pulling a pair of balls from a box with a simple twist I'm stuck on this question, and I don't know how to solve it.
In a box there are $3$ balls, numbered 1, 2 and 3. We randomly pull $2$ balls out of it, and we write the lower number on the higher number, and then return them to the box.. so for instance if (2,3) were pulled, we return (2,2). We repeat the process until all of the balls in the box are marked 1. If we get the same number twice, we just put them back to the box without doing anything.
a) If after the first pull, there is at least one ball with the number 2 on it, what are the odds that the process will end after only one additional ball pulling?
b) What are the odds that the process will be over after exactly two pulls?
My attempt:
a) To end the task as needed, we need something like this: we pull (1,3), we return (1,1) - the ball with 2 is left in the box - then we pull it out along with 1, i.e (1,2) and we retrieve (1, 1) as needed. so the probability should be $\frac{1}{3} \cdot \frac{2}{3}$
2) Because we are not limited to having just 2 inside after the first pull, we can repeat the process with pulling (1,2) returning (1,1), then pulling (1,3) and returning (1,1) as needed. Since it's the same process, it seems to be the same calculation times 2, i.e: $2 \cdot \frac{1}{3} \cdot \frac{2}{3}$
Is this calculation correct or should I use a different kind of calculation/ formula?
 A: a) Having a $2$ still in the box means that we didn't choose $(1,2)$ which has a $\frac23$ probability.
The probability that we end with another pull means we first pulled a $(1,3)$, a $\frac13$ probability, then next a $(1,2)$, a $\frac23$ probability.
Since the second case falls under the first, the conditional probability would then be $\frac{\frac13\cdot\frac23}{\frac23}=\frac13$
b) The first pull is not $(2,3)$ which gives a probability of $\frac23$, and the second pull is not $(1,1)$, which gives a probability of $\frac23$. So the probability is $\frac23\cdot\frac23=\frac49$, as you computed.
A: First, I assume that by 

'after the first pull, there is at least one ball with the number 2 on it' 

it is meant: 
'After pulling two balls from the box, writing the lower number over the higher, and returning them to the box, there is at least one ball with the number 2 in the box. At least, this is how you seem to interpret this as well.
OK, now, for a): you did not calculate the conditional probability of having all balls being $1$ after two pulls given that after the first pull (+write+return), there is at least one ball with a $2$ left in the box. Instead, you calculated the probability of getting all balls being $1$ after two pulls (+writes+returns) and having at least one ball with a $2$ left after the first pull.
That is, if we define:
$A$: the event of having all balls have a number $1$ after two pulls 
and 
$B$: the event of having at least one ball left with a number $2$ after the first pull (+write+return)
then you calculated:
$P(A \cap B)$
instead of
$P(A|B)$
