Getting a fair result from an unfair non-independent coin It's well known that if you have a biased coin, where the heads has probability p and tails q=1-p, you can get an unbiased toss by tossing the coin twice and taking the first toss if they differ and retossing if they are the same. This technique is called the von  Neumann extractor.
This, of course, assumes that the coin tosses are independent. I'm interested in the case where they are not independent.
In particular, let $p_H$ be the probability of throwing a head after getting a head, and $q_H$ be that of getting a tail after throwing a head. (So that $p_H + q_H = 1$) Define $p_T$ and $q_T$ similarly. 
Applying the von Neumann extractor to this generates a non-independant sequence. Intuitively I expect that if each throw is positively correlated with the previous one then the extracted sequence will be negatively correlated and vice-versa.
So my questions are


*

*Is my intuition about the correlation of the extracted sequence correct, and can the correlation of the extracted sequence be made more explicit?

*Is there a similar simple process able to convert such a correlated sequence into a fair, unbiased and uncorrelated one.
 A: To summarize the core idea from the comments; here is a simple way to extract a fair coin toss from the given situation:
Note that at any given moment (except possibly prior to the first toss) the coin is in one of two states, "Heads" or "Tails" according to whatever the prior toss was.  In either state the coin behaves like an ordinary weighted coin, so the standard method for extracting a fair toss applies. To be specific, toss the coin once to get it in one state or the other.  For the sake of definiteness, let's say it comes up H so the coin starts as "Heads".  Now toss it a lot of times.  Say we get the sequence: $$\{H,H,T,H,T,T,T,T,H,T,T,H,H,T,H,H,H\}$$  Split that in two, so we are recording the tosses of the Heads coin and the tosses of the Tails coin.  In this case we'd get (trusting that no errors were made): $$\mathscr H =\{H,H,T,T,T,H,T,H,H\}$$
$$\mathscr T=\{H,T,T,T,H,T,H,H\}$$
Now we apply the standard extractor to either of the coins.  If you had chosen the Tails coin, the first two tosses are $H,T$ so von Neumann would tell us we chose $H$.  Had you chosen the Heads coin the first two tosses are $H,H$ which we must discard.  The next pair is $T,T$ which we also discard.  The third pair is $T,H$ so we have selected $T$.
As @user21820 rightly observes in the comments, the efficiency of this method can be improved significantly (the outlined method generally requires you to discard a lot of tosses).  The reference given is clear and can be applied to the two sequences I have described.
Note I:  a variant question may be of some interest. Suppose the coin is not modeled by any sensible finite state system.  That is, suppose the next toss depends non-trivially on the entire history (suppose, of course, that the nature of this dependence is fully specified).  Now the "disentangling" method sketched above does not appear to apply. Can we still extract a fair toss?   (to be clear, I have not spent any time at all on this. )
Note II (sketch): as to the correlation.  The OP's intuition is correct but it doesn't appear to be trivial to see that.  To be precise, let $p_H$ denote the probability that the Heads coin comes up H and let $\pi_T$ denote the probability that the Tails coin comes up T (slightly different notation that appeared in the question).  We assume that $p_H,\pi_T > .5$ and we need to compute $\phi$, the probability that von Neumann returns H starting from the Heads coin, and $\psi$, the probability that von Neumann returns H starting from the Tails coin. This is simple, if messy, to do via recursions (just looking at the four possible outcomes of the first two tosses. We get that $$(1-p_H^2)\phi=p_H(1-p_H)+(1-p_H)\pi_T\psi$$ $$(1-\pi_T^2)\psi=(1-\pi_T)(1-p_H)+(1-\pi_T)p_H\phi$$  After some tedious algebra we get that $$\phi=\frac {p_H+\pi_T}{1+p_H+\pi_T}$$ And it is easy to see that this is greater than $\frac 12$.  By symmetry we also get $\psi<\frac 12$.  Thats all we need!  It means that, starting from Heads we are more likely to get H than T (which means we'll switch to the Tails coin) and from the Tails coin we are more likely to get T than H (which means we'll switch to the Heads coin).  There may, of course, be a more intuitive argument for this. 
