Intuitive explanation for dividing by n-1 when calculating sample variance? I understand how to mathematically show that the sample variance (that involves dividing by n-1) is an unbiased estimator of the population variance (which divides by n), and the mathematics has been shown many times here on Math.SE.
I am wondering however if there is an intuitive way to understand this result that I can use to easily explain why this is done to layman.  So far I have seen many derivations but I haven't seen an elegant intuitve explanation for the result.
 A: At an elementary level it is possible to give a couple of "reasons" for dividing
by $n - 1$. (At higher levels there are rationales that involve
discussions about n-dimensional vector spaces, but let's not go there now.)
"Reason 1." Suppose you are finding the sample variance of observations 2, 3, 1, 6.
Then you computations might look like this:
        x   x - 3  square
        -----------------
        2    -1     1
        3     0     0
        1    -2     4
        6     3     9
       ----------------
   Tot 12     0    15
 Mean 3          Var = 15/3

If somehow one of the four rows between dashed lines got smudged and was
unreadable, you would be able to reconstruct it from the rest of the
information. (2 + 3 + 'smudge' + 6 = 12; what is 'smudge'? Etc.) So in some sense, given the structure of the computation
you have only $n - 1 = 3$ rows that contain information. The jargon
for that is you have "degrees of freedom $DF = n - 1$."
"Reason 2." If you divide by $n - 1$ in the definition of the sample
variance $S^2$, then $E(S^2) = \sigma^2.$ In statistical terminology
this means "$S^2$ is an unbiased estimator of $\sigma^2.$" If you
divided by $n$ instead, then you would have an estimator of the population
variance that is too small. 
Note: Dividing by $n - 1$ is pretty much agreed upon, but reputable
authors in statistics and probability have proposed $n$, $n + 1$, and even
$n + 2$ as divisors--each giving a rationale aimed at a particular objective.
None of these alternative denominators has received wide acceptance.
But these discussions confirm that it is not a stupid question to ask
why we use $n - 1.$
$Addendum$ (Jan 25, '16): I have just read a latter by Jeffrey S. Rosenthall (U. Toronto) in the December '15 issue of the IMS Bulletin, arguing that in elementary statistics
courses it is OK to use $n$ as the denominator of the sample
variance. Briefly, his view is based mainly on arguments involving
mean square error (MSE). For example, with normal data, MSE for estimating $\sigma^2$ is minimized by denominator $n + 1$ instead of $n - 1.$
(See his letter on page 9 for details.)
However, in more advanced courses: as in my Comment below, a penalty for changing from $n - 1$ would be minor
confusion in getting confidence intervals for $\sigma^2$ and doing tests for $\sigma^2$ based
on the sample variance---mainly because $\sum (X_i - \bar X)^2/\sigma^2
\sim Chisq(df = n - 1).$
A: The sample variance is computed using deviations from the sample mean $\bar x$ instead of the population mean $\mu$, and this is the source of the bias.
When the squared deviations are accumulated, there is a deficit of $\bar x-\mu$ every time, so that the computed variance is too small.

Hint:
$$(x_i-\mu)^2=((x_i-\bar x)+(\bar x-\mu))^2=(x_i-\bar x)^2+2(x_i-\bar x)(\bar x-\mu)+(\bar x-\mu)^2.$$
When averaging over $i$, the double product in the middle vanishes due to the factor $(\bar x-\bar x)$, and you get
$$\sigma^2=\overline{(x_i-\mu)^2}=\overline{(x_i-\bar x)^2}+\overline{(\bar x-\mu)^2}=s^2+\frac{\sigma^2}N.$$
A: The sample mean is defined as $\bar{X} = \frac{1}{n}\sum_{i=1}^{n} X_i$, which is quite intuitive. But the sample variance is $S^2 = \frac{1}{n-1}\sum_{i=1}^{n} (X_i - \bar{X})^2$. Where did the $n - 1$ come from ? 
To answer this question, we must go back to the definition of an unbiased  estimator. An unbiased estimator is one whose expectation tends to the true expectation. The sample mean is an unbiased estimator. To see why: 
$$ E[\bar{X}] = \frac{1}{n}\sum_{i=1}^{n} E[X_i] 
= \frac{n}{n} \mu = \mu $$
Let us look at the expectation of the sample variance, 
$$ S^2 = \frac{1}{n-1} \sum_{i=1}^{n} (X_i^2) - n\bar{X}^2 $$
$$ E[S^2] = \frac{1}{n-1} \left( n E[(X_i^2)] - nE[\bar{X}^2] \right). $$
Notice that $\bar{X}$ is a random variable and not a constant, so the expectation $E[\bar{X}^2] $ plays a role. This is the reason behind the $n-1$.
$$E[S^2] = \frac{1}{n-1} \left( n (\mu^2 + \sigma^2) - n(\mu^2 + Var(\bar{X})) \right). $$
$$Var(\bar{X}) = Var(\frac{1}{n}\sum_{i=1}^{n} X_i) 
= \sum_{i=1}^{n} \frac{1}{n^2} Var(X_i) 
= \frac{\sigma^2}{n} $$
$$E[S^2] = \frac{1}{n-1} \left( n (\mu^2 + \sigma^2) - n(\mu^2 + \sigma^2/n) \right). = \frac{(n-1)\sigma^2}{n-1} = \sigma^2 \\
$$
As you can see, if we had the denominator as $n$ instead of $n-1$, we would get a biased estimate for the variance! But with $n-1$ the estimator $S^2$ is an unbiased estimator. 
A: It seems your question is related to the statistical concept of degrees of freedom.  The sum of deviations of n observations from their sample mean must be zero.  This means that if n-1 of the deviations are known, they completely determine the nth deviation.  It is the squared deviations from the mean that are used to construct the sample variance and hence we say that the sample variance has n-1 degrees of freedom.  Box, Hunter, and Hunter gives a more in depth explanation, and I'm sure many other statistics texts do as well.    
A: I think of it that you lose a degree of freedom in estimation of the mean, since to calculate variance you also need to estimate the mean. This is essentially what Bruce is saying in his comment regarding a line being smudged.
