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We have $n$ people: $\alpha n$ are boys and $(1-\alpha)n$ are girls. They are standing in a line in a random order. We pick up one boy also at random.

What can one say about the probability that there are more girls than boys before this randomly selected boy if $n\to \infty$?

Is it true that this probability is $O(1/n)$?


Yes, i meant $\alpha>0.5$. What is the correct approach to find the coefficient before $1/n$?

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No, because if $\alpha \lt \frac{1}{2}$ the probability goes to $1$. – Ross Millikan Apr 26 '11 at 21:00
You have a small issue that $\alpha n$ must be an integer, so $\alpha$ must be rational with $n$ a multiple of its lowest terms denominator. Recasting the question it becomes $b m$ boys and $g m$ girls, with $b$, $g$ and $m$ positive integers (and $b>g$), what happens as $m \to \infty$? – Henry Apr 27 '11 at 0:02
Henry, i agree but i will leave current setup – Roah Apr 27 '11 at 7:27

I suspect you forgot to specify $\alpha > 1/2$; otherwise the probability would tend to $1$, not to $0$. Yes, if $\alpha > 1/2$ the probability is $O(1/n)$, since the probability of each place in the line being chosen is the same and the probability of there being more girls than boys before a certain distance from the front doesn't depend on $n$ (or, to be more precise, depends on $n$ merely through the fact that knowing that there's one boy somewhere else slightly changes $\alpha$), and eventually falls of exponentially with the distance from the front, so the sum over the non-negligible contributions from the spots near the front eventually stops changing significantly, and from then on you're essentially just averaging over additional zeros, so the average will fall of with $1/n$.

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And this reasoning shows that $n$ times the probability converges to the finite $1+\sum\limits_{n\geqslant1}P(\mathrm{Bin}(n,\alpha)\lt\frac12 n)$. – Did Nov 27 '11 at 11:50

If we assume $n$ is large enough, the fact that you pick a boy will not change the population appreciably. I will use $a$ instead of $\alpha$ to save typing. If you pick the first in line, the chance of more girls than boys is $0$. If you pick the second, it is $(1-a)$. The third $(1-a)^2$. The fourth $(1-a)^3+3a(1-a)^2$. If you pick the $k^{\text{th}}$, it is $$\sum_{i=0}^{\lceil \frac{k}{2} \rceil -1}{{k-1}\choose{i}}a^{k-i}(1-a)^i$$ You can keep adding these up until the $(1-a)^i$ term gets small enough to get the coefficient for $\frac{1}{n}$

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Ross, i understand these probabilities but not how to get the coefficient before 1/n from them. Could you explain this idea? – Roah Apr 27 '11 at 7:29
Each of these accounts for one choice for who you picked,of which there are $n$ choices. So you should add them up and divide by $n$. – Ross Millikan Apr 27 '11 at 13:05

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