An urn of 4 balls with 2 colors. Pick 2 balls and place them back 4 times. What's the probability of picking 2 balls of the same color twice in a row? An urn of 4 balls with 2 colors. Pick 2 balls and place them back 4 times. What's the probability of picking 2 balls of the same color twice in a row?
So the probability of picking 2 balls of the same color is $2\choose1$$2\choose2$/$4\choose2$, but I don't know the probability of getting that twice in a row out of 4 draws
 A: Assumptions: 


*

*There are two balls of one color and two balls of a second color

*Suppose the colors are Green and Red. A successful outcome occurs in any of these cases when you choose two green followed by two green, two green followed by two red, two red followed by two green, or two red followed by two red.


The probability of picking two balls of the same color in a single draw is as you found $\dfrac{1}{3}$. A single draw will be labeled $S$ for both balls are the same color and $D$ for the balls are different colors. Here are the possible outcomes:
$$\begin{array}{c|c|c}\text{Draws} & \text{Probability} & \text{2-in-a-row?} \\ \hline DDDD & \left(\dfrac{2}{3}\right)^4\left(\dfrac{1}{3}\right)^0 & \text{No} \\ DDDS & \left(\dfrac{2}{3}\right)^3\left(\dfrac{1}{3}\right)^1 & \text{No} \\ DDSD & \left(\dfrac{2}{3}\right)^3\left(\dfrac{1}{3}\right)^1 & \text{No} \\ DDSS & \left(\dfrac{2}{3}\right)^2\left(\dfrac{1}{3}\right)^2 & \text{Yes} \\ DSDD & \left(\dfrac{2}{3}\right)^3\left(\dfrac{1}{3}\right)^1 & \text{No} \\ DSDS & \left(\dfrac{2}{3}\right)^2\left(\dfrac{1}{3}\right)^2 & \text{No} \\ DSSD & \left(\dfrac{2}{3}\right)^2\left(\dfrac{1}{3}\right)^2 & \text{Yes} \\ DSSS & \left(\dfrac{2}{3}\right)^1\left(\dfrac{1}{3}\right)^3 & \text{Yes} \\ SDDD & \left(\dfrac{2}{3}\right)^3\left(\dfrac{1}{3}\right)^1 & \text{No} \\ SDDS & \left(\dfrac{2}{3}\right)^2\left(\dfrac{1}{3}\right)^2 & \text{No} \\ SDSD & \left(\dfrac{2}{3}\right)^2\left(\dfrac{1}{3}\right)^2 & \text{No} \\ SDSS & \left(\dfrac{2}{3}\right)^1\left(\dfrac{1}{3}\right)^3 & \text{Yes} \\ SSDD & \left(\dfrac{2}{3}\right)^2\left(\dfrac{1}{3}\right)^2 & \text{Yes} \\ SSDS & \left(\dfrac{2}{3}\right)^1\left(\dfrac{1}{3}\right)^3 & \text{Yes} \\ SSSD & \left(\dfrac{2}{3}\right)^1\left(\dfrac{1}{3}\right)^3 & \text{Yes} \\ SSSS & \left(\dfrac{2}{3}\right)^0\left(\dfrac{1}{3}\right)^4 & \text{Yes}\end{array}$$
Adding this up, I get:
$$3\left(\dfrac{2}{3}\right)^2\left(\dfrac{1}{3}\right)^2+4\left(\dfrac{2}{3}\right)^1\left(\dfrac{1}{3}\right)^3+\left(\dfrac{2}{3}\right)^0\left(\dfrac{1}{3}\right)^4 = \dfrac{7}{27}$$
