1-Variance $\chi^2$ Test - Odd behavior Every resource I have found states that the sample variance of a normal random variable follows a $\chi^2$-distribution such that $(n-1)\dfrac{s^2}{\sigma^2} \sim \chi^2_{n-1}$.
I am confused about how to find the $P$-value for a two-tailed test when the $s \approx \sigma$. Suppose $s$ is very slightly less than $\sigma$. I would expect that the $P$-value of a two-tailed test could be found by multiplying the left-tail area by $2$. However, this produces a $P$-value that is greater than $1$.
Many technologies instead use the right-tail area (which is less than $0.5$) and multiply that by $2$, but this does not sit well with me because (a) the sample variance is less than the population variance, and (b) this makes the upper limit for a $P$-value strictly less than $1$.
Am I missing something here? Is there a good way to calculate the $P$-value for this type of test that does not run into these issues?

Edit:
To possibly clarify: my issue is that, intuitively, it seems like $s = \sigma$ should be the least extreme result, but $P(s^2\rm{\ more\ extreme\ than\ }\sigma^2) \ne 1$ using this approach. It seems like the distribution is implying that the least extreme result is $s = \sigma\sqrt{\frac{\rm{Med}(\chi^{2}_{n-1})}{n-1}} < \sigma$.
 A: Note that the distribution is skewed. 
The mistake is to take the mean ($E(s^2) = \sigma^2$) as the dividing line. If you want equal probability in each tail - as your post suggests by the mentioning of doubling - then you would look to see which side of the median you're on, rather than the mean. [If the left tail area exceeds 0.5 you're in the right tail. Given the "doubling", the rest isn't something you can just choose to change because you feel like it]
The odd behaviour you encounter is a direct result of not looking at the relationship to the median. 
Now you can choose a different* measure of "at least as extreme"** from the one that results in equal-tail-probabilities; an appropriate choice for that may give something near to what you expect in terms of how you consider which "tail" you're in when looking at $s$ vs $\sigma$ but you'll instantly lose "double the tail area".
* perhaps the next most common choice would set the heights of the density in each tail to be the same (which gives shortest CIs) -- but this will make the effect you complain about even stronger, since then it's the relationship to the mode that says which tail you're in.
** (see the definition of a p-value)

Example
Let us explore a definition of "more extreme" that would imply $\sigma$ as the dividing line for which tail you're in.
Consider a definition where we make more extreme be based off the ratio $s/\sigma$. We now have to figure out which values of $s>\sigma$ are equally extreme to values where $s<\sigma$.  
We could, for example say that $s/\sigma$ for $s>\sigma$ corresponds to $\sigma/s$ for $s<\sigma$. This would, for example say that $s=2\sigma$ is exactly as extreme as $s=\frac12\sigma$. Equivalently, we may say that "extremeness" is based on $|\log(s/\sigma)|$.
In our example, let's compare that with two other definitions of "more extreme":
Imagine we have $n=6$, so we have $5$ df. The mean of the chi-squared distribution with $\nu=5$ is $5$, the mode is $3$ and the median is about $4.35146$.
Now imagine we observe $s=2.8$ ($s^2=7.84$) and we're testing $\sigma=3$ ($\sigma^2=9$). What's the two-tailed p-value? 
Under the null, $E(s^2)=9$, the mode of the distribution of $s^2$ is $5.4$ and the median is about $7.8326$.


*

*"more extreme" means "smaller tail area" (perhaps the most common approach in elementary texts): 
$s^2=7.84$ is at the $50.056$ percentile of the distribution of $s^2$ under the null (we're to the right of where there's 50% in each tail), so smaller tail area is to its right (not its left). The area to the right is 0.49944 and so the two-tailed p-value is 0.99888

(the gap between the left and right tails is slightly exaggerated here; they're closer together than they appear in the diagram)

*"more extreme" means "lower density" 
Since the highest density (the mode) is to the left of our sample vale, we're in the right tail, and again there's 0.49944 area to the right. The density at $s^2=7.84$ (above the mode) is the same as the density at $s^2\approx 3.5279$ which has a left tail area of $0.14534$ for a two-tailed p-value of $0.64478$


*"more extreme" means "$\log(s^2/\sigma^2)$ is further from $0$"
Here $\log(s^2/\sigma^2)$ is $-0.13799$ (i.e. we're in the left tail), and equally extreme the other side would be $9/7.84 = 1.14796$, and we want the area above 9/7.84*5 on a $\chi^2_5$. The two tail areas are now $0.50056$ and $0.33237$ giving a two-tailed p-value of $0.83293$

Note that in each case, if you're consistent in your calculations (by sticking to the definition of more extreme rather than swapping from one to the other), no p-value can exceed 1. 
[You may like to investigate likelihood ratio tests and ponder what the correct approach would be for that test in this case, with a small sample such as my example. In very large samples it will make no difference.]
