Mathematical riddle

take a room
measure distances in the room, all of them (theoritically an infinity for argument sake)
from zero meters to say 15 meters (again for argument sake)
how do u calculate the average measurement knowing that there is an infinity of measurements

your question is dumb by definition

calculus

By the power of Calculus!

Let S be the set of all line segments contained in the room.  You are trying to calculate (1/|S|) integral_S |x| dx, where x is a line segment and |x| is its length.  The problem you are faced with is that you need to know what the hell dx and |S| are, and for this you need to define a metric space whose elements are line segments.  Since there's no real standard way of doing that, you can just make one up.  You could say that S is a subset of R^3 x R^3 (aka R^6) and use the Euclidean metric, integrating both starting and ending points of a line segment over the volume of the room.  Go nuts, I just solved the problem for you (because integration is the easy part, no?)

beginning example:
say the room is 15mx15m. you have defined it by a unit standard, separating it into 15^2 square meters, each a cell. determine the lengths of all possible distances; corner to corner, cell to itself, one side to the other, all of them. get your average.

now do it again, but with another unit standard you make up. make this one smaller. when you have your average, use your new well-defined unit measure and convert to meters. if the new unit was smaller than a meter, your new result will be more precise than the old one.

continue in this fashion until your unit is an arbitrarily small one, or of "infinitely small" length, so that for 15 meters, 1500 meters, or .000001 meters, converting to your arbitrarily small measure always nets you an arbitrarily large number of units (an infinity of infinitely small lengths defining a non-infinite length). Your last answer is infinitely accurate, and the only precise one.

Achieve this BY THE POWER OF CALCULUS

That integral can be approximated numerically, but can it be  given in closed form? This problem reminds me of the needle-dropping method of calculating pi (see http://mste.illinois.edu/reese/buffon/buffon.html)

One intuitive way to do this is to pick 2 random points, find their distance, and integrate over all possible pairs of points (with a probability associated with each pair).

integrate[distance(x1,x2)P(x1,x2),x1,x2], where
integrate[P(x1,x2),x1,x2] = 1
Since we want a uniform distribution, P = 1/L^2n

In the case of measuring distances on a 2 dimensional floor of size L x L,

integrate[sqrt[(x1-x2)^2 + (y1-y2)^2] (1/L^4),x1,x2,y1,y2, each from 0->L] =
L integrate[sqrt[(x1-x2)^2 + (y1-y2)^2] ,x1,x2,y1,y2, each from 0->1] =
(3 hrs on mathematica)
L(2 + Sqrt[2] + 5 ArcSinh[1])/15
~ .521405L

In case you're wondering, ArcSinh[1] = Ln(1 + Sqrt[2])

what >>8 said

calculate the expected value of the distance between two random points in your room