A coin is tossed continuously until 2 heads or 2 tails appear respectively. Let the result of first toss is tail. The game is over when we get 2 heads respectively.
Determine the transition probability matrix.
Check my answer is correct or not.
Let $X_n$ denote the number of tail that appear. Let the state $S=\{0,1,2\}$.
So, the transition probability matrix is $$P= \begin{bmatrix} 1&0&0\\ 0&\dfrac{1}{2}&\dfrac{1}{2}\\ 0&0&1 \end{bmatrix}. $$
On
I think in this case you need to have 4 states: state 2 tails, state tail, state head and state 2 heads which I will call states 1,2,3 and 4 respectively.
$$P= \begin{bmatrix} 1&0&0&0\\ \dfrac{1}{2}&0&\dfrac{1}{2}&0\\ 0&\dfrac{1}{2}&0&\dfrac{1}{2}\\ 0&0&0&1 \end{bmatrix}. $$
If I understand you correctly you keep playing until you get either two heads or two tails. Then, states 1 and 4 are absorbing states.
Then when you are in state 2 (got a tail) then you can jump after the coin flip to either state 1 (two tail flips) or state 3 (one head).
If you start with a tail then your initial state probability vector $\pi$ would be
$$\pi= \begin{bmatrix} 0\\ 1\\ 0\\ 0 \end{bmatrix}. $$
Notice that $a_{i,j}$ element in matrix describes the probability of transition from state $i$ to $j$. Your matrix describes process in which '$0$' is absorbing state (which is obviously not true).
If the matrix shall describe number of consecutive tails (and the two tails is absorbing state) then the matrix is as follows: $$P= \begin{bmatrix} \dfrac{1}{2}&\dfrac{1}{2}&0\\ \dfrac{1}{2}&0&\dfrac{1}{2}\\ 0&0&1 \end{bmatrix}. $$