$2$ players take turns and draw from a box containing $1000$ balls, $3$ of them are black. I'm not sure how to tackle this question. Assume a box containing $1000$ balls, $3$ of them are black and the rest are white. $2$ players $A_1$ & $A_2$ take turns and draw from the box without replacement. $A_1$ starts first. What's the probability that player $A_2$ draws a black ball given that player $A_1$ has already drawn a black ball?
If the box was containing total of $10$ balls, I would've just constructed the probability tree (a large one) and listed the whole probability space and computed the associated probabilities. But $1000$ balls !! It's impossible to handle using the tree approach.
I would really appreciated if someone could direct me onto how to tackle such situation.
Thanks.
 A: For an approximation, which is rather good because there are a lot of white balls, each black ball when drawn has $\frac 12$ chance to go to either player.  The winning sequences are $AAB, ABB, ABA, BAB$, so the chance of your event is $\frac 12$.  The correct value will be a little higher because $A$ is more likely than $B$ to get the first black ball.
A: Imagine each sequence of the balls laid out in a separate line. Since balls have no preference for positions, each black ball is equally likely to occupy odd/even positions.
$A_2$ wins if they occupy (in order) odd-even-odd, odd-odd-even, or odd-even-even positions, and loses if they occupy odd-odd-odd positions
Thus $P(A_2\text{wins}) = \dfrac34$
The above is an approximation.
Exact computation
If you want an exact value, there are $500$ possible starting points for $A_1$ to get the first black ball. Suppose it gets it at position $k, 500-k$ odd positions, and $1001-2k$ total positions remain. Also, the last chance for $A_1$  to win is starting with $495$.
$$\text{Putting it all together,} Pr = (1/495)*\sum_{k=1}^{495}1-\binom{500-k}{2}/\binom{1001-2k}{2}$$
$P(A_2 wins)$  wolframalpha = $\approx 0.7535$
A: You can use conditional probability here. P(A2(B) given A1(B) is = P(A2BintersectionA1B)/P(A1B). From here you know P(A1B)=3/1000. This would simply come out to be 2/999. Do you have a way of checking this answer is correct or not?
A: Convert $A_1$ to $A$ and $A_2$ to $B$ for ease of use.
As I understand the question, $A$ has drawn a single black ball and $B$ has drawn no black balls. The question asks what is the probability that $B$ will eventually draw a black ball.
Using Ross's notation, $AAA$ is the only failure condition.  $AAB, ABA, ABB$ are the possible win conditions.  This means $B$ has a $75%$ chance to draw a black ball eventually if $B$ draws the same number of balls as $A$ after $A$ draws his first black balls.
If $A$ draws the first black ball on turn $1000-N$ and $N$ is even, the $75%$ approximation is accurate as they draw the same number of balls.
If $N$ is odd. $B$ draws one more ball than $A$.  If the first one he draws is black, he won.  If it is white, the situation is idential to $N$ being even.  He, therefore, has a $100*(.75*(N-2)+2)/N=(75+50/N)%$ chance.
