Gender Island

[md65536]

There is an isolated island on which all parents want a single baby girl. Every couple on the island will try to conceive as many children as needed until one of them is a girl, at which point they'll stop.

 

 

New question (after reading zapatos' answer): What is the average or expected proportion of boys to girls on this island?

 

You may assume an absence of any factors that unnecessarily complicate the puzzle, such as multiple births, infertility, deaths etc.

 

THE CHALLENGE is to figure out and justify the answer without using any maths, only logical reasoning. This is a lot easier if you already know the math solution, so don't peek!

 

The question is adapted from here: http://www.wired.co....-work-at-google where there is also the math answer, as well as some other mental exercises.

 

Note: The original question was: Would you expect there to be more girls or boys born on the island? But I think this is harder to answer with logic only!

Edited May 14, 2012 by md65536

[zapatos]

Ok, I have a guess.

 

 

I would guess an equal number of boys and girls, or more girls, depending on the last round.

 

Every round of births should result in an equal number of boy and girls.

 

When you are down to only one couple having a baby, if it is a boy they will try again and have a girl (assuming you have an exactly even split in each round). That means an equal number of boys and girls in each round, then 1 boy, then one girl. It is an even split.

 

If we are down to one couple having a baby and it is a girl, then they stop. That means and equal number of boys and girls in each round, then 1 girl. The girls win by one.

 

 

[md65536]

Ok, I have a guess.

Interesting.

 

I think I've made an error. I think what I really wanted to know is the original wired.com question: "What is the proportion of boys to girls in this country?" I think my modified question has a different answer!

 

 

 

Interesting to consider the problem in terms of "rounds". I hadn't thought of that.

I don't think it's true that each round has an equal number of girls and boys, except on average... which still makes a reasonable argument. I also don't think that the "with one couple it would alternate boy and girl" reasoning is correct.

 

But it does hilight an error I've made. If there was only one couple, they're as likely to have a girl right away (ie. more girls born on the island), as not, however if not then they're as likely to have a girl next (equal number of boys and girls), and then in all other cases more boys. "Are there more girls or boys?" isn't the same as "what's the average proportion?"!

 

... Anyway I think your answer is correct, and the reasoning is good, but I'm not sure if the logic is 100% perfect. I also think that your "depending" can be restated as depending on if there are an even or odd number of babies born!

 

 

Edited May 14, 2012 by md65536

[zapatos]

Interesting.

 

 

 

Interesting to consider the problem in terms of "rounds". I hadn't thought of that.

I don't think it's true that each round has an equal number of girls and boys, except on average... which still makes a reasonable argument. I also don't think that the "with one couple it would alternate boy and girl" reasoning is correct.

 

But it does hilight an error I've made. If there was only one couple, they're as likely to have a girl right away (ie. more girls born on the island), as not, however if not then they're as likely to have a girl next (equal number of boys and girls), and then in all other cases more boys. "Are there more girls or boys?" isn't the same as "what's the average proportion?"!

 

... Anyway I think your answer is correct, and the reasoning is good, but I'm not sure if the logic is 100% perfect. I also think that your "depending" can be restated as depending on if there are an even or odd number of babies born!

 

 

 

 

 

Even though this was set up as a one time event, I kind of looked at it as an average so that I could draw some conclusions. That is why I assumed equal number of boys and girls in each round, and alternating boys and girls at the end. I felt if I didn't then there was no way to answer.

 

I agree with your "depending" restatement. Not being a math guy I missed that.

 

I'll see if I can come up with something on your new question.

 

 

[md65536]

I think that you've answered the question for a specific set of cases... a "proof by example". It's not obvious (at least to me!) that the example applies to the general case. But I think the reasoning is the same or similar for your case and the general case. Also, I did say you could make assumptions to avoid complications, and I didn't limit that, so I suppose it's an acceptable answer!

 

The new question can be answered more simply, but not necessarily more easily reasoned about???

Edited May 14, 2012 by md65536

[michel123456]

Take ten couples.

