Calculating the probability mathematically 
A football player's performance is recorded as 'well' or 'bad'.



*

*The probability of the football player performing well after the day he performed well is $3/4$.

*The probability of him performing bad after the day he performed bad is $1/2$.

*Given that this player has performed bad on Monday, what is the probability of him performing well on Friday ? $\left(~4\ \mbox{days later}~\right)$.



I was able to solve this problem using a large tree diagram. But using a tree diagram for these type of question is not the best option as it is easy to make mistakes in the middle and it takes a lot of time.

How to I set this situation mathematically ?.
 A: This is a Markov chain entirely described by the matrix
$$ P=\begin{pmatrix}\frac{3}{4} & \frac{1}{4} \\ \frac{1}{2} & \frac{1}{2}\end{pmatrix} $$
and the wanted probability is 
$$ \begin{pmatrix} 0 \\ 1 \end{pmatrix}^T P^4 \begin{pmatrix} 1 \\ 0 \end{pmatrix} =\color{red}{\frac{85}{128}}$$
A: You can set this up as a recurrence:
$$P_{n+1}(well)=\frac34 P_n(well)+\frac12 P_n(bad)$$
$$=\frac12 + \frac14 P_n(well)$$
So you're given $P_0(well)=0$ and you want to find $P_4(well)$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{m: \mathsf{m}\mbox{onday}\,,\ tu: \mathsf{tu}\mbox{esday}\,,\ w: \mathsf{w}\mbox{ednesday}\,,\ th: \mathsf{th}\mbox{ursday}\,,\ f: \mathsf{f}\mbox{riday}}$. $\ds{P_{d}}$ is the probability of performing
  $\ds{\underline{well}}$ in day $\ds{d = m, tu, w,th, fr}$. Note that
  $\bbx{\ds{P_{m} = 0}}$.

\begin{align}
P_{f} & = P_{th}{3 \over 4} + \pars{1 - P_{th}}{1 \over 2} =
{1 \over 4}\,P_{th} + {1 \over 2}\implies
P_{f} - {2 \over 3} = {1 \over 4}\pars{P_{th} - {2 \over 3}}
\\[5mm] \mbox{Similarly},&
\\
P_{f} - {2 \over 3} &= \pars{1 \over 4}^{2}\pars{P_{w} - {2 \over 3}} =
\pars{1 \over 4}^{3}\pars{P_{tu} - {2 \over 3}} =
\pars{1 \over 4}^{4}\pars{P_{m} - {2 \over 3}} = -\,{1 \over 384}
\end{align}

Then,
  $\ds{P_{f} = {2 \over 3} - {1 \over 384} = \bbx{85 \over 128} \approx 0.6641}$.

