Need someone to explain answer to question without using Markov chains. Consider the following question here:
Probability of two consecutive head or tail or any one of them in a row?

Question. Fair coins are tossed and when either four consecutive heads and tails appear the process will be stopped. What is the probability of two consecutive head or tail or any one of them in a row?
Answer. Because heads and tails play symmetric roles in the stopping criterion (and presumably have equal chances in each Bernoulli trial), it suffices to find the probability of getting two consecutive heads before the process stops.
If the process stops with four consecutive heads, obviously that means also we got two consecutive heads before the process stops.  So we can focus on the probability $p$ that we get two consecutive heads before the process stops with four consecutive tails.
One way to compute this is by defining a Markov chain with two absorbing states, a) two consecutive heads and b) four consecutive tails.  Finding $p$ amounts to finding the probability of reaching the first of these absorbing states (two consecutive heads).
The Wikipedia write-up of 
  absorbing Markov chains
  may be overly concise, so here are some details.  Ordering the 
  transient states before the absorbing states gives a
  probability transition matrix having block structure:
$$ P = \begin{bmatrix} Q & R \\ 0 & I \end{bmatrix}$$
so that $Q$ gives the transition probabilities between the
  transient states and $R$ the transition probabilities from
  transient to absorbing states.
We define the fundamental matrix
  $N = \sum_{k=0}^\infty Q^k = (I-Q)^{-1}$, and the product
  $NR$ then gives the probabilities of eventually reaching an
  absorbing state starting from a transient state.
In the case at hand it is convenient to use four transient
  states (consecutive runs of one Head or of one to three
  Tails, resp.) and two absorbing states (two consecutive
  Heads or four consecutive Tails).  Taking the states in
  just this order gives:
$$ P = \begin{bmatrix} 0 & \frac{1}{2} & 0 & 0 & \frac{1}{2} & 0
  \\ \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 & 0
  \\ \frac{1}{2} & 0 & 0 & \frac{1}{2} & 0 & 0
  \\ \frac{1}{2} & 0 & 0 & 0 & 0 & \frac{1}{2}
  \\ 0 & 0 & 0 & 0 & 1 & 0 
  \\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} $$
$$ N = (I-Q)^{-1} = \begin{bmatrix}
  \frac{16}{9} & \frac{8}{9} & \frac{4}{9} & \frac{2}{9}
 \\ \frac{14}{9} & \frac{16}{9} & \frac{8}{9} & \frac{4}{9}
 \\ \frac{4}{3} & \frac{2}{3} & \frac{4}{3} & \frac{2}{3}
 \\ \frac{8}{9} & \frac{4}{9} & \frac{2}{9} & \frac{10}{9}
 \end{bmatrix} $$
$$ NR = \begin{bmatrix} \frac{8}{9} & \frac{1}{9}
 \\ \frac{7}{9} & \frac{2}{9}
 \\ \frac{2}{3} & \frac{1}{3}
 \\ \frac{4}{9} & \frac{5}{9} \end{bmatrix} $$
For economy I omitted an "empty" state, so we imagine our
  Markov process to initialize in state one Head with probability
  $1/2$ and likewise in state one Tail with probability $1/2$.
  From the above computation it follows that our chance of
  reaching two consecutive Heads before four consecutive Tails is:
$$ (1/2)\frac{8}{9} + (1/2)\frac{7}{9} = \frac{5}{6} $$
This is the same as our chance of reaching two consecutive
  Tails before four consecutive Heads.

I was wondering if someone could give an explanation of this answer that doesn't end up using fancy stuff like Markov chains? This is a practice question for the Math GRE subject test, so there probably is a way to do this that doesn't involve fancy stuff.
 A: If I understand correctly, you want to compute the probability of getting 2 heads before 4 tails (ie ending the process). Let 


*

*$p$ be the probability of landing heads (here $p=1/2$)

*$P(A \mid 0)$ be the probability of 2 consecutive heads at the beginning of the sequence

*$P(A\mid 1)$ be the probability of 2 consecutive heads given last toss was tails

*$P(A\mid 2)$ be the probability of 2 consecutive heads given last 2 tosses were tails

*$P(A\mid 3)$ be the probability of 2 consecutive heads given last 3 tosses were tails

*$f$ be the probability of 2 consecutive heads given last coin was heads.


Then note that


*

*$P(A\mid 3) = 0(1-p) + f p$

*$P(A\mid 2) = P(A\mid 3)(1-p) + fp$

*$P(A\mid 1) = P(A\mid 2)(1-p) + fp$

*$P(A\mid 0) = P(A\mid 1)(1-p) + fp$

*$f = P(A\mid 1) (1-p) + p$.


So 


*

*$P(A\mid 2) = fp(2-p)$

*$P(A\mid 1) = fp(2-p)(1-p) + fp = fp(3-3p+p^2)$

*$f = fp(3-3p+p^2)(1-p)+p = 7f/16 + 1/2 \implies f = 8/9 \implies P(A\mid 1) = 7/9$

*$\therefore P(A\mid 0) = (7/9)/2 + (8/9)/2 = 5/6$.

