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## Question

In a village in each family they give birth to children till they

get a boy. IF girl child they try again. What is the ratio of boys to

girls.

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umm...

1to1

some will get a boy on the first try and stop, others will get multiple girls first

im probably missing some thing since im not really sure

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In a village in each family they give birth to children till they

get a boy. IF girl child they try again. What is the ratio of boys to

girls.

Just to clarify, Are you saying that once they give birth to a boy they don't have any more children? If that's the case there should be more girls born than boys.

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With this tradition, the village could never survive. The number of men in every generation will never increase. Since it is a village, and it's just the way life is, some of the males will die before being able to reproduce until eventually all the males will be killed.

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umm...

1to1

some will get a boy on the first try and stop, others will get multiple girls first

im probably missing some thing since im not really sure

It can't be 1 to 1...let's give 4 instances:

Family A - boy first, no girls

Family B - 1 girl first, then boy

Family C - 2 girls first, then boy

Family D - 3 girls first, then boy

Just in that small sample, there's 6 girls, and 4 boys.

Edited by GIJeff
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eventually the ratio will be zero and no family will exist because each male will extinct.

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for primative man to survive, you need 5k ratio 2/1 in favor of females. Villages won't survive on the basis of birth control after a male is born, perhaps though that is why the ratio is in favour of females and we have this malevolence in our DNA. Too many men = confrontation. we have different reasons now of course. Mostly due to inflicting our beliefs on someone else!

Make love not war - i'm ready, willing and able.

Edited by Lost in space
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Well, it depends on the type of the community. In a polygamy society, a man can have multiple families. There might even instances that the number of boys will be more than that of the girls.

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The ratio will be exactly the same as it would be under normal circumstances (say 1:1 but it probably isn't). Each baby has a chance of being male or female which is the same regardless of what has gone before, or at what point the parents decide to stop having more.

If I toss a coin until I get heads, then stop, you might expect to get more tails than heads on average, since I can only have one heads but many tails. But this is not true. What would happen if I repeat the experiment over and over again? It would be the same as if I just kept tossing a coin without stopping, so the ratio must be 1:1.

Respect to naveed for a nice counterintuitive puzzle!

Edited by octopuppy
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Nice Logic i am convinced

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My Solution

the probablity of having a boy or having a girl is 1/2.

Lets assume that n villagers have the first baby ...

Number of boy = n/2 and numbers of girls = n/2

Now the n/2 who had girl have second baby ....

Number of boys in them = n/4

Total number of boys = n/2 + n/4 and number of girls = n/2 + n/4

Similarly after third baby

Number of boys will be n/2 + n/4 + n/8 and number of girls will be n/2 + n/4 + n/8

Ratio will be 1

there fore the ratio is 1

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The ratio will be exactly the same as it would be under normal circumstances (say 1:1 but it probably isn't). Each baby has a chance of being male or female which is the same regardless of what has gone before, or at what point the parents decide to stop having more.

If I toss a coin until I get heads, then stop, you might expect to get more tails than heads on average, since I can only have one heads but many tails. But this is not true. What would happen if I repeat the experiment over and over again? It would be the same as if I just kept tossing a coin without stopping, so the ratio must be 1:1.

Respect to naveed for a nice counterintuitive puzzle!

I'd be interested in seeing a computer simulation of this. (I'm not disagreeing at all.) Right a program that will randomly choose a 1 or 2. Let 1 be a girl and 2 be a boy. Let each choice pick again if it is a 2. Run a certain number of trials, n. Of course the number of boys will be n. I would bet the number of girls is less. Since I am not a programmer at all, I used a coin with 20 "families":

1 B

2 B

3 B

4 B

5 B

6 G B

7 G G G G B

8 B

9 G B

10 B

11 B

12 G G G B

13 B

14 B

15 G G G G G B

16 B

17 G B

18 B

19 G B

20 B

My initial thoughts were that there would be more girls since it is impossible for a family to have more than one boy. Then I realized that, according to the parameters of the problem, every family MUST have a boy. Things I'd like input on:

Odds of a family having 0 girls: 1/2?

