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I am not expecting a "realistic" answer to my question, since it is based on an impossible scenario. What I'm waiting for is a theoretic explanation/interpretation so that I can sleep at night :)

Let's take n reals from the interval (0; 1), evenly distributed. What is the probability of the sum of squares being less than 1? With a geometric approach it is relatively easy to see that the solution is $$P\left(\sum_{i=1}^nx_i^2<1\right)={\pi^{\frac n 2}\over 2^n\cdot{\frac n 2}!}=\frac {volume \;of\;hypersphere}{volume\;of\;hypercube}$$ This solution works fine for all nonnegative integer n (for odd n we compute the factorial using the $\Gamma$ function).

But what if n could be something else, namely a real from (0; 1)? In that case the resulting probability is greater than 1, (properly) indicating that something went wrong. Is there a way to make sense of this result?

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    What do you mean by non integral $n$? What is \12 of a number? What is $\sqrt 2$ of a number?2012-11-09
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    You may as well choose $2$ out of $-1$ objects in ${{-1}\choose 2}=-1$ ways...2012-11-09
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    @tomasz You have to be careful there since $\binom{-1}{2}$ involves (-1)! and (-3)!, which cannot be defined. You can take a limit though, which gives +1. How did you compute that -1 ways?2012-11-09
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    Dave: For every n, n!=n(n-1)! hence (-1)!=(-1)(-2)(-3)!=2(-3)!, whatever (-3)! is.2012-11-10
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    @Dave: ${\alpha \choose k}=\frac{\alpha^{\underline k}}{k!}$ (see e.g. [Wikipedia](http://en.wikipedia.org/wiki/Binomial_coefficient#Generalization_and_connection_to_the_binomial_series)). It's an extrapolation that, at first glance, doesn't make much combinatorial sense – just like yours. Although the generalization I mentioned actually does make sense in context of infinite series expansions, which I wouldn't be too sure about in your case. That's how I computed it, although I made a mistake, it's actually $1$, not $-1$. :) ${-1 \choose 1}=-1$, though...2012-11-10

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