ set of families from generation 1, and the new generation, which are represented by 25% of society who are families with 1 girl and now 1 boy (which families will now stop growing), and 25% of society who are families with 2 girls – these are the only families that will continue to expand.

At this point 75% of society is made up of families with a boy. The proportion of society made up by parents goes down (to 58%) because there are now more children than before, female children make up 1/4 of the families with 1 boy 1 girl and 1/2 of the families with 2 girls, which works out to about 19%, with boy children up to 23%.

Now 75% of society is fixed, the remaining 25% of society get jiggy with it and have another child, again 50/50 have a boy/girl…

Follow this logic through and ultimately 27.3% of society is a boy child, of which the heterosexually inclined have to compete to attract the 18.2% of society who are girl children, although that could now get really complicated…

tell me where i’ve gone wrong, please

I think your problem is that you do not realise that “There is no such thing as society”

”the new generation, which are represented by 25% of society who are families with 1 girl and now 1 boy”. Here is where you’ve gone wrong. I think you’re mixing up percentages counted as “percentage of families” and percentages counted as “percentage of total people in the society”, and that’s why your percentages counted for boy and girl children are off from generation 2 onwards.

For clarity, let’s say there are 100 people originally, 50 men and 50 women. Generation1: 50 men, 50 women, 25 boys, 25 girls. Of these, 50% of the population “stop growing”, that is, 25 men, 25 women and 25 boys stay as 25 happy families.

So generation 2 is where your percentages go off. Of these 25 families, half have now a boy, half a girl (say 12 boys and 13 girls). But the total number of people in the population is 175:

25 men + 25 women + 25 boys (the boy-as-first-child families) = 75 people

12 men + 12 women + 12 girls + 12 boys (the boy-as-second-child families) = 48 people

13 men + 13 women + 13 girls + 13 girls (the oh-we’re-not-done-yet families) = 52 people

So as you see, the families with one boy and one girl are not “25% of society”, but slightly more than that, even though they are 25% of families. And if you calculate here the percentage of society members who are boys, you come up with (25+12)/175=0.21, and girls similarly (12+13+13)/175=0.22, the difference here being just because I didn’t want to include a hermaphrodite child.

Hope this helps!

The question is about the percentage of *children* that are male, so we can ignore the parents.

After two generations, we have:

(a) 50% of families have one boy,

(b) 25% of families have one girl and one boy, and

(c) 25% of families have two girls.

Suppose we have n families. Then the expected number of boys is, from (a) n/2, and from (b) n/4. Total: 3n/4. And the expected number of girls is, from (b) n/4, and from (c), 2*(n/4) = n/2. Total: 3n/4. Equal numbers of boys and girls, as it must be given that “[w]ith every birth there is a 50% chance its a boy and a 50% chance it is a girl.”

After two generations, 4/7 ~= 57.1% of society is made up by parents. I don’t see where you have got 19% and 23% from, for the girls and boys.

The puzzle is spurious.

There is no way to tell unless you assume an infinite amount of time has passed…and how pray tell do you do observe a temporal snapshot at the point of infinity?

The problem is, true random numbers (assumed here) are random. As you approach an infinite amount of time, and therefore children, you will get infinitely close to 50%. But at any specific snapshot in time, there could be any ratio of boys to girls.

It’s certainly possible that after 100,000,000 years, every child could be female–not likely, but possible. Speaking from a quantum point of view, there are infinite ratios of boys to girls, until the point at which you “observe” a ratio–similar to the cat being both alive AND dead, until you open the box. Asking for a specific ratio forces the box to open and what you have inside is most likely very close to a 50% ratio, but what you would actually get could be anything.

This is what I don’t understand about mathematicians.

‘It’s certainly possible that after 100,000,000 years, every child could be female–not likely, but possible.’

How meaningful is that? I mean, how likely is it? And at what point does it cease to become possible?

You say that ‘…what you would actually get could be anything.’

