Better choice of default calibration_min_idx

[tbenthompson]

Justify the choice based on binomial variance?

minimum K = 8192
p = 0.025
std = sqrt(np(1-p)) = 14.13

The minimum of $n$ N(0,1) random variables is approximately $\sqrt{2 \log{n}}$.

Heuristically... If we have 1 million tiles, $\sqrt{2 \log{n}}$ is ~5. So, we go consider about $5\sigma$. So, we'd like to consider about 5*14 = 70 as the minimum calibration index.

Related to this issue is doing a better job choosing between deepening and refinement for a tile where the calibration is less than the minimum calibration index.