Intuition behind Gambler's Ruin Problem Solution

In case of gambler's ruin (fair) the probability that a process starting at state $j$ eventually will reach state $N$ before state 0 is $\frac{j}{N}$. I understand that it should be proportional to $j$. But, intuitively, why $\frac{j}{N}$ ?. I am asking intuitive answer. Formal proof is available in any standard text book.

• Once you understand that the probability is $\sim j$ (and in fact that is the less trivial part), then it is clear that the proportional factor is such that the probability is $0$ if $j=0$ and $1$ if $j=N$. The proportionality itself folows from tha fact that $f(j)$ should be the average of $f(j-1)$ and $f(j+1)$. – Hagen von Eitzen Nov 2 '13 at 18:05

The expected value of your position is always $j$. So that means the probability of getting to $N$ times the "weight" of going to $N$ has to equal the probability of getting to 0 times the "weight" of going to 0. In symbols:

$$P(N) * (N-j) = P(0) * j$$

And since $P(0) + P(N) = 1$, we get that $P(N) = j/N$.

here's an intuitive way to arrive at P(j)= j/N.

the game is fair, so the avg gain per game should be zero.
Avg gain = ( (Prob to win) x gain per won game) - ((Prob to lose) x loss per lost game) = (Prob to win) x (N-j) - (Prob to lose) x (j) = 0

But (Prob to win + Prob to lose) = 1,
so (Prob to lose) = (1 - (prob to win) ).

so (Prob to Win) x (N-j) - (1 - Prob to Win) x j = 0
(Prob to Win) x N = j