Odds of a family having 1 girl: 1/4?

Odds of a family having 2 girls: 1/8?

Odds of a family having 3 girls: 1/16?

If you have 16 families with the above odds, you will have 1 boy per family, 8 families with 0 girls, 4 families with 1 girl, 2 families with 2 girls, and 1 family with 3 girls for a total of 11 girls for a ratio of 11 girls / 16 boys.

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The ratio will be exactly the same as it would be under normal circumstances (say 1:1 but it probably isn't). Each baby has a chance of being male or female which is the same regardless of what has gone before, or at what point the parents decide to stop having more.

If I toss a coin until I get heads, then stop, you might expect to get more tails than heads on average, since I can only have one heads but many tails. But this is not true. What would happen if I repeat the experiment over and over again? It would be the same as if I just kept tossing a coin without stopping, so the ratio must be 1:1.

Respect to naveed for a nice counterintuitive puzzle!

If the chances of having a boy or a girl are 50/50, then 50% of the households in the village will have had boys as the first child and 50% will have girls. The 50% that had girls will try again and 50% of them will have boys and 50% will have another girl - and will try again. So 25% of the village will try for a 3rd child resulting 50% of them having boys and 50% having girls...

So you end up with:

100% of the households have an eldest child: 50% boys 50%

50% have a second child: 50% boys 50% girls

25% have a third child: 50% boys 50% girls

...

but always a 50/50 split - the ratio of boys to girls is 1:1

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I think, since there is more girls than boys as if someone keeps getting girls then they will continue trying until they get a boy. If someone gets a boy on their first try then they won't try again lowering the ratio of boy even more. The ratio probably looks like this - 4:1. (Girls to Boys.)

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Odds of a family having 0 girls: 1/2?

Odds of a family having 1 girl: 1/4?

Odds of a family having 2 girls: 1/8?

Odds of a family having 3 girls: 1/16?

If you have 16 families with the above odds, you will have 1 boy per family, 8 families with 0 girls, 4 families with 1 girl, 2 families with 2 girls, and 1 family with 3 girls for a total of 11 girls for a ratio of 11 girls / 16 boys.

You are on the right track. But you stopped short. You accounted for 8 + 4 + 2 + 1 families. But that is only 15 families. The last family is has 4 girls (with a 50% chance that they have even more).

But this raises an interesting point. When you work it out with acutal figures, you always end up with at least 1 indeterminate family which may have a boy (giving 1 more boy than girls) or a girl (meaning they will try for more kid(s) and end up equalising or exceeding the number of boys).

Village of 128 families:

64 have B

32 have G B

16 have G G B

8 have G G G B

4 have G G G G B

2 have G G G G G B

1 has G G G G G G B

1 has G G G G G G G +they will try again and have a (B or G)

This givs us a:

50% chance the village will have 128 boys, 127 girls (last family is G G G G G G G + B).

25% chance the village will have 128 boys, 128 girls (last family is G G G G G G G + G B).

12.5% chance the village will have 128 boys, 129 girls (G G G G G G G + G G B)

6.25% chance the village will have 128 boys, 130 girls (G G G G G G G + G G G B)

3.125% chance the village will have 128 boys, 131 girls (G G G G G G G + G G G G B)

etc.

The odds that the ratio will favour boys is 1:2

The odds that the ratio will favour girls is 1:4

The actual odds that the ratio will be 1:1 are only 1:4.

Thanks Naveed - now my brain is going to hurt all the rest of the day.

Edited by cpotting
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All the families will have one boy.

1/2 will have no girls

1/4 will have 1 girl

1/8 will have 2 girls.

1/16 will have 3 girls.

...

the geometric infinite series 0 + 1/4 + 2/8 + 3/16 + ... + (n-1)/2n + ... = 1

The boy:girl ratio is 1:1.

Kudos, however, to octopuppy for the best answer IMHO.

He made it immediately obvious.

naveed is a close 2nd. Again, IMHO.

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the geometric infinite series 0 + 1/4 + 2/8 + 3/16 + ... + (n-1)/2n + ... = 1

The boy:girl ratio is 1:1.

