As in title: you're flipping a coin indefinitely (let's say heads gives 1, tails gives 0). E.g.:
${0, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1\dots}$
How would you try proving that lengths of series of zeros or ones are of geometric distribution, that is of form:
$\mathbb{P}(\text{length} = k) = (1-p)^{k-1}p$.
Thanks for tips.
I will answer my own question correctly (my comments can be unclear), thanks to @saulspatz:
Let's say we tossed heads, and that begins the run. So we already have one heads. To get $k$-tosses-long run of heads finally, we need $k-1$ more heads and one time tails to end that run. So the probability is:
$\mathbb{P}(k) = \underbrace{0.5^{k-1}}_{k-1\text{ heads}} \cdot \underbrace{0.5}_{\text{1 tails}} = (1 - 0.5)^{k-1} \cdot 0.5$.
And because the geometric distribution is of form: $(1-p)^{k-1}p$, we proved that for a fair coin $\mathbb{P}(k)$ is geometric.