Intricate Markov Bayesian Theorem Probability Problem. Their exist a fraudulent gambling den that uses $2$ kinds of dices. A fair dice which has a $\frac16$ probability of rolling any number, their also exist a loaded dice that has $0.5$ possibility to roll a $6$ and $0.1$ possibility to roll any other number. The possibility that the fraudulent gambling den switches from the fair to loaded dice is $0.01$ and the probability of switching back from loaded to fair is $0.2$. 

Sequentially record dice by $q_{1},q_{2},q_{3}...,$ and numbers rolled by $o_{1},o_{2},o_{3}..$ for example $:q_{1}= F, o_{1}=2$ means the first die is Fair and by rolling the first die the number is observed is $2$, moreover $p(q_{1}=F)=P(q_{1}=L)=0.5.$
Derive the following possibility that : $P(o_{100}=4|q_{99}=F), P(q_{3}=L|o_{1}=1,o_{2}=3),$
$P(o_{100}=4 ,q_{99}=L| q_{98}=F), P(o_{15}=3,o_{16}=6|q_{14}=F), P(o_{1}=4,o_{2}=2,q_{2}=F), $
$P(o_{1}=4,o_{2}=2,o_{3}=4,q_{3}=F)$
One must use the theorem $$P(A_i | B) = \frac{P(B|A_i) * p(A_i)}{P(B)}$$
I must assume that $P(B|A) = 0.016666$ and $p(A_i) = 0.5$ $p(B) = 0.1$
Then $$P(A_i | B) =\frac{0.016666 * 0.5}{0.1} = 0.08$$ 
My question for this problem is there any steps that I may be missing. I feel that their must be a second step in this problem. Also the Markov chain must play in important part in solving this enigma.
 A: *

*At each step, there are 2 events : the dice change and the dice throw.

*The dice change is a markovian process to hold the states Fair and Loaded


We have a vector repartiton $\begin{pmatrix} F \\ L \end{pmatrix}$ and a transformation matrix $ T = \begin{pmatrix} 0.9 & 0.1 \\ 0.2 & 0.8 \end{pmatrix}$. At each step, the probabilities to be in the states Fair and Loaded are reflected by the components of the vector. The initial states must be set. for example $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ when starting with a fair dice or $\begin{pmatrix} 0.314 \\ 0.686 \end{pmatrix}$ if you want to start from such state.
The probability $\vec P$ to get a single result at a step $s$ after $\vec V$ has been set is :
$\vec P = \begin{pmatrix} P_1 \\ P_2 \\ P_3 \\ P_4 \\ P_5 \\ P_6 \end{pmatrix} = \vec V \times T^{s} \times D = \begin{pmatrix} F \\ L \end{pmatrix} \times \begin{pmatrix} 0.9 & 0.1 \\ 0.2 & 0.8 \end{pmatrix}^{s} \times \begin{pmatrix} 0.2 & 0.1 \\ 0.2 & 0.1 \\ 0.2 & 0.1 \\ 0.2 & 0.1 \\ 0.2 & 0.1 \\ 0.2 & 0.5\end{pmatrix} $
the $P_i$ are the probability to get the outcome $i$ accordingly to the initial conditions and the number of steps.
Now for you exercise : each sub question data must give 3 arguments, the number of steps , the initial setting and the target outcome(s). Let's see 2 computations you have to do :


*

*$P(o_{100}=4|q_{99}=F)$ : I understand this one : $\vec P = \vec V \times T^{100-99} \times D$ with $\vec V = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ , $s=100-99$ and the probability being the 4th row of the resulting vector.

*$P(q_{3}=L|o_{1}=1,o_{2}=3)$ : I don't understand it. Outcomes are on a step without initial conditions and knowing that the next state in Loaded doesn't help ( me ). Perhaps, it is a mismatch. 
Anyway, I hope that the above informations will help and/or that they will incitate others to better answers.
