in a country in which people only want boys, every family continues to have children until they have a boy. if they have a girl, they have another child. if they have a boy, they stop. what is the proportion of boys to girls in the country?
Saturday, December 8, 2007
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3 comments:
My friend very deligently came up with a wonderful simulation on Excel to show that this would be a 1:1 ratio.I appreciate his effort
Assume there is an equal probability of the birth of a boy or girl in each individual case.
You could reason then, that no matter when families decide to stop having babies, there
will end up being an equal number of boys and girls. Not convinced? See the more rigorous
proof:
B is boy
G is girl
Notation used:
P(x) is the probability of a certain family birth pattern.
Prob(x) is the overall probability (but not normalized), sum of partial probabilities.
First assume there is no limit to family size (up to an infinite number of kids).
P(B) = 1/2 (probability of having one boy first)
P(G,B) = 1/4
P(G,G,B) = 1/8
P(G,G,G,B) = 1/16
Let series S = 1/2 + 1/4 + 1/8 + 1/16 + ...
S/2 = 1/4 + 1/8 + 1/16 + ... -> S = 1/2 + S/2 -> S = 1 [Equation Line 1]
Relative number of boys, Prob(B) = 1/2 + 1/4 + 1/8 + 1/16 + ... = 1
Relative number of girls, Prob(G) = (0 x 1/2) + (1 x 1/4) + (2 x 1/8) + (3 x 1/16) + (4 x 1/32) + ...
Prob(G) = S/2 + S/4 + S/8 + S/16 + S/32 + S/64 + S/128 + ...
[remember that value of series "S" = 1]
Prob(G) = 1
Result: If there is no limit to family size, the ratio of boys to girls would be 1.
------
If there is a limit to family size, these are the ratios that would result:
Family size limit of 1: Prob(B) = 1/2, Prob(G) = 1/2; ratio = 1
Family size limit of 2:
P(B) = 1/2
P(G) = 0 (they would try again for a boy)
P(G,B) = 1/4
P(G,G) = 1/4
P(B,x) = 0 (they would stop at 1 boy)
Prob(B) = 1/2 + 1/4 = 3/4
Prob(G) = 1/4 + 2/4 = 3/4; ratio(B/G) = 1
Family size limit of 3:
P(B) = 1/2
P(G) = 0 (they would try again for a boy)
P(G,B) = 1/4
P(G,G) = 0 (they would try again for a boy)
P(B,x) = 0 (they would stop at 1 boy)
P(G,G,B) = 1/8
P(G,G,G) = 1/8
Prob(B) = 1/2 + 1/4 + 1/8 = 7/8
Prob(G) = 1/4 + 2/8 + 3/8 = 7/8; ratio(B/G) = 1
Family size limit of 4:
Prob(B) = 1/2 + 1/4 + 1/8 + 1/16 = 15/16
Prob(G) = 1/4 + 2/8 + 3/16 + 4/16 = 15/16; ratio(B/G) = 1
Family size limit of 5:
Prob(B) = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 = 31/32
Prob(G) = 1/4 + 2/8 + 3/16 + 4/32 + 5/32 = 31/32; ratio(B/G) = 1
To generalize this, see my other documents as proof that ratio is 1 for all family sizes.
-Franz
Proofs from a mathematical geneous
Schiffmann, Franz
Assume there is an equal probability of the birth of a boy or girl in each individual case. You could reason then, that no matter when families decide to stop having babies, there will end up being an equal number of boys and girls. Not convinced? See the more rigorous proof:
B is boy
G is girl
Notation used:
P(x) is the probability of a certain family birth pattern.
Prob(x) is the overall probability (but not normalized), sum of partial probabilities.
First assume there is no limit to family size (up to an infinite number of kids).
P(B) = 1/2 (probability of having one boy first)
P(G,B) = 1/4
P(G,G,B) = 1/8
P(G,G,G,B) = 1/16
Let series S = 1/2 + 1/4 + 1/8 + 1/16 + ...
S/2 = 1/4 + 1/8 + 1/16 + ... -> S = 1/2 + S/2 -> S = 1 [Equation Line 1]
Relative number of boys, Prob(B) = 1/2 + 1/4 + 1/8 + 1/16 + ... = 1
Relative number of girls, Prob(G) = (0 x 1/2) + (1 x 1/4) + (2 x 1/8) + (3 x 1/16) + (4 x 1/32) + ...
Prob(G) = S/2 + S/4 + S/8 + S/16 + S/32 + S/64 + S/128 + ...
[remember that value of series "S" = 1]
Prob(G) = 1
Result: If there is no limit to family size, the ratio of boys to girls would be 1.
------
If there is a limit to family size, these are the ratios that would result:
Family size limit of 1: Prob(B) = 1/2, Prob(G) = 1/2; ratio = 1
Family size limit of 2:
P(B) = 1/2
P(G) = 0 (they would try again for a boy)
P(G,B) = 1/4
P(G,G) = 1/4
P(B,x) = 0 (they would stop at 1 boy)
Prob(B) = 1/2 + 1/4 = 3/4
Prob(G) = 1/4 + 2/4 = 3/4; ratio(B/G) = 1
Family size limit of 3:
P(B) = 1/2
P(G) = 0 (they would try again for a boy)
P(G,B) = 1/4
P(G,G) = 0 (they would try again for a boy)
P(B,x) = 0 (they would stop at 1 boy)
P(G,G,B) = 1/8
P(G,G,G) = 1/8
Prob(B) = 1/2 + 1/4 + 1/8 = 7/8
Prob(G) = 1/4 + 2/8 + 3/8 = 7/8; ratio(B/G) = 1
Family size limit of 4:
Prob(B) = 1/2 + 1/4 + 1/8 + 1/16 = 15/16
Prob(G) = 1/4 + 2/8 + 3/16 + 4/16 = 15/16; ratio(B/G) = 1
Family size limit of 5:
Prob(B) = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 = 31/32
Prob(G) = 1/4 + 2/8 + 3/16 + 4/32 + 5/32 = 31/32; ratio(B/G) = 1
To generalize this, see my attached documents as proof that ratio is 1 for all family sizes.
A different format:
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