1. They all conceive at the same time

The first couple makes a girl, bingo!

But the other 9 pregnant cannot stop, they make children.

So that makes an equal average of boys & girls.

1.2. If no-one makes a girl, which is higly improbable, its a boy win for all other rounds.

 

2. they do not conceive at the same time, but a 9 months interval so that they can see the result of each try.

2.1. the first couple makes a girl: bingo! Girls win.

2.2. the first couple makes a boy: the 2nd couple go to work (note that it doesn't matter how many couples)

2.2.1. the 2nd couple makes a girl, bingo but its fifty-fifty with the boy of point 2.2.

2.2.2. the 2nd couple makes a boy. its a boy win and for all the successive tries.

 

So globally I think the odds are for the boys.

[John Cuthber]

Well, here's my take on it.

 

 

50% of babies will be girls. What the parents want doesn't affect this.

For each couple who "gets lucky" at the first attempt, there will be a couple who doesn't.

Ditto for each subsequent child.

 

I think the average still runs to 50% because they don't (and can't) do anything to stop boys being born or to ensure that girls are born.

or to put it another way, the population is composed of people who are born. Half the people born are female. Therefore half the population are female.

 

 

Edited May 14, 2012 by John Cuthber

[michel123456]

There is an isolated island on which all parents want a single baby girl. Every couple on the island will try to conceive as many children as needed until one of them is a girl, at which point they'll stop.

(...)

 

That's the point. without the "at which point they'll stop" it would be a regular population.

 

-----------

hum, I suddenly realized I misunderstood the question.

"every couple"...

Edited May 14, 2012 by michel123456

[md65536]

hum, I suddenly realized I misunderstood the question.

"every couple"...

Sorry for the ambiguity. I meant that each couple individually keeps trying for a girl. A couple only stops when they have their own girl.

[John Cuthber]

It doesn't matter that they stop.

 

 

 

If there's no child then that couple doesn't affect the stats. A first child of any couple has a 50: 50 chance of being a girl.

A second child either isn't begotten or has a 50:50 chance. A third, or subsequent, child is in the same position. No child is ever born with anything other than a 50:50 chance.

The overall outcome is 50:50

No action (such as a decision to stop having children) taken after a child is born can stop that child having a 50:50 chance.

 

You can point at each child and say "you had a 50% chance of being a boy". Adding a whole lot of 50% chances gives an average chance of 50%

 

 

Edited May 14, 2012 by John Cuthber

[md65536]

Well, here's my take on it.

 

 

50% of babies will be girls. What the parents want doesn't affect this.

For each couple who "gets lucky" at the first attempt, there will be a couple who doesn't.

Ditto for each subsequent child.

 

I think the average still runs to 50% because they don't (and can't) do anything to stop boys being born or to ensure that girls are born.

or to put it another way, the population is composed of people who are born. Half the people born are female. Therefore half the population are female.

 

 

 

Yes, this is the answer that I was going for, explained more simply than I could!

 

 

 

Note that the answer is the same whether you allow complicated situations or not (twins, deaths, parents stopping after 5 attempts, etc) as long as none of the factors bias the outcome of an individual birth.

 

 

 

[michel123456]

i stupidly thought they wanted a queen on the island...

[md65536]

Note: The original question was: Would you expect there to be more girls or boys born on the island? But I think this is harder to answer with logic only!

Does this question have a different answer then? If so, why?

Or to be clearer: If there is an unequal number of girls and boys, would it be more likely that there are more girls or boys?

 

 

 

 

Note: I think it is a bad assumption that the process can end successfully (with each couple having a girl) after any finite number of births, because that can affect the results. I think if one assumes that there is some reasonable limit to the number of children born, then it is not guaranteed that every couple will have a girl, no matter how high you make that limit.

 

 

Edited May 14, 2012 by md65536

[michel123456]

from up to down it goes like this for each couple.

At the first attempt, the odds for boy & girl are equals for each couple.

Here below a couple who gets a girl at the fifth attempt.

