Probability if no change occurs Suppose you have 2 red balls and 2 green balls.  You take out 1 ball,  and replace it with a red ball irrespective of its colour.
After 3 trials,  what is the probability that all the balls are red? 
Calculation shows 9/32.  
But if we take out a red ball and then replace it again with red ball,  then the system is similar to what it was before.  So we can consider taking out a red ball as doing nothing but just touching one of the balls in the box. Since there are only 2 green balls and 3 trials,  so one of the trials will get a red ball out. So we can consider this trial of taking out the red ball as doing nothing,  because the system was same as before. So there are only two effective trials. 
If I follow this strategy,  then the probability comes to be 1/8
Where is my logic going wrong? 
 A: The $3$-move sequences that result in $\text{RRRR}$ are:
\begin{align*}
&
{\phantom{\text{RRGG}}}
\left({\small{\frac{1}{2}}}\right)
{\phantom{\text{RRGG}}}
\left({\small{\frac{1}{2}}}\right)
{\phantom{\text{RRRG}}}
\left({\small{\frac{1}{4}}}\right)
{\phantom{\text{RRRR}}}
\;\;=\;\;{\small{\frac{1}{16}}}
\\[-2pt]
&\text{RRGG}\;\,{\Large{\to}}\;\,\text{RRGG}\;\,{\Large{\to}}\;\,\text{RRRG}\;\,{\Large{\to}}\;\,\text{RRRR}\\[12pt]
&
{\phantom{\text{RRGG}}}
\left({\small{\frac{1}{2}}}\right)
{\phantom{\text{RRRG}}}
\left({\small{\frac{3}{4}}}\right)
{\phantom{\text{RRRG}}}
\left({\small{\frac{1}{4}}}\right)
{\phantom{\text{RRRR}}}
\;\;=\;\;{\small{\frac{3}{32}}}
\\[-2pt]
&\text{RRGG}\;\,{\Large{\to}}\;\,\text{RRRG}\;\,{\Large{\to}}\;\,\text{RRRG}\;\,{\Large{\to}}\;\,\text{RRRR}\\[12pt]
&
{\phantom{\text{RRGG}}}
\left({\small{\frac{1}{2}}}\right)
{\phantom{\text{RRRG}}}
\left({\small{\frac{1}{4}}}\right)
{\phantom{\text{RRRG}}}
\;\left(1\right)\;
{\phantom{\text{RRRR}}}
\;\;=\;\;{\small{\frac{1}{8}}}
\\[-2pt]
&\text{RRGG}\;\,{\Large{\to}}\;\,\text{RRRG}\;\,{\Large{\to}}\;\,\text{RRRR}\;\,{\Large{\to}}\;\,\text{RRRR}\\[12pt]
\end{align*}
Summing the probabilities, we get
$$\frac{1}{16}+\frac{3}{32}+\frac{1}{8}=\frac{9}{32}$$
As regards the error in your attempt . . .

You computed the probability of success as
$$(1)\left({\small{\frac{1}{2}}}\right)\left({\small{\frac{1}{4}}}\right)={\small{\frac{1}{8}}}$$
where the factor $(1)$ represents the probability for the selected red ball, and the factors $\left(\frac{1}{2}\right)$ and $\left(\frac{1}{4}\right)$ represent the probabilities for the selected green balls.

Agreed that the first green ball to be selected will be selected with probability $\frac{1}{2}$, and the second green ball to be selected will be selected with probability $\frac{1}{4}$, but for the actual $3$-move sequence of selections, the probability for selecting the red  ball depends on when it's selected, and those probabilities don't add to $1$ (they add to more than $1$).
A: Taking a red ball out does not mean doing nothing: It gives you another chance to get a green ball out. But you have just one such extra chance.
