Let ~P(s, i)~ be the probability that we have collected exactly ~i~ critical bars after ~s~ plays of the song. So ~P(0, 0) = 1~, and for ~i > 0~, ~P(0, i) = 0~. We can then compute ~P(s, i)~ for ~s > 0~ and ~0 \le i \le 20~ recursively as follows:

$$\displaystyle P(s, i) = \sum_{j=0}^i (P(s-1, j) \times C(20-j, i-j) \times p^{i-j} \times (1-p)^{20-i})$$

(Note that ~C(n, k)~ is the binomial coefficient, or "combinatoric choose" function.)

The intuitive explanation for the above formula is that we want to have exactly ~i~ critical bars after ~s~ playthroughs, and we consider all values ~j \le i~ such that we had exactly ~j~ critical bars right before our ~s^\text{th}~ playthrough. The probability of that having been the case is ~P(s-1, j)~, the number of ways to select exactly ~i-j~ of the remaining ~20-j~ sections (on which to get new critical bars) is ~C(20-j, i-j)~, and these values are multiplied by both the probability of getting critical bars on ~i-j~ bars, and by the probability of not getting critical bars on the remaining ~20-i~ bars.

For ~i > 0~, let ~Q(i)~ be the probability that we receive our ~20^\text{th}~ critical bar on the ~i^\text{th}~ playthrough. For ~i > 0~, we can compute the following:

$$\displaystyle Q(i) = P(i, 20) - P(i-1, 20)$$

For example, if there's a ~50\%~ chance that we have all of the bars after one playthrough, and a ~60\%~ chance that we have them all after two playthroughs, then there must be a ~60\% - 50\% = 10\%~ chance that it will take us exactly two playthroughs to get all of the bars.

We can then compute the expected number of playthroughs, ~E~, with the following infinite sum:

$$\displaystyle E = \sum_{i=1}^\infty (i \times Q(i))$$

Practically though, we only need to evaluate this sum for small values of ~i~, up to some value ~L~. ~i~ increases linearly, but ~Q(i)~ falls off exponentially, so their product also decreases exponentially. Since we only need 5 decimal places of precision, it's safe to stop computing the remainder of the sum once this product is reasonably small (say ~10^{-9}~ for example). For the lower bound in this problem, ~p = 0.01~, giving ~L~ a value of ~5\,000~ or so is more than sufficient. Computing ~P(s, i)~ for ~0 \le s \le L~ and ~0 \le i \le 20~ takes on the order of ~L \times 20^2~ operations.