Videogame (Dota 2) probability In Dota 2 after you launch a spell you usually have a cooldown time you need to wait before being able to launch that spell again.
In the Year Beast Brawl anyway powerful beasts can have Refresher Aura:
Refresher Aura: When an allied hero casts a spell within 900 range of the Beast,
that spell has a 50% chance to be instantly refreshed. 

Given that my hero has Refresher Aura active,the probability of having my spell instantly refreshed(no cooldown) $n$ times in a row is: $\frac{1}{2}^n$.
Now my question is:
What is the probability of having my spell instantly refreshed $n$ times in a row if I have Refresher Orb (Orb gives the ability to reset the cooldowns of all your items and abilities) given that Refresher Orb can be instantly refreshed by Refresher Aura too?
Edit to make the question more clear:
What is the probability of being able to cast the spell(Orb has no cooldown or spell has no cooldown or both have no cooldown) after $n$ consecutive spell casts?  
 A: You can model this as a markov chain. We start with our ability on cooldown, with both the refresher orb and the aura available, denote this state 2. If we obtain the aura, with chance 0.5, we stay within state 2. If we do not get the aura, with chance 0.5, we use the orb, which means the orb is on cooldown, but the aura is again available, denote this state 1. If then we again dont get the auro we have both the aura and the orb on cooldown, denote this state 0. We get the following matrix
\begin{pmatrix}
1 & 0 & 0\\
\frac{1}{2} & 0 & \frac{1}{2}\\
0 & \frac{1}{2} & \frac{1}{2}
\end{pmatrix}.
You can obtain the chance of not running out of cooldowns after $n$ steps by taking the $n$th power of the matrix and adding the right bottom and right middle number.
Below you can see a picture of how this chance develops over the first 25 turns.

A: Thanks to @Marc I was able to come up with my own answer to the question:
We have 4 possible states:
$s_1$:Both orb and spell are ready.
$s_2$:Orb is ready,spell is on cd(cooldown).
$s_3$:Spell is ready,orb is on cd.
$s_4$:Both orb and spell are on cd.
Given that Refresher Aura is active and working for both the spell and the orb,we get the following transition matrix:
$$
P=\begin{bmatrix}
   0.5 & 0.5 & 0 & 0 \\
  0.5 & 0 & 0.5 & 0 \\
  0 & 0 & 0.5 & 0.5 \\
 0 & 0 & 0 & 1
  \end{bmatrix}
$$
The probability we are looking for is the probability of being in $s_{1}$ or $s_{2}$ or $s_3$ after $n$ steps and starting from $s_1$.
This probability is given by $p_{11}^n + p_{12}^n + p_{13}^n$ where $p_{ij}^n$ is the $ij$th entry of the $P^n$ matrix.
E.g. for $n=3$ we get :
$$
P^3=\begin{bmatrix}
   0.375 & 0.25 & 0.25 & 0.125 \\
  0.25 & 0.125 & 0.25 & 0.375 \\
  0 & 0 & 0.125 & 0.875 \\
 0 & 0 & 0 & 1
  \end{bmatrix}
$$
and so $p_{11}^3 + p_{12}^3 + p_{13}^3=0.875=$ probability of being able to cast the spell after $3$ consecutives casts.
