How to interpret the results of a chi square and p-values of a distribution uniformity test I am reading about a method called consistent hash designed to distribute load among servers. 
The best case scenario would be a discrete uniform distribution where each server would get the same amount of requests.
In a specific implementation I am considering using there is a table of describing the chi squared and p-values test results:

I am having difficulties understanding how to interpret the results: From what I understand a Chi square test tests a null hypothesis and that the p value is the probability that we would accept the null hypothesis even if it is wrong.
What I don't understand is how to interpret a row in the results table.
How do I associate the chi-squared test results and the value in the p-value[0.05/0.01] values to the probability that the distribution is not uniform ? 
 A: They're showing that the distribution of load is statistically indistinguishable from the uniform distribution, by (I'm guessing) partitioning the servers into 2 through 18 sets and conducting a chi-squared goodness of fit test, comparing the observed requests for each bucket with expected requests.
$\chi^2$ goodness of fit test
A $ \chi^2$ test comes up anytime you have a test-statistic that follows the $\chi^2$ distribution, i.e. the statistic is the sum of squared standard normal random variables.
One classic such situation is the chi-squared goodness of fit test of whether an observed sample came from a particular distribution.


*

*Let $X$ be some random variable.

*Let's partition the range of $X$ into $r$ buckets. (This partition is arguably somewhat arbitrary.)


If we assume that $X$ follows some distribution for our null hypothesis, we can calculate the probability $p_i$ that $X$ falls into bucket $i$ for each bucket. If we have $n$ observations, we'll expect $np_i$ entries in bucket $i$ under the null hypothesis. Let $v_i$ be the actual number of observations in bucket $i$. 
As Dmitry Panchenko's lecture notes show here, the following test statistic converges to the chi-squared distribution with $r-1$ degrees of freedom.
$$ \sum_{i=1}^r \frac{\left(v_i - np_i \right)^2}{np_i} \rightarrow \chi^2_{r-1} $$ 
A: The table reports the chi-square statistic from the test.  The values under "p-value = 0.05" are the chi-square value you would need to get to get a p-value of 0.05.  So, for two buckets you would need a chi-square value of at least 3.84 to get a significant result, and they got a chi-square value of only 0.5.  So obviously, the p-value is quite greater than 0.05 for this test.
We used to use tables of p-value thresholds like this before we all had computers that could tell us the precise p-value.
