TITLE: $2$ players take turns and draw from a box containing $1000$ balls, $3$ of them are black.
QUESTION [1 upvotes]: 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.

REPLY [0 votes]: 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.
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