Heavy Bias

Two initial heads end the process in two flips with chance 1 in 10,000. In general one can overview the problem with the probability tree above. Filled disks and circles represent heads and tails respectively in the last two flips. Note two distinct non-stopping states (START and $x$ ) which can be captured with a system of equations. Expected number of flips:

$\mathbb{E}[N] = p^2 \cdot 2 + p \, q \cdot (2 + x) + q \cdot (1 + \mathbb{E}[N])$

$x = p + q \cdot (1 + \mathbb{E}[N])$

Probability of heads is $p$ , and of tails is $q = 1 - p$ . Plug $x$ into the first equation to get:

$\mathbb{E}[N] = \frac{1 + 2 p - p^2}{p^2 (2 - p)} = \frac{1019900}{199}$

With $p = 0.01$ . This makes the answer $\boxed{1020099}$ .