But each branch shows what happen for all the island couples.

Say they are 100 couples.

At the first attempt, they get 50 boys, 50 girls.

the 50 girls-awarded couple stop, the other 50 continue.

At the second attempt, 25 couples get a girl, 25 get a boy.

At the third attempt, 12,5 couples get a girl (how is that possible?) and 12,5 get a boy.

At the fourth attempt, 6,25 couple get a boy, 6,25 get a girl.

And so on.

So this approach tells there is a 50/50 probability, but the amount of tentatives is a factor of the population (for a population of a million there are more branches than for a hundred), which must be wrong.

IMHO it is higly improbable not to get a girl after the 5th or 6th tentative. Girls should win.

Edited May 14, 2012 by michel123456

[John Cuthber]

Does this question have a different answer then? If so, why?

Or to be clearer: If there is an unequal number of girls and boys, would it be more likely that there are more girls or boys?

 

 

 

 

Note: I think it is a bad assumption that the process can end successfully (with each couple having a girl) after any finite number of births, because that can affect the results. I think if one assumes that there is some reasonable limit to the number of children born, then it is not guaranteed that every couple will have a girl, no matter how high you make that limit.

 

 

Since the original question says

"You may assume an absence of any factors that unnecessarily complicate the puzzle, such as multiple births, infertility, deaths etc."

The spoiler doesn't apply

[md65536]

Since the original question says

"You may assume an absence of any factors that unnecessarily complicate the puzzle, such as multiple births, infertility, deaths etc."

The spoiler doesn't apply

 

Do you mean that allowing such an assumption makes the puzzle easier? I disagree that it doesn't apply, because it changes the puzzle and changes the answer... I think if it's a complication it's a necessary complication.

 

I think the spoiler you quoted does apply, because it was used in zapatos' reasoning (that the process had been completed) and in michel123456's (that improbable cases that don't end successfully can be ignored).

 

 

And yet... I believe it is not always equally likely that there are more girls, as that there are more boys...

 

So this approach tells there is a 50/50 probability, but the amount of tentatives is a factor of the population (for a population of a million there are more branches than for a hundred), which must be wrong.

For a mathematical reasoning, see question 5 in the linked article: http://www.wired.co....-work-at-google

 

 

 

----

 

Here's another question that should be easier to differentiate from the thread's main (corrected/"new") question:

Would you expect there to be a greater number of families with more girls, or with more boys?

I'm also curious about what your first instinct might be.

Edited May 14, 2012 by md65536

[zapatos]

Here's another question that should be easier to differentiate from the thread's main (corrected/"new") question:

Would you expect there to be a greater number of families with more girls, or with more boys?

I'm also curious about what your first instinct might be.

 

 

I'd expect a greater number of families with more girls.

 

In the first round, 1/2 of families have a girl first and stop trying, so all of these families have more girls than boys.

 

It is possible for a family that had a boy first to then have a girl and stop trying. That family will have an even number of boys and girls. If this happens even once, the number of families with more boys than girls will be less than 50%.

 

 

[John Cuthber]

I don't think that one of the requirements of the first puzzle changes the first puzzle. It defines it.

If there are no deaths and no cases of infertility etc then people will (in an idealised way) go on having kids forever until they (and as I read it that means each couple) gets a girl.

You don't need to worry about how many children they are actually likely to have.

That simplifies the problem enormously.

 

 

 

 

Never mind the maths of summing an infinite series.

No child is ever born to whom you could not say "your chances of being a boy were 50%.", so the odds are 50%. So , by simple logic, half the population will be boys and the other half will be girls.

 

 

[md65536]

I don't think that one of the requirements of the first puzzle changes the first puzzle. It defines it.

If there are no deaths and no cases of infertility etc then people will (in an idealised way) go on having kids forever until they (and as I read it that means each couple) gets a girl.

You don't need to worry about how many children they are actually likely to have.

That simplifies the problem enormously.

 

 

 

 

Never mind the maths of summing an infinite series.

