Q:

Probability of non-uniform draw without repeats.

I'm programming a machine that places one of eleven balls in sequence. The probabilities of the machine picking ball $i$ is given by $p_i$, and not necessarily does $p_i=p_j$ (this is what I mean by non-uniform).
The machine first picks a ball $i$, and notes its number. If $i$ is then picked again, it redraws until $j$ is picked.
Given ball $i$ is drawn first, what is the probability of ball $j$ being drawn second? (And then of ball $k\neq i,j$ being drawn third.)
My conjecture is that given ball $i$ is first, I can simply multiply all $p_{j\neq i}$ with a scalar $\alpha$ so that $\sum \alpha p_{j\neq i}=1$. I think I can prove this works with $p_i=p_j$, but I don't know if it generalizes.  

A:

Here is how I am interpreting the question:  You are trying to draw three distinct balls from the collection.  Each time you draw a ball you note its number and replace it.  Should you draw that number again, you put it back and repeat the draw.
Under this scenario, the probabilities just rescale by a constant factor, as imagined in the original post.  To see that, let's suppose you first draw ball $1$.  Let $\hat {p_2}$ be the probability that the next draw is ball $2$.  Indeed, three things can happen on the next draw:  you get ball $2$, you get one of $3,\cdots, 11$, you get ball $1$ again.  Thus $$\hat {p_2}=p_2\times 1+ (p_3+\cdots +p_{11})\times 0 + p_1\times \hat {p_2}=p_2+p_1\times \hat {p_2}\implies \hat {p_2}=\frac {p_2}{1-p_1}$$
Nothing special about ball $2$ of course, all the probabilities just scale by $\frac 1{1-p_1}$ and we are done.

