Player 1 does not choose Heads or Tails but {HT} or {TH}. Remaining one (HT / TH) is for Player 2.

Assuming that observations are independent, we have the following probabilities:
P{HT} = p(1-p)
P(HH) = p^2
P(TT) = (1-p)^2
P(TH) = (1-p)p

In other words, first we flip twice if we get {HT} or {TH} then we have a winner; otherwise we flip (twice) again.

http://www.codingthewheel.com/archives/the-coin-flip-a-fundamentally-unfair-proposition

May or may not be relevant. I suspect it is relevant because the disk is weighted.

-Doug