Shuffling items between two boxes I have two initially identical boxes. They each contain 1 red ball and 3 green balls.
From the first box, I randomly take out a ball and put it into the second box.
From that second box, I randomly take out a ball and put it into the first box.
Finally, I randomly take out a ball from the first box.
What's the probability that the said ball is red?
I would appreciate any tips on getting started.
 A: The simplest solution is to note that in expectation, the two boxes are again identical after you've placed a ball from the second box back into the first, so the answer is $\frac{1}{4}$. Without any information, your best guess prior to drawing a ball a third time is that the boxes are the same as they were before you started the process. This is a consequence of the law of iterated expectations.
Here's a hint on how to work through it the long way and prove it to yourself:
There are three ways you can pull the red ball:

*

*The first ball was red and the second red.

*The first ball was not red and the second ball was red.

*The first ball was not red and the second ball was not red.

If the first ball was red and the second not red, there will be $0$ red balls in the first box, so it would be impossible to pull a red ball.
For the first option, there is a $\frac{1}{4}$ chance of pulling the red ball first. After moving it, there are now $2$ red balls and $3$ green balls in the second box. So, there's a $\frac{2}{5}$ chance of pulling a red ball from the second box. There are now $1$ red and $3$ green balls in the first box, so we have a $\frac{1}{4}$ chance of pulling the red ball from it. Overall, this sequence has probability $\frac{1}{4}*\frac{2}{5}*\frac{1}{4} = \frac{1}{40}$.
You should get that the second sequence of events has probability $\frac{3}{40}$, and the third has probability $\frac{3}{20}$, giving you a grand total of $\frac{1}{4}$.