But if you had to bet on it, I’m guessing you wouldn’t choose anything other than 50-50.

@Jeremy: What makes you think Guy is a mathematician? His “Speaking from a quantum point of view…” suggests to me a limited mathematical and scientific background.

The question asks, “what percent of the children are male?” Technically, it’s impossible to know exactly–it depends on the roll of the dice, so to speak.

But, statistically, the number is likely to be close to 50%, as the odds of having a boy with each birth is 50-50. Decisions to stop having children for any reason whatever HAS NO EFFECT on the sex of children already born or the children yet to be born. It’s that simple. I don’t know why a computer simulation is required to show this, but I’m glad it verifies this obvious fact.

Another way of looking at it: Suppose every mother goes to the hospital to have her child. We ask the hospital how many births were boys and how many were girls? Is it conceivable that the answer would depend on the fact that some of these mothers have already had a few girl children and some have had no children at all? No. The odds of a boy are 50-50 for all mothers, so we expect the overal number of boys to be about 50%.

Unbelievable how much debate there is for such a simple question. Some people are trying to be far too clever for their own good. Of course the answer is 50%. I’ve done it with a probability tree, and a a spreadsheet up to 100 girls/1 boy.

Of course, in the “real word”, there won’t be the exact number of boys as girls, due to random fluctuations and the fact that the answer relies on the sum of an infinite series, but that’s not the point of the problem. The point is to show that, despite the family planning policies of individual families, the proportion of boys and girls born is unaffected overall.

So I guess Richard is never going to confirm the answer.

i am disavowed of my previous confusion.

The child population is 50/50, for many of the reasons given above, and the air is sweet.

Don’t ask whether the population is getting larger though…

It can not be 50:50.

Each individual birth has 1:1 odds. However, for example, lets say that the odds were against them per se. So in the first round there are 0 boys born – equivalent to all tails. The next round, there is only 1 boy born. Lets say the population has only 100 families – so we now have 1:199, next round again only 1 male is born – 2:297, each family is independent of the others, so the answer approaches 0% not 50%. The problem is that you are looking simply at the group odds, and not recognizing that the percentage of total male births in each go round is not going to be necessarily 50% – some rounds may be all girls. Each coin toss has a 50/50 chance but not all rounds will be 50 heads and 50 tails, There may be only 1 head and all others land on the tail side up.

It is 50:50.

You said it yourself: “Each individual birth has 1:1 odds.” So in “…a country with a lot of people” the number of boys born will equal the number of girls born, and so the percent of boys is 50%.

Sure, if you focus on any one possible case, you can get weird percentages such as your (unlikely but possible) example in which a lot of girls are born and the answer approaches 0%. But then you have to look at an equally unlikely but possible scenario where all families have a boy as their first child and the percent of boys is 100%. Which example gives the correct answer: your example of a lot of girls or my example of a lot of boys? Neither! Nothing makes your example more likely than mine and vice versa. It is true that if you run an experiment (flip coins, etc) that “total male births in each go round is not going to be necessarily 50%” because sometimes it could be less than 50% and sometimes more than 50% (where you simply focused on it being less than 50%). If you accept that some rounds may be all girls you also have to accept that some rounds may be all boys.

This problem rests on one of the basic principles of probability: the Law of Large Numbers. If each coin toss has a 50/50 chance, it doesn’t matter that not all rounds will be 50 heads and 50 tails, it doesn’t matter than in one set of tosses they may all be tails. If you flip the coins a lot of times and track the total number of heads/tails, the actual percent of heads will be 50%.

I’m not seeing anything after the add on that video…..

The question states two important things:

1) There are a lot of families, not an infinte number, who can continue to have children for as long as they want.

2) What percentage of the CHILDREN from these families, once they have all finally had a son, are male?

So summing for population after population is incorrect because the question is only asking about the percentage of male children once all the families have had a son. At that point the puzzle is ‘over’ and we count the numbers.