No child is ever born to whom you could not say "your chances of being a boy were 50%.", so the odds are 50%. So , by simple logic, half the population will be boys and the other half will be girls.

 

 

I see what you mean now.

The interesting thing is that I think if you remove the "simplification", the answer is the same and your reasoning still holds. But if you (ie. others) try to simplify the problem with bad assumptions, it doesn't just simplify it but changes it. I should have been more precise about what may be assumed.

 

 

 

 

 

I'd expect a greater number of families with more girls.

 

In the first round, 1/2 of families have a girl first and stop trying, so all of these families have more girls than boys.

 

It is possible for a family that had a boy first to then have a girl and stop trying. That family will have an even number of boys and girls. If this happens even once, the number of families with more boys than girls will be less than 50%.

 

 

Yes, I agree.

 

 

Similarly, for all the possibilities that might happen on the island (given some specific number of couples, or considering all possible number of couples, too), there are a greater range of possibilities that involve more boys (there can be infinite boys, but never more girls than the number of couples), but these possibilities are rarer. By asking "are there more girls or boys" I threw away the influence of the rare cases where there are *many* more boys than girls, which have a greater relative impact on the proportion.

 

Or in other words, if you ran a simulation of this many times, using any number of couples you wanted, you should find that the average proportion of girls to boys tends toward 0.5, but that there are more cases where there are more girls, and a few cases where there are a lot more boys.

 

I think...

 

 

 

[michel123456]

Do you mean that allowing such an assumption makes the puzzle easier? I disagree that it doesn't apply, because it changes the puzzle and changes the answer... I think if it's a complication it's a necessary complication.

 

I think the spoiler you quoted does apply, because it was used in zapatos' reasoning (that the process had been completed) and in michel123456's (that improbable cases that don't end successfully can be ignored).

 

 

And yet... I believe it is not always equally likely that there are more girls, as that there are more boys...

 

 

For a mathematical reasoning, see question 5 in the linked article: http://www.wired.co....-work-at-google

 

 

 

(...)

 

From your link

5. Imagine a country where all the parents want to have a boy. Every family keeps having children until they have a boy; then they stop. What is the proportion of boys to girls in this country?

 

Ignore multiple births, infertile couples, and couples who die before having a boy. The first thing to realize is that every family in the country has, or will have, exactly one boy when they're done procreating. Why? Because every couple has children until they have a boy, and then they stop. Barring multiple births, "a boy" means one boy exactly. There are as many boys as completed families.

 

A family can have any number of girls, though. A good way to proceed is to take an imaginary census of girl children. Invite every mother in the country to one big room and ask on the public-address system: "Will everyone whose first child was a girl please raise her hand?"

 

Naturally, one-half of the women will raise a hand. With N mothers, N/2 would raise their hands, representing that many firstborn girls. Mark that on the imaginary tote board: N/2.

 

Then ask: "Will everyone whose second child was a girl please raise -- or keep raised -- her hand?"

Half the hands will go down, and no new hands will go up. (The mothers whose hands were down for the first question, because their first child was a boy, would not have had a second child.) This leaves N/4 hands in the air, meaning there are N/4 second-born girls. Put that on the tote board.

 

"Will everyone whose third child was a girl raise or keep raised her hand?" You get the idea. Keep this up until finally there are no hands still up. The number of hands will halve with each question. This produces the familiar series

 

(1/2 + 1/4 + 1/8 + 1/16 + 1/32 + . . .) xN

 

The infinite series sums to 1 (x N). The number of girls equals the number of families (N) equals the number of boys (or very close to it). The requested proportion of boys to girls is therefore 1 to 1. It's an even split after all.

 

Bolded mine.

 

The question is reversed, boys for girls and girls for boys.

 

As I said before, it is very very unlikely that a couple need to fornicate infinitely before getting the expected result (girl in the OP, boy in the link). After 5 or 6 attempts the probability of succeding is already very close to 1. And anyway the population is not infinite, so it is not an infinite serie, but a finite one.

