# Simplest possible (uniform) sampling of the interval [0,1] with n points

The simplest possible sampling of a function in the region [0,1] for the purpose of finding the area under the curve, for instance, could be just take the left offsets of the bars. That is, you have divided your [0,1] into n vertical bars and sample f(left border of every bar). If we take only 2 points, they will be 0 and .5. This has unsatisfactory shift to the left.

To fix it, you decide to sample in the centers of the bars. That is, you have the same 2 sample points, you take them in .25 and .75 -- right in the middles of your two bars. Looking at it, you immediately realize that it is not better -- you sample too closely to the 0 and 1 but leave a big gap in between .25 and .75. You realize that fair sampling would be to have 3 subintervals instead of two and you divide [0,1] into n+1 sections by exactly n points. Now, two points will sample at .33 and .66 -- equidistantly both from the interval borders as well as from each other. How do you call this kind of uniform sampling?

Carry me over to the math please if you think that deterministic sampling that I address here is not suitable in probability/statistic area.

For a uniform distribution, if the range is unknown prior to sampling, a least variance unbiased range, from sampling data, is from $\frac{n-1}{n}\text{min}(X)$ to $\frac{n+1}{n}\text{max}(X)$ for random samples. For math see. The least variance unbiased estimate of location of samples is the mean of extreme values, $\frac{\text{min}(X)+\text{max}(X)}{2}$, with (much) less variance than the mean or median.
With respect to binning, a minimum variance measure of location of bin contents would not actually be at the mid-interval point unless the bin contents are indeed uniform distributed, which is not the case for only one observation in that bin. For a single value within a bin, the best measure of location is the $x$-coordinate of that observation.