Let N be the number of families having children. Take an individual family from this group and look at the results they may have.

s <– they have a son

ds <– they have a daughter then a son

dds <– they have 2 daughters then a son

ddds <— they have 3 daughters then a son

…

Dds 0 to multiple instances of D=0 to keep the balance. For a finite group of N however, this is not possible, so the distribution is relevant.

I can’t work out how to go from here but I am seeing probability curves and offsets in my head, and I am seeing the male children comprising a touch less than 50% because there is a statistical likelihood of D>0 in a finite group.

It is the same as saying “I toss a fair coin until it comes up heads, and I note how many tails there were. I then repeat this process 1000 times. At the end, I will have counted 1000 heads and a number of tails. How many tails would I expect to see, and so what percentage of the tosses were heads?”

[repost since the previous post was corrupted because symbols were incorrectly interpreted as html]

The question states two important things:

1) There are a lot of families, not an infinte number, who can continue to have children for as long as they want.

2) What percentage of the CHILDREN from these families, once they have all finally had a son, are male?

So summing for population after population is incorrect because the question is only asking about the percentage of male children once all the families have had a son. At that point the puzzle is ‘over’ and we count the numbers.

Let N be the number of families having children. Take an individual family from this group and look at the results they may have.

s (they have a son)

ds (they have a daughter then a son)

dds (they have 2 daughters then a son)

ddds (they have 3 daughters then a son)

…

Dds (they have D daughters then a son)

The higher D becomes, the less likely it is to exist in the group of N (one family may have D=500 daughters then a son but this is much less likely than a family which has D=3 daughters then a son, for example).

The correct sum is therefore one which captures the probability of D occuring, as the value of D increases, for a generalised group of N instances. The percentage of male children will drop out of this sum.

If N was infinite then it would be 50% because you could match all instances of D is greater than 0 to multiple instances of D=0 to keep the balance. For a finite group of N however, this is not possible, so the distribution is relevant.

I can’t work out how to go from here but I am seeing probability curves and offsets in my head, and I am seeing the male children comprising a touch less than 50% because there is a statistical likelihood of D is greater than 0 in a finite group.

It is the same as saying “I toss a fair coin until it comes up heads, and I note how many tails there were. I then repeat this process 1000 times. At the end, I will have counted 1000 heads and a number of tails. How many tails would I expect to see, and so what percentage of the tosses were heads?”

[Now that I think about it more I have a nagging feeling that it IS 50% even in a finite group, with a tiny bias to just below 50% caused by the probability of the last family being unmatched and having 2 or more daughters. Damn, I’m not supposed to debate this with myself! I’m off to bed!]

“I toss a fair coin until it comes up heads, and I note how many tails there were. I then repeat this process 1000 times. At the end, I will have counted 1000 heads and a number of tails. How many tails would I expect to see, and so what percentage of the tosses were heads?”

Answers:

1000 tails (on average)

50% (approximately)

You can simulate this relatively easily. Flip a coin and record each flip until you get 1000 heads. So, for example, just now I went to random.org and generated this list of 20 digits:

11010001111001011011

Assume each 1 is a head, we can then separate out each trial:

1 1 01 0001 1 1 1 001 01 1 01 1

Trial 1 had a head immediately

Trial 2 also an H

Trial 3 started with a tail and then a head was thrown

Trial 4 was 3T and then H

and so on…

At the end we have 8T and 12 H so the percent is 60%

The value is relatively far from 50% because we have a small sample.

But do this over and over with thousands or millions of tosses (or thousands or millions of families) and the average result will be 50%

Here’s a link for 2000 random 0s and 1s:

http://www.random.org/integers/?num=2000&min=0&max=1&col=20&base=10&format=plain&rnd=new

what if the number of boys is N and the number of girls is 4N so that

the ratio is 20%

seriously though, without overcounting 50 % seems (late night and all) correct because the following graph

N/1—-N/2—-N/4—-N/8—-N/16—-N/32 ….

| | | | | |

each —- adds girls half as many as there are families and each | adds boys half as many as there families.