[md65536]

As I said before, it is very very unlikely that a couple need to fornicate infinitely before getting the expected result (girl in the OP, boy in the link). After 5 or 6 attempts the probability of succeding is already very close to 1. And anyway the population is not infinite, so it is not an infinite serie, but a finite one.

Ah yes, but the math is still sound. If you ignore the infinite series after some arbitrary cut-off because it becomes improbable, you must accept that a couple might reach that cut-off and not have a girl. Say you choose 20. If a couple has 19 boys, it is a mistake to think that the next one will surely be a girl because 20 boys in a row is improbable. The probability of the next child being a girl is still 50% (assuming this is still just a random case).

 

Such a case with 20 boys should happen once in roughly a million cases. So especially if it's a big island, it can happen.

As for 5 or 6... 1/32 and 1/64 couples respectively will have that many boys.

 

Regardless of any actual case, the precise proportion of girls to boys expected is 1:1.

You can ignore the improbable unlikely cases, and actually compute the epsilon error in your approximation. The higher you choose your cut-off, the smaller epsilon you'll have, and the closer your estimate will get to 0.5.

 

 

 

The moral of the story is that if there is a simple reason that some maths will work out some specific way, then it isn't a coincidence that they work out. Even if it involves adding up some complicated infinite series or considering every possible case separately, it will add up to an equivalent simple answer.

 

 

For example, if you change the puzzle to make it more reasonable and say "After 6 boys, a couple will stop trying", then you've removed the infinite series, but the proportion is still 0.5.

--- Oops I mixed up an infinite series of babies and one of population. But hopefully the point still makes sense.

Edited May 15, 2012 by md65536

[michel123456]

Ah yes, but the math is still sound. If you ignore the infinite series after some arbitrary cut-off because it becomes improbable, you must accept that a couple might reach that cut-off and not have a girl. Say you choose 20. If a couple has 19 boys, it is a mistake to think that the next one will surely be a girl because 20 boys in a row is improbable. The probability of the next child being a girl is still 50% (assuming this is still just a random case).

 

Such a case with 20 boys should happen once in roughly a million cases. So especially if it's a big island, it can happen.

As for 5 or 6... 1/32 and 1/64 couples respectively will have that many boys.

 

Regardless of any actual case, the precise proportion of girls to boys expected is 1:1.

You can ignore the improbable unlikely cases, and actually compute the epsilon error in your approximation. The higher you choose your cut-off, the smaller epsilon you'll have, and the closer your estimate will get to 0.5.

 

 

 

The moral of the story is that if there is a simple reason that some maths will work out some specific way, then it isn't a coincidence that they work out. Even if it involves adding up some complicated infinite series or considering every possible case separately, it will add up to an equivalent simple answer.

 

 

For example, if you change the puzzle to make it more reasonable and say "After 6 boys, a couple will stop trying", then you've removed the infinite series, but the proportion is still 0.5.

--- Oops I mixed up an infinite series of babies and one of population. But hopefully the point still makes sense.

It is exactly like coin tossing. Head and tails are boys & girls.

Mathematics aside, you can make the experiment (I did)

How many tossing do you need to get Head?

One, two, three?

In my small experiment, over 10 tossings, I got

Head at the first toss 5 times,

Head at the 2nd toss 4 times

Head at the 3rd toss 1 time.

 

I wonder how many tosses do I have to endure before getting Head at the 4th toss, how many ours to get Head at the 5th toss, how many days before getting Head at the 6th toss. Not speaking about getting Head after the 20th toss (or after infinite tosses).

Edited May 15, 2012 by michel123456

[John Cuthber]

"I wonder how many tosses do I have to endure before getting Head at the 4th toss, how many ours to get Head at the 5th toss, how many days before getting Head at the 6th toss. Not speaking about getting Head after the 20th toss (or after infinite tosses). "

 

I will try to save you the trouble of all that tossing.

I presume that you mean getting a head at the 4th toss, but not before.

the odds should be 1 in 2 for the first toss, 1 in 4 for the second, 1 in 8 for the third 1 in 16 for the 4th and so on.