Ok, bonanova - I don't get it. The above is only true for n = inifinity. For any finite number (as the number of families in the village must be) the sum is less than 1. This would mean that there would always be fewer girls than boys.

However, the example I gave earlier showed that having more boys than girls occurs only 50% of the time.

Is my math wrong?

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Ok, bonanova - I don't get it. The above is only true for n = inifinity. For any finite number (as the number of families in the village must be) the sum is less than 1. This would mean that there would always be fewer girls than boys.

However, the example I gave earlier showed that having more boys than girls occurs only 50% of the time.

Is my math wrong?

You're taking it as given that exactly 64 couples will have a boy on their first try, that exactly 32 couples will have a boy on their second try, and so on, with variability only introduced at the last stage, because you can't have exactly 1/2 a couple getting a boy on their 8th try. But this isn't quite right: the very fact that your methodology would indicate "exactly 1/2" except for your ad hoc workaround at that point indicates the need for reformulation. If you use a limited population but account for the probabilities that "the last" couple could have any number of girls, you really ought to account for the probability that not exactly 1/2 of them will have a boy on the first try, and so on. After you account for all those probabilities, the odds that the ratio will favor boys and the odds that the ratio will favor girls will be precisely equal.

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Ok, bonanova - I don't get it. The above is only true for n = inifinity. For any finite number (as the number of families in the village must be) the sum is less than 1. This would mean that there would always be fewer girls than boys.

However, the example I gave earlier showed that having more boys than girls occurs only 50% of the time.

Is my math wrong?

Well they don't tell us how many are in the village, so it's tough to take that into account.

The series is taken to infinity to compute odds. It doesn't imply there are an infinity of families or children.

We're also not certain about anything: perhaps every family had a boy as a first child, and there are no girls at all.

It's possible.

The more I think about this problem, the more it simply comes down to the given condition that the two sexes are equally likely.

Nothing about deciding to quit or have more children changes that.

Certainly, I don't mean to suggest your math is wrong; I am just adding my \$0.02.

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It all boils down to the fact that if the odds of having a boy or girl are 50/50, then no matter when a family decides to stop having children, they should have pretty close to the same amount of boys and girls. The "stop when they have a boy" line was just a very nice way of putting a spin on the stopping point. It's the same as saying "I'm going to flip a coin and stop after 10 tries." You would expect 5 heads and 5 tails since there is a 50/50 ratio. Now if you said, "I'm going to stop after I get 5 tails." You would still expect to have 5 heads and 5 tails. Of course it won't always be exact, but take enough examples and it would get closer and closer to an equal 50/50 like bonanova's infinite series shows.

Now, given that the sex of the child is determined by the father's input, it is very possible that a man or two is unable to donate that chromosome needed to produce a male and would therefore be blessed (or haunted) with having many many daughters. This could greatly tip the scales, but it is outside the scope of the riddle and makes it less fun.

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Actually the odds of boy or girl in reality are not equal. There is a slightly greater chance of women being born. That is a medical fact!

<_<

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Actually the odds of boy or girl in reality are not equal.

There is a slightly greater chance of women being born.

That is a medical fact! <_<

True. But in the spirit of the riddle, equal likelihood is generally assumed.

In this particular riddle, that's not stated.

However, the "village" in the OP might have its own statistics.

Maybe it's near Bikini, and there are radiation effects.

Your point is well taken.

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Actually the odds of boy or girl in reality are not equal. There is a slightly greater chance of women being born. That is a medical fact!

<_<

What happens in the case of twins when a boy comes out first?

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In a village in each family they give birth to children till they

get a boy. IF girl child they try again. What is the ratio of boys to

girls.

Forget the ratio... i feel these villagers should get an education.

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That was a nice one!!!

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What happens in the case of twins when a boy comes out first?

i assume you mean boy/girl twins.

as vime pointed out, these villagers should get an education. They are loaded up on facts about why they have broken the rules and statistics showing what they have done to this community, and then sent on with a stern warning and a good smack upside the head.

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