Given enough time, the girls will be close to 100 % !!!!

Just did several simulations in Excel with more then a 1,000,000 rounds of births and got exactely (very very close) to 2/3 girls and 1/3 boys

Then it’s time to debug your simulation! 🙂

I checked it and It is not bugged, it is prety simple to do in Excel, and each iteration of 1,000,000 rounds of births is giving 1/3 boys and 2/3 girls. It does make sense that there will be more girls then boys. For each round of births there will be only 1 boy but there can 0 or several girls.

The number of boys per birth round is exactely 1

The number of girls is exactely 2:

1 50.0000% 0.5

2 25.0000% 0.5

3 12.5000% 0.375

4 6.2500% 0.25

5 3.1250% 0.15625

6 1.5625% 0.09375

7 0.7813% 0.0546875

8 0.3906% 0.03125

9 0.1953% 0.017578125

10 0.0977% 0.009765625

11 0.0488% 0.005371094

12 0.0244% 0.002929688

13 0.0122% 0.001586914

14 0.0061% 0.000854492

15 0.0031% 0.000457764

1.999481201

50% for 1 girls + 25% for 2 girls + 12.5% for 3 girls … This serie euqal exactely 2.

Try it manually with a small number of families (say, 4-8) and a fair coin.

My table did not look right:

1 X 50.00000% = 0.500000

2 X 25.00000% = 0.500000

3 X 12.50000% = 0.375000

4 X 6.25000% = 0.250000

5 X 3.12500% = 0.156250

6 X 1.56250% = 0.093750

7 X 0.78125% = 0.054688

8 X 0.39063% = 0.031250

9 X 0.19531% = 0.017578

10 X 0.09766% = 0.009766

11 X 0.04883% = 0.005371

12 X 0.02441% = 0.002930

13 X 0.01221% = 0.001587

14 X 0.00610% = 0.000854

15 X 0.00305% = 0.000458

Your error is that you are counting some girls multiple times.

For example, when you count 50% of families as having 1 girl, some of those families are counted again with 2 girls, and some counted a third time with 3 girls, and so on.

How many families have only 1 girl? Half the families start with 1 girl, but then half of those have a boy and half go on to have more than 1 girl. So only 25% of families have ONLY 1 girl.

Of the 25% that have more than one girl, half have 2 girls and then a boy and half go on to have more girls. So only 12.5% have ONLY 2 girls.

Etc.

So your table is really this:

1 x .25 = .25

2 x .125 = .25

3 x .0625 = .1875

4 x .03125 = .125

…

The series equals 1. This is the expected number of girls per family. The expected number of boys per family is also 1.

The ratio of boys to girls is thus 1:1, and the percent of boys is 50%.

Sorry I do not agree. We are looking at all the families 50% will have 1 girl, 25% 2 girls, 12.5% 3 girls… for an average of 2 girls

Let’s look at it without the math. First child 50% boy 50% girl. If it is a boy we stop if it is a girl we continue then we will have more girls then boys. If the rule was only one child per family then we will obviously have 50/50. But the rule is we continue until the family has a boy. Then to the probability of 50% girl on the first try we add the probability of a girl on the second try if the family had a girl on the first try and so on until the family has a boy.

And as I said earlier I did a simulation in Excel and it always gives a ratio very close to 2 girls for 1 boy (I did it for a population of 1,000,000).

Here’s a simulation with 8 families. All I did was alternate boys and girls.

Family: Children

1: B

2: G,B

3: B

4: G,G,B

5: B

6: G,B

7: B

8: G,G,G,B

Thus, 25% of families have exactly 1 girl and 12.5% have exactly 2 girls, as Kevin notes above.

If you’re not happy with alternating boys and girls, try it yourself with a fair coin!

