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A society that prizes girls


BMAD
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Average family size tends to 2 for a large number of families and we know that each family has 1 and only 1 girl this means the other family member must be a boy

However as having an infinite number of families would be a nightmare for local councils there is a 50% chance that there is an extra girl. It is also an imposition to have an infinitely large family.

What would be the ratio be If all families give after 10 attempts at having a girl given that the number of families is n?

Since it's solved, I won't spoiler this.

The answer is not complex. Every birth, regardless of anything that is said, done, legislated, prohibited, presupposed, calculated, imagined or hoped for, has equal chances of being a boy or girl.

If you flip a coin an odd number of times, then, yes, the number of heads and tails cannot be equal. But the question is for large numbers of conceptions and births - for a society - where, even though the numbers of boys and girls may not be exactly equal at every moment - each birth changes the running total - the chances of an excess of boys exactly equals the chances of an excess of girls. The question asks whether a birth control strategy can affect the society's overall gender balance in a systematic way. The answer is no, it can not.

No calculus needed. ^_^

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In a society in which people only want

boys, every family continues to have children while they have a boy. If they have a boy, they are enthusiastic and keep making babies. If they have a girl, they stop having sex in despair. What is the proportion of girls to boys in this society?
there is a 50% chance of having a boy or girl in this society since it wasn't implied. I should had stated it as being stacked in the favor of boys to make the ratio more fun, but oh well Edited by BMAD
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The number of people in the society is not specified either. I assume, there are n married couples.

There is exactly one girl per each family. I estimate the average number of boys per family is:


1 + 1/2n * (n - 1/2).
So, just a bit more boys than girls. When they reach marrying age, I say, send the extra boys to war abroad. Make sure all the rest are married and procreate in the prescribed fashion to maintain the population.

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This topic was posted here a few years back.
At the time, I summed the series and in parallel forwarded it to a friend.


His reply was this.

There is a 50% chance if having a boy or girl in this society.

He simply highlighted in my email the words that corresponded to that statement.
It's like asking, after 999 consecutive coin flips that came up HEADS, the chance that the next flip will be HEADS.

Tossed coins and flailing sperm alike have no knowledge of the state of the outside world, nor its future or history
.

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...but I get log(2), approx .693.

The number of people in the society is not specified either. I assume, there are n married couples.

There is exactly one girl per each family. I estimate the average number of boys per family is:

1 + 1/2n * (n - 1/2).

So, just a bit more boys than girls. When they reach marrying age, I say, send the extra boys to war abroad. Make sure all the rest are married and procreate in the prescribed fashion to maintain the population.

I think there is a discrepancy with bonanova's answer because...

There is a difference between the proportion of boys to girls in a society and the average proprotion of boys in a family. Consider a society of only 2 families,

Family A: 0 boy, 1 girl

Family B: 2 boy, 1 girls

The proportion of boys in society is (2/4) = 1/2

The average proportion of boys in a family is (1/2)( 0/1 + 2/3) = 2/6

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The proportion is roughly the same, but not exactly. Since there are only

n married couples (where n happened to be a power of 2,) the following marriage procedure must be instituted.
First, the girls who have no brothers (1/2 of the families) pick one boy each from the remainnig half of the families. Then 1/4 of the girls whose only brother was taken pick their husband from the other 1/4 of the families. That leaves 1/8 of the families where all boys were taken.
And so the procedure of choosing mate must continue. In the end, one girl would remain without a match. She must go to war abroad. The number of couples in the next generation would decrease by 1.

The above procedure for choosing mate may be a hint at how to construct a closed-form expression for the series.

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...but I get log(2), approx .693.

>The number of people in the society is not specified either. I assume, there are n married couples.

There is exactly one girl per each family. I estimate the average number of boys per family is:

1 + 1/2n * (n - 1/2).

So, just a bit more boys than girls. When they reach marrying age, I say, send the extra boys to war abroad. Make sure all the rest are married and procreate in the prescribed fashion to maintain the population.

I think there is a discrepancy with bonanova's answer because...

There is a difference between the proportion of boys to girls in a society and the average proprotion of boys in a family. Consider a society of only 2 families,

Family A: 0 boy, 1 girl

Family B: 2 boy, 1 girls

The proportion of boys in society is (2/4) = 1/2

The average proportion of boys in a family is (1/2)( 0/1 + 2/3) = 2/6

Sure. The gender distribution within families is terribly skewed.

  1. A family can have an unlimited number of boys; a family can never have more than one girl.
  2. A family can have a girl with no brothers [half!]; a family can never have a boy with no sisters.
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Average family size tends to 2 for a large number of families and we know that each family has 1 and only 1 girl this means the other family member must be a boy



However as having an infinite number of families would be a nightmare for local councils there is a 50% chance that there is an extra girl. It is also an imposition to have an infinitely large family.

What would be the ratio be If all families give after 10 attempts at having a girl given that the number of families is n?
Edited by phaze
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