<PRE><B>Question 125652</B>
This is a one dimensional random walk with absorbing barriers, more commonly known as the gambler&#39;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&#39;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, *[tex p_i] be my winning probability from that position and *[tex q_i] be your winning probability. Let *[tex p] be the probability I win a game 9/20 and *[tex q] 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

*[tex p_i = pp_{i+1} + qp_{i-1}]

If we have 0 coins we can&#39;t win, sp *[tex p_0=0] and if we have 30 coins we have won so *[tex p_{30}=1]. Our task is to find *[tex p_{20}].

I can write *[tex p+q=1] so I can write *[tex p_i = pp_i+qp_i]. Substituting this in and rearranging gives

*[tex (p_{i+1}-p_i) - \frac{q}{p}(p_i - p_{i+1})]

Now let me define *[tex z_i = p_{i+1}-p_i]. This leads to the recurrence *[tex z_{i+1}=\frac{q}{p}z_i] This has the well known solution *[tex z_n = z_0\left(\frac{q}{p}\right)^n]. The proble is that we don&#39;t know *[tex z_0]. Let&#39;s move on though and consider

*[tex \sum_{i=1}^{n-1} z_i]

Notice that it is a collapsing sum and actually evaluates to *[tex p_n - p_1]. Doing the actual sum gives a standard geometric series which I&#39;m sure you can evaluate. This is great because we know *[tex p_{30}=1] so using this we can calculate *[tex p_1]. Given this we can either use the p recurrence, or the sum of the z recurrence (we now know *[tex z_0]) so we can calculate *[tex p_{20}]. I get it as

*[tex \frac{11^{20}9^{10} - 9^{30}}{11^{30}-9^{30}}]

Which I calculate to be about 13.2%</PRE>