Weighted Variance, one more time Unbiased weighted variance was already addressed here and elsewhere but there still seems to be a surprising amount of confusion.  There appears to be a consensus toward the formula presented in the first link as well as in the Wikipedia article.  This also looks like the formula used by R, Mathematica, and GSL (but not MATLAB).  However, the Wikipedia article also contains the following line which looks like a great sanity check for a weighted variance implementation:

For example, if values {2,2,4,5,5,5} are drawn from the same
  distribution, then we can treat this set as an unweighted sample, or
  we can treat it as the weighted sample {2,4,5} with corresponding
  weights {2,1,3}, and we should get the same results.

My calculations give the value of 2.1667 for variance of the original values and 2.9545 for the weighted variance.  Should I really expect them to be the same?  Why or why not?
 A: Yes, you should expect both examples (unweighted vs weighted) to give you the same results.
I have implemented the two algorithms from the Wikipedia article.
This one works:

If all of the $x_i$ are drawn from the same distribution and the integer weights $w_i$ indicate frequency of occurrence in the sample, then the unbiased estimator of the weighted population variance is given by:
$s^2\ = \frac {1} {V_1 - 1} \sum_{i=1}^N w_i \left(x_i - \mu^*\right)^2,$

However this one (using fractional weights) does not work for me:

If each $x_i$ is drawn from a Gaussian distribution with variance $1/w_i$, the unbiased estimator of a weighted population variance is given by:
$s^2\ = \frac {V_1} {V_1^2-V_2} \sum_{i=1}^N w_i \left(x_i - \mu^*\right)^2$

I am still investigating the reasons why the second equation does not work as intended.
/EDIT: Found the reason why the second equation did not work as I thought: you can use the second equation only if you have normalized weights or variance ("probability/reliability") weights, and it is NOT unbiased, because if you don't use "occurrences/repeat" weights (counting the number of times an observation was observed and thus should be repeated in your math operations), you lose the ability to count the total number of observations, and thus you can't use a correction factor.
So this explains the difference in your results using weighted and non-weighted variance: your computation is biased.
Thus, if you want to have an unbiased weighted variance, use only "occurrences/repeat" weights and use the first equation I have posted above. If that's not possible, well, you can't help it.
For more theoretical details, here is another post about unbiased weighted covariance with a reference about why we cannot unbias with probability/reliability type weights and a python implementation.
/EDIT a few years later: there is still some confusion as to why we cannot unbias probability/reliability weights.
First, to clarify, the difference between probability/reliability weights and repeat/occurrences weights is that probability/reliability weights are normalized, whereas repeat/occurrences weights are not, so you can get the total number of occurrences by just summing the latter but not the former. This is necessary to unbias because otherwise you lose the ability to know what I would call the statistical magnitude, what other calls polarization.
Indeed, it's like anything else in statistics: if I say that 10% of my subpopulation have X disease, what does it mean for the broader population? Well it depends on what is my subpopulation: if it's only 100 people, then my 10% figure doesn't mean much. But if it's 1 million people, then it may faithfully represent the whole population. Here it's the same, if we don't know the total N, we can't know how representative of the whole population our metric is, and hence we cannot unbias. Unbiasing is exactly the process of generalizing to the broader population.
