The average number of rolls is approximately 5.294 for the specific case in the question.

The probability distribution for the specific case is:


$$
p(1) = \frac{1}{20}
$$

for \$n=1\$, and

$$
p(n) = \frac{n}{20} \prod_{i=1}^{n-1} \left( \frac{20-i}{20} \right)
$$

for \$n>1\$, and \$p(n)\$ indicates the probability that the success occurs on the n'th roll.


The general structure is

$$
p(n) = q_n \prod_{i=0}^{n-1} (1-q_i)
$$
where \$0 \le q_i \le 1\$ is the probability of success on the i'th roll, with the convention \$q_0=0\$ (you can't have succeeded before your first roll)


You can think of the process in terms of this sort of tree:

<pre>

        (1-q_0)=1
       /   \
     q_1  (1-q_1)                    # first roll
            /  \
           q_2 (1-q_2)               # second roll
                /   \ 
               q_3  (1-q_3)          # third roll
                      ...
</pre>

Where the branches to the left indicate success on that particular roll, and the branches to the right indicate failure on this particular roll.  Evaluating the probability of success on a given roll is just taking the product of the factors along the path to the "success" branch on that level.

For the specific case called out in the question:
<pre>

        start=1
       /   \
     1/20  19/20                    # first roll
            /  \
           2/20 18/20               # second roll
                /   \ 
               3/20  17/20          # third roll
                      ...
</pre>                  


From here we can tabulate the probabilities for the specific case:

| n  | p(n)| \$n p(n)\$|
|----|-----|---|
| 1  | 0.05 | 0.05| 
| 2 | 0.095 | 0.19|
| 3  | 0.12825 |  0.38475 |
| 4 | 0.14535 | 0.5814 |
| 5 | 0.14535 | 0.72675 |
| 6 | 0.130815| 0.78489 |
| 7 | 0.106832|0.747824
| 8 | 0.0793611| 0.6348888 |
| 9 | 0.0535687| 0.4821183 |
| 10 | 0.0327364| 0.327364 |
| 11 | 0.0180050| 0.198055 |
| 12 | 0.00883884 | 0.10606608 |
| 13 | 0.00383016| 0.04979208 |
| 14 | 0.00144368|0.02021152 |
| 15 | 0.000464039| 0.006960585 |
| 16 | 0.000123743| 0.001979888 |
| 17 | 2.62956e-05| 0.0004470252 |
| 18 | 4.17635e-06| 7.51743e-5 |
| 19 | 4.40937e-07| 8.377803e-6 |
| 20 | 2.32020e-08| 4.6404e-7 |

From, this you can get the average of 5.294 as the sum of \$n p(n)\$.

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