Probability of a fair sequence of tosses ending on two successive tails given the first toss was a head? Suppose a coin is tossed repeatedly until either two successive heads appear or two successive tails appear. Then, assume that the first coin toss results in a head. I would like to find the probability of this tossing ending on two successive tails. One way I know of doing this is by creating a Markov chain and then doing first step analysis. I was wondering if there was perhaps an easier way to do this type of problem using some other principles. thank you.
 A: Let $p$ be the probability you want.
Look at the possible sequences of three tosses (including the initial $H$).  We have $\{HH*,HTT,HTH\}$  Conditional on the first being $H$, the probability of $HH*$ is $\frac 12$, and the probability of the other two are each $\frac 14$.  The first is a win for $H$, the second is a win for $T$, and the third restarts the game.  Thus $$p=\frac 12\times 0+\frac 14 \times 1 +\frac 14\times p\implies p=\frac 13$$
Note:  if you want to allow for a weighted coin, the same calculation works but we need to correct for those initial probabilities.  Thus assume that the coin comes up $H$ with probability $\phi$.  Then $HH*$ has probability $\phi$, $HTT$ has probability $(1-\phi)^2$, and $HTH$ has probability $(1-\phi)\phi$.  So the recursion is now $$p=\phi \times 0+ (1-\phi)^2 \times 1 + (1-\phi)\phi \times p\implies p=\frac {(1-\phi)^2}{1-(1-\phi)\phi}$$  Comparison shows that this matches the result obtained by looking at infinite sums, as in the posted solution of @BCLC (with different notation, unfortunately).
A: Let $p$ be the probability of heads.
It seems to me that we have the following possible scenarios with the following probabilities:
HH --> $p^2$
H T HH --> $p(1-p)p^2$
H T HT HH --> $p(1-p)[p(1-p)]p^2$
H T HT HT HH --> $p(1-p)[p(1-p)]^2p^2$
H T HT HT ... HT HH --> $p(1-p)[p(1-p)]^np^2$
H TT --> $p(1-p)^2$
H TH TT --> $p[(1-p)(p)](1-p)^2$
H TH TH TT --> $p[(1-p)(p)]^2(1-p)^2$
H TH TH ... TH TT --> $p[(1-p)(p)]^n(1-p)^2$
So we have:
$$P(H...TT) = \sum_{n=0}^{\infty} p[(1-p)(p)]^n(1-p)^2$$
$$ = p(1-p)^2 \sum_{n=0}^{\infty} [(1-p)(p)]^n$$
$$ = p(1-p)^2 \frac{1}{1-(1-p)(p)}$$
If we have $p=1/2$, then $P(H...TT) = 1/6$.
Similarly $P(H...HH or HH) = 1/3$
Of course $P(T) = 1/2$.
Check if it adds up: $P(T) + P(H...HH or HH) + P(H...TT) = 1/6 + 1/3 + 1/2 = 1$.

Just realised (before lulu pointed out) that I didn't condition on $H...$ so just divide all (or most? eg $P(T)$?) above by p or something. Anyway:
$P(T|H...) = 0$
$P(H...HH or HH|H...) = \frac{1/3}{1/2} = 2/3$
$P(H...TT|H...) = \frac{1/6}{1/2} = 1/3$.
Check if it adds up: $P(T|H...) + P(H...HH or HH|H...) + P(H...TT|H...) = 0 + 2/3 + 1/3 = 1$.