OK I was wrong, all the way wrong.

My simulation was wrong I have now a 50% boy and 50% girl, as you said Nick “Then it’s time to debug your simulation!”

And the mathematical reasoning should be:

0 x 0% = 0

1 x 25% = .25

2 x 12.5%= .25

3 x 6.25% = .1875

4 x 3.125% = .125

…

Giving us an average of 1 girl on average per family.

Cool!

Two parents (1 boy + 1 girl) have a boy. This means 66% are male.

Two parents (1 boy + 1 girl) have a girl; then they have a boy (on average); this means 50% are male.

So is the answer between 50% and 66%… 7/12ths male?

Its great to see all the mathematical theory in play, but I would say it really is too random to answer. In theory, every baby has an equal chance to be a boy or a girl, however, it doesn’t always mean if you had 2 babies, one will be a boy and one will be a girl. In theory, you can flip a coin, a process with only two potential outcomes, and come up heads 100 times in a row. Its an unanswerable question.

Sorry, I would like to clarify that my response is based on the question asked, “What percent of the children in that country are male?”. Its not asking for probabilities or theories, but a definitive number. A number that is impossible to get.

Touché. Maybe it should have asked for an “expected” percentage.

So far, Keenan is the only one that’s correct. This is asking a specific percentage of males and females. All the probabilities and mathematical equations, cannot account for actual births of boys or girls. When looking at mathematical equations, you must have a known variable to have any actual findings. But because you are looking at unknown variables in the form of chance, you are left with only probabilities. The other factor is that of time. At what point and time are you going to do the count. If it really is 50%, the next time there is a birth, your percentage is now off. Sorry folks, but math can’t find the answer for everything.

Aw shucks. I’m really glad I happened upon this blog, I look forward to continued puzzles, illusions, and debate. Will Mr. Wiseman ever tell us what he thinks (or knows) the answer to be?

If you’re not going to post the answer don’t post the puzzle-if it can’t be answered you should say so,or better still not post the alleged puzzle.

If you start with 16 couples, you get 8 m 8f, the couples keep reproducing: 4m, 4f, &next time round: 2m, 2f, next time: 1 m 1f. So far so good, we now have 30 children, 15 m, 15 f, a neat 50/50 ratio. But the one couple with a female child will keep reproducing.

As long as there now no longer exists other couples to “offset” the gender of the last couple´s efforts, the outcomes suddenly start getting slanted. We have three scenarios.

The last couple produce a male, setting the male ratio of children to (previous males)/(number of children)+ 1/(total number of children)= 15/31 + 1/31 = 16/31 male. A little over 50 %, depending on total population size.

Second scenario: they produce a female, AND then a male, attaining the 50/50 ratio. It is possible to argue that the puzzle implies this, as every second born is male, every second female. But this is a question of how you interpret the actual words in the puzzle: “With every birth there is a 50% chance its a boy and a 50% chance it is a girl.” Interpreted as a coin toss, my example would be slanted to begin with, and the genders would _approach_ 50/50, getting ever closer depending on the size of the starting population. Interpreted absolutely, the last couple would produce either a male first (reaching scenario 1), or a female first, then a male (reaching this scenario, scenario 2: 50/50).

Interpreted loosely, the last couple could conceivably toss a “tail” (for “femail” ;D) over and over again in their ongoing attempts to produce a son, in principle swarming the population with females until infinity, reaching a female ratio that is anywhere between a little over 50% up until 99.9999 etc. This scenario 3 would be stretching the wording of the puzzle a little, but it really isn´t ruled out by it.

Methinks! 😀

The question asks “what percentage of the children in the country are male”.

As it is perfectly possible, albeit very unlikely, for a couple to have a very large number of girls before having a boy. One could argue that in this situation, the oldest daughters are no longer children, in which case the result would slightly bias towards male children.

You imagined, so the answer depends on what you think/imagine