Is the distribution of the ratio of the sample variance to the populaton variance from a normal population exactly or approximately Chi Square? The reason I ask is this:  Suppose you draw a large sample from normally distributed population.  And just by chance all samples have exactly the same value--not very likely but not impossible either.  The sample mean is of course the identical value of each sample drawn.  But the sample variance is zero.  So the ratio of the sample variance to the population variance is also zero.  At zero the Chi Square distribution is also zero suggesting that there is no chance of drawing such a sample...yet I just did it.
Where is the disconnect??  What is wrong with this reasoning??  Is the ratio of sample variance to population variance only approximately Chi Square distributed?  
Many of the posted answers indicate that it is impossible to draw the exact same value from the same normal distribution more than once.  Does this mean that the “disconnect” that I mention in my original question arises from trying to apply a continuous distribution to a discrete population?
Consider the following example. Non-Senior Executive Service white-collar Federal employees are paid on the GS scale.  It consists of 15 grades with 10 steps within each grade for a total of 150 possible pay levels.  Assume for this discussion that pay is distributed normally with the exception that it is grouped into these 150 pay levels.  Clearly it is possible—not very likely but possible—to draw a large sample from this population where the same value is obtained for each observation. (Even if the sample size is large enough to exceed the number of population members in the largest pay group this could occur if sampling is done with replacement—you could get the same guy multiple times.)  As a consequence, a zero sample variance could occur.
So, I have some questions:
1.   As noted in the original question, a zero sample variance would imply a zero probability of occurrence because the ratio of the sample variance to the population variance times n-1 is Chi Square distributed.  Yet it is possible to get such an outcome.  Does this mean that because the population is not exactly normal—because of the 150 groups—that the distribution of the ratio of the sample variance to the population is only approximately Chi Square?


*Much of the world is discrete.  When dealing with discrete populations how can one proceed to estimate the mean and variance and establish confidence intervals for each without the possibility of stumbling into this sort of situation?  Is there a way to design your sampling process to avoid this?

 A: 
Suppose you draw a large sample from normally distributed population. And just by chance all samples have exactly the same value--not very likely but not impossible either. The sample mean is of course the identical value of each sample drawn. But the sample variance is zero. So the ratio of the sample variance to the population variance is also zero. At zero the Chi Square distribution is also zero suggesting that there is no chance of drawing such a sample...yet I just did it.

I think the issue here is that you are confusing impossibility with probabilistic irrelevance.  If you even take just two independent normal random variables, the probability that they will be equal to the each other is zero.  It is possible they are the same, but the probability of this event is zero.  No matter how you set the variance ratio in this instance,$^\dagger$ this event occurs with zero probability, so it is effectively irrelevant.  As you point out, the chi-squared distribution might give zero probability to this event.  That would not make it wrong --- it would make it correct.

$^\dagger$ This gives you a ratio $0/0$ which is what we call an indeterminate form.  You could set this ratio to some value by convention, and it need not be set to zero.
A: For title, the answer is "exactly". For last question, the answer is the ratio of sample variance to population variance is only exactly Chi Square distributed, if chi-square distribution has no other name.
Suppose You have $X_1,..., X_n$ from Normal distribution and $X_1=x$, then $Pr(X_1=X_2=...=X_n=x) = \int_x^x\int_x^x...\int_x^xf(x_2)...f(x_n)dx_2,...dx_n = 0 $
assuming $X$s are independent. The condition of independent is just for convenience of writing. For dependent situation, $Pr =0$ is also true.
So probability that sample variance being 0 is zero.
Let $Y$ follows chi-square distribution with pdf g(y). Then $Pr(Y=0)=\int_0^0g(y)dy=0$
So nothing is wrong.
