The least positive integer n such that $$1 - \frac{2}{3} - \frac{2}{3^2} - \ldots - \frac{2}{3^{n-1}} \lt \frac{1}{100}$$, is:

$$1 - \left( \frac{2}{3} + \frac{2}{3^2} + \dots + \frac{2}{3^{n-1}} \right)$$

First term ($$a$$) = $$\frac{2}{3}$$, Common ratio ($$r$$) = $$\frac{1}{3}$$, Number of terms ($$k$$) = $$n-1$$

Using the sum formula $$S_k = \frac{a(1 - r^k)}{1 - r}$$:

$$S_{n-1} = \frac{\frac{2}{3} \left( 1 - (\frac{1}{3})^{n-1} \right)}{1 - \frac{1}{3}} = \frac{\frac{2}{3} \left( 1 - \frac{1}{3^{n-1}} \right)}{\frac{2}{3}} = 1 - \frac{1}{3^{n-1}}$$

$$1 - \left( 1 - \frac{1}{3^{n-1}} \right) = \frac{1}{3^{n-1}}$$

$$\frac{1}{3^{n-1}} < \frac{1}{100}$$

$$3^{n-1} > 100$$

Least value of n = 6