If I read this correctly:

1)

Every time you draw a white ball, you draw a ball from that bin, so the number of white balls with a certain bin number on it is always equal to the number of balls in that bin. (You start out with 27 of both).

So, the two steps together are equivalent (probability-wise) to just drawing a ball at random from all the balls left in the bins.

The probability of drawing the winning ball, given that this is your n-th try, is 1/(2700+1-n)

2)

When you draw a black ball, you're done. Therefore, all the other dotted balls except the winning ball are, so far, equivalent to any other black ball.

The chance of the winning ball being in a certain bin has no correlation to the number of red balls you have already pulled out of it, as far as I can see - however the chance of drawing it, if it should happen to be there, obviously increases as the number of red balls shrinks.

Imagine if there were only two bins, one untouched and one with only black balls left. The probabilities for each bin having the winner ball is 1/2. For the first bin, you have a probability of 1/2*1/100=1/200 of winning. For the other, it's 1/2*1/3=1/6.

So, which should you pick? If you pick from the almost-empty bin, your chance of winning is 1/6 but your chance of the game being over is also 5/6. If you pick from the other, the chance of winning is 1/200, but the chance of the game ending is also a lot less - 1/2*2/100 + 1/2*3/100 = 5/200.

Notice that the ratio of probabilities of winning to "game over" is a constant 1:5 (because there are six black balls and one has a white dot). I'm not sure about this, but my intuition is that the red balls are just "stalling" and won't matter in the end - the probability of reaching a decision increases, but the chances of winning don't, so you might as well just pick one at random.

3: I'm not sure if I get this question, but if you assume that you're back to random drawing, plus the game is over as soon as you draw any black ball (dots or no dots):

Eventually if you go on long enough, you will draw a black ball. When this happens, your average return on your investment will be

(1,000,000/300)$, approx. 33,333$. (Most of the time it will be 0, but if you do it enough times, this is what the average will tend to).

The probability of drawing a black ball increases as you go along. For n tries, the probability of drawing a black ball at some time during those is 1-(the probability of drawing n red balls in a row) = 1 - (2600/2700)*(2599/2699)...(2600-n+1)/(2500-n+1)= 1 - 2600!(2500-n)!/(2600-n)!2500!. In terms of binomial coefficient, that's (2600 pick n)/(2500 pick n).

My guess is that you're looking for the number of tries where the ratio of (this number times the average prize) to n*1000$ is the best.