Overall equal probabilities when replacing choices Explanation:
I want to choose marbles from a bag x times. Once chosen, the marbles are not put back in the bag. After each choice, some number of new marbles are added to the bag. x is less than the total number of marbles, i.e. not all marbles will be chosen. My goal is to make it so every marble has an equal chance of being chosen overall (or as close as possible).
Example 1 (x = 2):
Start with marbles A and B. Choose one at random. Let's say B was chosen. Now marble C is added to the bag and the second marble is chosen. A has already had a 50% chance of being chosen, so giving A and C the same odds would mean A was more likely than C. Is there any weighting possible to make it such that A, B, and C were all equally likely in the beginning? I assume it is not possible because C was not available in the beginning. I believe the next best thing would be to make it so A and C were equally likely from the beginning. My guess is that this can be done by making C twice as likely as A for the second draw (A should have 1/3 chance and C should have 2/3 chance). However, I am not sure how to prove it.
Example 2 (x = 3):
Start with marbles A and B. Again let's say B is chosen. Now Marbles C and D are added and a second is chosen. Again, I want to say C and D should be twice as likely as A. Assuming I was correct before, A should have a 1/5 chance and C and D should both have a 2/5 chance. Let's say C was chosen. Now marbles E and F are added and so the odds of each marble for the third draw should be 0.0909 for A, 0.1818 for C, 0.3636 for E and 0.3636 for F.
Is this correct? How can it be shown mathematically? Does my general approach of marbles added after the last draw being twice as likely as the marbles in the draw before that which are twice as likely as the marbles in the draw before that... work?
 A: There are actually three parameters here: the number of starting balls $s$, the number added each time $t$, and the total number of draws to be undertaken $x$
You can say that the total number of balls involved is $xt + s-t$ and if $x$ are drawn then the average overall probability a particular ball is drawn at some stage is $\frac{x}{xt+s-t}$.  For the balls added after the penultimate draw, this must be the probability they are individually drawn in the final round and so the combined probability they are drawn that round is $\frac{xt}{xt+s-t}$, leaving a combined probability of $\frac{s-t}{xt+s-t}$ for the all the other balls in that round.  
This puts a constraint on the parameters: you need $s\ge t$ (or $x=1$) to avoid an impossible situation.  
I think you can then work backwards, finding weights for the balls introduced after the antepenultimate draw which keep to same probabilities, and so on.  In your examples:


*

*$s=2,t=1,x=2$ imply $\frac{x}{xt+s-t} =\frac{2}{3}$ so that means you want $C$ to have a probability of $\frac{2}{3}$ in the second round and so the older remaining ball to have probability $\frac13$.  This then gives each of the original balls an overall probability of being chosen $\frac12+\frac12 \times \frac13=\frac23$, so this works

*$s=2,t=2,x=3$ imply $\frac{x}{xt+s-t} =\frac{1}{2}$ so that means you want $E$ and $F$ to each have a probability of $\frac{1}{2}$ in the third draw and so the older remaining balls to have probability $0$. That shuts out $C$ and $D$ from the third draw so you want them to each have a probability of $\frac{1}{2}$ in the second draw, shutting $A$ and $B$ out from the second and third rounds.  This works, just, since each ball has an overall probability of $\frac12$ of being drawn, but does not allow a finite weighting ratio between new and remaining balls
