SOLUTION: A urn has 9 white balls and 11 black balls. A ball is drawn and then replaced. If you draw a white ball you win 5 cents, if you draw a black ball you lose 5 cents. You have a do

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This is a one dimensional random walk with absorbing barriers, more commonly known as the gambler's ruin problem.
At this point note that there are 3 outcomes I win, you win, game goes on indefinitely, so P(I win)+P(you win) does not necessarily add to 1.
Let's think of this in terms of 5 cents. to make it easier. Initially I have 1 dollar(20x5cents) and you have half of that (10x5cents). The game is over when one of us has nothing and the other has 30x5 cents.
Let i be the number of 5 cents someone has, be my winning probability from that position and be your winning probability. Let be the probability I win a game 9/20 and be the probability you win a game 11/20.
Now, the probability I win with i coins is the probability I win this game, and that I win with i+1 coins, or that I lose this game and win with i-1 coins. (Not these two events are mutually exclusive) So

If we have 0 coins we can't win, sp and if we have 30 coins we have won so . Our task is to find .
I can write so I can write . Substituting this in and rearranging gives

Now let me define . This leads to the recurrence This has the well known solution . The proble is that we don't know . Let's move on though and consider

Notice that it is a collapsing sum and actually evaluates to . Doing the actual sum gives a standard geometric series which I'm sure you can evaluate. This is great because we know so using this we can calculate . Given this we can either use the p recurrence, or the sum of the z recurrence (we now know ) so we can calculate . I get it as

Which I calculate to be about 13.2%