What reasons could cause variance to be underestimated to a large degree? I ran a simulation from probabilities estimated in some survival models (i.e., Cox regression with 2 causes).  $n$ samples of size $m$ were simulated, and sample means and sample variances were calculated. 
When I use the average of sample variances divided by $m$ to estimate the variance of the sample mean, the two numbers don't match. The former is only 2% of the latter. 
What could cause this underestimation?
 A: The sample variance of a pooled sample is not equal to the average of the sample variances of the individual samples.  You are underestimating because you have ignored differences in the sample means when aggregating the variances.  It also sounds like you have divided by $m$ twice; once to get the average and then again for some reason!  
To see how to pool sample moments, consider two samples with respective sizes $n_1$ and $n_2$ and subscript their sample means $\bar{x}_1$ and $\bar{x}_2$ and sample variances $s_1^2$ and $s_2^2$ to indicate which sample they are from.  Then the sample variance of the pooled sample is related to the sample moments of the individual samples by the rule:
$$(n_1 + n_2 - 1) s_\text{pooled}^2 = (n_1-1) s_1^2 + (n_2-1)s_2^2 + d_{12}^2 \quad \quad \quad d_{12}^2 \equiv \frac{n_1 n_2}{n_1 + n_2}(\bar{x}_1 - \bar{x}_2)^2.$$
This pooled sampling rule is taken from O'Neill (2014) (p. 283, different notation).  You can obtain the sample variance of a pooled sample that is constructed from a larger number of individual samples by using iterative application of this rule.

O'Neill, B. (2014) Some useful moment results for sampling problems. The American Statistician 68(4), pp. 282-296 (edited by corrigendum).