The odds on only getting a head at the 20th toss are close to one in a million, but I suspect that MD65536 could tell you the exact odds (or at least for the 16th).

 

None of this makes any difference to the population of the island.

 

Each and every islander had a 50% chance of having been born male so half of them are male, and the other half are female.

No maths required.

 

It also doesn't matter if and when (or why) the couples stop having children.

Every baby has a 50 % chance of being male (at least in terms of maths problems, actually the numbers born are not exactly 50:50).

All the population were once babies.

The population is 50% male (on average).

Edited May 15, 2012 by John Cuthber

[md65536]

I wonder how many tosses do I have to endure before getting Head at the 4th toss, how many ours to get Head at the 5th toss, how many days before getting Head at the 6th toss. Not speaking about getting Head after the 20th toss (or after infinite tosses).

 

This video explains some of the probability stuff:

The probability of an arbitrary sequence of coin tosses being eg. T,T,T,H is the same as for T,T,T,T since H and T are equally probable.

 

 

You should average nearly 2 tosses per attempt (just like the average number of children is 2 per couple).

If you can toss once every 3 seconds, that would be about 102 minutes for a sequence of 10 (though that's faster than the videos).

For a sequence of 20, working 8 hours a day, I figure 218.45 days!

3s * 2 tosses * 2^20 / 3600 s per hr / 8 hr per day

But as explained in the video, that would be the average time it takes to get an average of 1 success.

 

I suspect that MD65536 could tell you the exact odds (or at least for the 16th).

Haha yes, that would be... 1 in uh... some really big number that I can't quite seem to remember...

Edited May 16, 2012 by md65536

[michel123456]

This video explains some of the probability stuff:

The probability of an arbitrary sequence of coin tosses being eg. T,T,T,H is the same as for T,T,T,T since H and T are equally probable.

You should average nearly 2 tosses per attempt (just like the average number of children is 2 per couple).

If you can toss once every 3 seconds, that would be about 102 minutes for a sequence of 10 (though that's faster than the videos).

For a sequence of 20, working 8 hours a day, I figure 218.45 days!

3s * 2 tosses * 2^20 / 3600 s per hr / 8 hr per day

But as explained in the video, that would be the average time it takes to get an average of 1 success.

 

 

Haha yes, that would be... 1 in uh... some really big number that I can't quite seem to remember...

 

bolded mine.

 

I saw this argument also in post #21

 

Say you choose 20. If a couple has 19 boys, it is a mistake to think that the next one will surely be a girl because 20 boys in a row is improbable. The probability of the next child being a girl is still 50% (assuming this is still just a random case).

 

And in wiki also (looking for the quote, I lost it maybe in the french wiki)

 

It is clear that each toss has a 50/50 chance to happen, and since each toss is indenpendent from the precedent one, the chance for any result to happen shouldn't depend from the precedent ones.

Shouldn't.

But if you make the experiment, does that happen?

 

If you make T,T,T, (after about 30 tosses, I got it), I can bet a lot that the next toss will be H. (in my small experiment I won my bet)

Or in the unlikely event you get T,T,T,T,T,T,T,T,T,T,T,T,T,T,T,T,T,T,T, I can bet Greece's debt that the next toss will be H (assuming this is still just a random case).

 

I suppose that the explanation resides in that we expect the result to happen in a particular way about a phenomena that is random by nature.

Inside randomness, the T,T,T,T,T,T,T,T,T,T,T,T,T,T,T,T,T,T,T, configuration exists somewhere, but WE give it a spectacular value. So is the T,H,T,H,T,H,T,H,T,H,T,H,T,H,T,H configuration but nobody cares about it.

When WE expect a serie to happen (T,T,T), this serie may get a long time to come, and the next serie (T,T,T,T) may get longer. I suspect that the probability to get the whole serie is the same that to get the final toss, IOW the probability to get a particular final toss must be the same as the probability to get the whole serie.