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emailable at "t.mathoverflow@[Google-email-entity]" for any MO, math.SE, or SE-related correspondence.

Jun 23 |
awarded | Nice Answer |

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Sep 24 |
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Nov 19 |
awarded | Good Answer |

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Jan 14 |
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Apr 16 |
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Feb 26 |
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Jan 4 |
comment |
Google question: In a country in which people only want boys
I'm not assuming any specific model, but pointing out that differences between your answer and 1/2 arise from artificial (i.e., gender-asymmetric) conditioning of the problem. Assuming $k$ families as in Zare's model or your present suggestion, is equivalent to assuming "at most $k$ boys in population", or exactly $k$ boys if it is also assumed the families complete their reproduction. No such asymmetric conditioning was part of the Google problem. Your calculations show that a symmetrical distribution can be approximated by asymmetric ones, not that the Google problem is asymmetric. |

Jan 4 |
comment |
Google question: In a country in which people only want boys
Steven, that is incorrect. The issue is whether the proportion of boys, denoted $f(B,G)$ above, is convex as a function of two variables so that $E[f(B,G)] > f(E[B],E[G]) = 1/2$. It isn't convex, as simple calculations demonstrate. (See the comments on convexity and Jensen's inequality under Douglas Zare's posting). It is convex if you condition on B (i.e., restrict f to lines B=constant), and concave if you condition on G. Such conditioning is foreign to the Google problem and imposed artificially in Doug's model. |

Jan 4 |
comment |
Google question: In a country in which people only want boys
Ratios versus differences doesn't address the main point, which is whether the family reproduction rule can break boy/girl symmetry in the underlying distribution of $(B,G)$. [It does gender-asymmetrize the allocation of boys and girls into sets called "families", but this extra structure does not play a role in the calculation requested by Google.] If the distribution is symmetrical then the proportion of girls will have expected value 1/2, because the random variables "proportion of girls" and "proportion of boys" will have the same probability distribution, and their sum is equal to 1. |

Jan 3 |
comment |
Google question: In a country in which people only want boys
@DZ: the statement "[allowing unfinished families] ... there is a higher proportion of girls when the population is larger" is in general false. It is true only in your a priori asymmetrical model conditioned on the number of families. The asymmetry arises not from the stopping rule, but because the stopping rule allows phrasing of boy/girl asymmetric conditions ("the number of boys is at most $k$") in equivalent terms without direct reference to $B$ or $G$ (i.e., "the number of families is $k$", as in your model allowing unfinished families). This asymmetry is foreign to the Google puzzle. |

Nov 17 |
awarded | Nice Question |

Sep 18 |
comment |
Values of zeta at odd positive integers and Borel's computations
As a point of comparison, there was recently a 60-70 year old who solved a notorious open problem in graph theory, the Road Coloring Problem. However, this man held a doctorate and a professorship in mathematics, and although he may not have solved any major problems before age 60 (or maybe he did, I don't know), he was known, as evidenced by publications, to have been working on the problem for years earlier with partial results, and on other mathematical problems. He did not suddenly take up engineering after no sign of interest in 30 yrs and claim to build an ultra-efficient automobile. |