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Boy Girl Paradox |
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| May5-08, 05:37 AM | #1 |
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Boy Girl Paradox
Here is a very famous problem:
A random two-child family with at least one boy is chosen. What is the probability that it has a girl? An equivalent and perhaps clearer way of stating the problem is "Excluding the case of two girls, what is the probability that two random children are of different gender?" |
| May5-08, 08:37 AM | #2 |
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Options: Older is a boy, younger is a boy - 1/4 Older is a girl, younger is a girl - 1/4 Older is a boy, younger is a girl - 1/4 Older is a girl, younger is a girl - 1/4 So, removing the 4th case from possibility, the odds of having two different gendered children, without knowing which is older is left as 2 of the 3 remaining possibilities. Hence, since each of the possibilities is equally probable, it's 2/3. DaveE |
| May5-08, 09:12 AM | #3 |
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Though this might be counter-intuitive (which is why it is a famous problem) I don't see the paradox.
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| May5-08, 11:19 AM | #4 |
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Boy Girl ParadoxA king of a certain distant country has decided that he wants more men in the population for military purposes. He thus decides of a new law enforcing that a couple is allowed to have only one girl. What is the boy/girl ratio resulting in the population ? (assuming we wait for several generations, after the last person born before the law has deceased, to get a stable asymptotic result for instance). A family could have say 7 boys, the eighth kid being a girl preventing any further child in the family. |
| May5-08, 12:01 PM | #5 |
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Technically, I think we need more information regarding how many kids the parents typically want. That is, parents may not WANT (or be able to have!) 6 children. So even though they might theoretically get 5 boys and then 1 girl, they may stop after (say) the 3rd son, and lower the ratio because they didn't keep procreating until they got to their natural limit. DaveE |
| May5-08, 12:04 PM | #6 |
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| May5-08, 12:13 PM | #7 |
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If parents who give birth on Tuesdays are not allowed to have more children, it would have no effect on the ratio. If you pick parents at random and tell them that them must stop having children, it would have no effect on the ratio. All of these things just reduce the number of people who are allowed to have more children. None of them has any effect on the ratio. |
| May5-08, 01:00 PM | #8 |
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Admin
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Code:
#include <stdlib.h>
#include <stdio.h>
typedef int BOOL;
int main(int argc, char* argv[])
{
int iTotal = 0, iBoys = 0;
while (1)
{
BOOL bBoy;
do
{
bBoy = rand() % 2;
if (bBoy) iBoys++;
iTotal++;
} while (bBoy);
printf("%lf\n",double(iBoys)/iTotal);
}
return 0;
}
![]() Borek -- http://www.chembuddy.com http://www.ph-meter.info |
| May5-08, 01:13 PM | #9 |
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http://www.fertilityfriend.com/Faqs/...es-Method.html and here. http://www.physicsforums.com/showthr...light=Shettles |
| May5-08, 01:21 PM | #10 |
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Now let me post my argument (I don't know if it's correct):
The chance of a family having exactly n boys when the chance for a boy or a girl is 1/2 every time (as in, binomially distributed): [tex]\left( p_\mathrm{boy} \right)^n \cdot p_\mathrm{girl}[/tex]. So expectation value of the number of boys in the family, assuming every family will have 0, 1, 2, ... boys until they eventually get a girl: [tex]E = \sum_{n = 0}^\infty n \left( \frac{1}{2} \right)^{n + 1} = 1[/tex] So we expect every family to have 1 boy, before they get a girl. That means that 1/2 of all families, and therefore of the population, will be male. |
| May5-08, 04:10 PM | #11 |
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There are several ways to answer more or less rigorously as always. A pedestrian way would be the following. The possible family configurations are obviously (1) G (2) BG (3) BBG (4) BBBG ... (N) BBBBB...G (N boys) It is not obviously trivial from this that the ratio 1:2 is conserved. It seems, and that is the initial goal of the king by issuing the law, that there will be more boys. It is of course very easy to count them and find out that it does not work. The probability for line (N) to occur is (1/2)^N, so the probability weighted total number of individual in this table is [tex]T=\sum_{n=1}^{\infty}\frac{n}{2^{n}}[/tex] while the weighted total number of boys is [tex]B=\sum_{n=2}^{\infty}\frac{n-1}{2^{n}}=\frac{1}{2}\sum_{n=1}^{\infty}\frac{n}{2^{n}}[/tex] from which we find the desired result, that the ratio will indeed be 1:2. |
| May5-08, 06:20 PM | #12 |
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| May6-08, 03:36 AM | #13 |
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Look at it this way. In the first year, every couple has a child. Half the kids are boys, half are girls. Now don't let the parents of girls have any more children. Of those parents that cannot have more children half (0) have boys and half (0) have girls. Of those parents that are allowed to have children half have boys and half have girls. Keep on doing this. Every year, the same number of boys are being born as girls. The only thing that was accomplished is that fewer and fewer people are allowed to have children. How this increases the number of boys is a mystery known only to the king.
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| May6-08, 03:41 AM | #14 |
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| May6-08, 04:01 AM | #15 |
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Recognitions:
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| May6-08, 06:13 AM | #16 |
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Math works. But are we applying it correctly? |
| May6-08, 07:19 AM | #17 |
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