1 Original and reasons for this game
This game is actually changed from Geoffrey and David's Book [1] (section 11.2, page 445) and its accompanying solution manual [2]. The book gives the result without saying why it is that, which is what I am interested in. So I try to figure out the reason for the equation, but I failed to do so. And hope to get help from any of you.
2 The game and unexplained result from book
2.1 Game
Suppose you have a box, and there are $i$ red balls and $j$ lemon balls in it. You are standing at the original point with coordinate $0$. You then repeatedly draw balls from the box without replacement (which means each drawn ball will be threw away, and totally you have $i+j$ drawings). If you get a red ball, you move right one step. And if you get a lemon ball, (1) you move left with one step conditioning on you are not at the original point; or (2) do nothing if you are at the original point. If define the random variable $X$ as the coordinate that you are when the box is empty, then what is the probability that $X=n$.
2.2 Unexplained result
$\pi(n;i,j)\triangleq Prob\{X=n\}=\frac{\binom{i}{n}}{\binom{j+n}{n}}-\frac{\binom{i}{n+1}}{\binom{j+n+1}{n+1}}$.
3 My current progress for this question
Above is what I got from the book. After some guess and try, I found $\frac{\binom{i}{n}}{\binom{j+n}{n}}$ is actually like the CCDF (complementary cumulative distribution function) of random variable $X$. If define $F(n;i,j)\triangleq Prob\{X \geq n\}$, we have $F(n;i,j)=\frac{\binom{i}{n}}{\binom{j+n}{n}}$. And it naturally follows that $\pi(n;i,j) = F(n;i,j) - F(n+1;i,j)$. It seems to think with $F(n;i,j)$ is easier than $\pi(n;i,j)$. However, after several days of thinking and discussion. Still, I cannot explain either $F(n;i,j)$ or $\pi(n;i,j)$.
4 Some hints I can give
(1) This actually comes from the book[2]. The $\pi(n;i,j)$ has recursion formula as $\pi(n;i,j) = \frac{i}{i+j}\pi(n-1;i-1,j)+\frac{j}{i+j}\pi(n+1;i,j-1)$. It can be derived if you consider the last step of drawing the ball (either red or lemon). But I am not sure if it is helpful in getting the result that I am interested in.
(2) Where you can end with depends on the order of all drawings. The sequence of red balls and lemon balls with the $i_{th}$ element as the the $i_{th}$ step that you draw determines where you are. And there are totally $\binom{i+j}{i}$ different sequences.
(3) Therefore we have $\pi(n;i,j) = \frac{\#\,of\,sequences\,that\,take\,you\, ended\,at\,coordinate\,n}{\binom{i+j}{i}}$.
(4) Meaningful $n$ is at range $max\{0,i-j\}\leq n \leq i$.
4.1 Following hint may misleading, read with criticism
(5) It seems $\frac{\binom{i+j}{j+n}}{\binom{i+j}{i}}$ will give you right answer for $F(n;i,j)$, but I cannot explain the meaning of $\binom{i+j}{j+n}$. A wrong way (but first glance seems right) to explain it is to pick $j+n$ positions from a $i+j$ sequence, and fill these positions with $j$ lemon balls in first $j$ positions and $n$ red balls in following positions; and for the remaining $i-n$ positions in original sequence, fill them with remaining red balls. With this construction, you will get a sequence that get you at least to coordinate $n$. However, this construction counts some sequences (end with more than n red balls) more than once and miss some sequences(there are lemon balls in the later part of sequence), and these the number of these two groups are just coincidentally the same. Following this clue, it seems there is a way to construct the sequence correctly.
5 Reference
[1]Grimmett G, Stirzaker D. Probability and random processes[M]. Oxford university press, 2001.
[2]Grimmett G, Stirzaker D. One thousand exercises in probability[M]. Oxford University Press, 2001.
Red and blue function the same as PUSH and POP operations for a computer stack. Push adds an object to the stack, Pop removes the top item in the stack unless the stack is empty.
The quantities in the book's answer appear to be equivalent to showing that the sequence of Push/Pop instructions is uniquely determined by the $i-n$ times, from $i+j-1$ possibilities, of Pop instructions that are applied to a non-empty stack. Thus the number of sequences with $X=n$ would be $\binom{i+j-1}{i-n}$.