Q:

Drawing balls from a pocket

I have a pocket of N=4 balls of different colors, say green, red, blue and yellow. 

You can draw as many times as you want
After each draw, the ball will be put back to the pocket so that the chance of drawing each ball is always equal 
You need to pre-determine how many times you want to draw and must draw that many times even if you already drew a green ball

To win the game, your draws have to have at least one green ball and no red balls. How many times should you draw to maximize your winning chance?

A:

You have the best chances if

 you draw 2 balls. The probability is $\frac{5}{16}$

To calculate the probability, given $n$ draws you need to know that:

There are $3^n$ ways to get no reds
There are $2^n$ ways to get no reds and no greens
There are $4^n$ possible draws.

So the probability of drawing at least one green, but no reds is:

 \begin{equation}\frac{3^n-2^n}{4^n}\end{equation}

This is optimized with the result given.
Results for the first few $n$ are:

 \begin{eqnarray} n=1 &\longrightarrow \frac{1}{4} &= 0.25 \\ n=2  &\longrightarrow \frac{5}{16} &= 0.3125\\ n=3 &\longrightarrow \frac{19}{64} &\approx 0.297\\ n=4 &\longrightarrow \frac{65}{256} &\approx 0.2539\\ n=5 &\longrightarrow \frac{211}{1024} &\approx 0.206 \end{eqnarray}

A:

The chance not to draw a red ball in $n$ drawings is

 $\left( \frac{3}{4} \right)^n$.

If you don't draw a red ball, the chances of drawing at least one green ball are

 $1 - \left( \frac{2}{3} \right)^n$.

So the total winning chance is

 $\left( 1 - \left( \frac{2}{3} \right)^n \right)\left( \frac{3}{4} \right)^n$,

which is maximal for

 $n = 2$. Then the probability of winning is $\frac{5}{16}$.

