Spoiler for Suggestive proof
The results of a biased coin can be separated into fair-coin flips and two-headed coin flips.
Suppose p(H) = 0.6, so that pmin = 0.4.
Representative flips are T T T T H H H H H H.
Partition them as: T T T T H H H H || H H.
2pmin (80%) of the flips are "fair" flips (equal chance of T and H.) The others are "two-headed" flips.
The fair flips reach H=T with p=1; the others reach H=T with p=0.
Thus, with probability 2pmin, H=T will eventually be reached.
Putting it differently: a biased coin acts with probability 2pmin like a fair coin.