Finding the value of an biased coin flip game with stop loss Question:
I have a biased coin, with 70% H and 30% T. I receive \$1 each time I flip a H and lose \$1 for T. 
I am going to flip the coin 100 times, without any constraints, the expectation of money I will received will be \$40. 
Now, what is the expectation if I stop once lose \$10?  
My thoughts:
I model this as a random walk, $S_t = \sum_{i=1}^t Z_i$. And $\tau$ be the stopping time when $S_t$ reaches -10 for the first time. Our game stops at $\tau \wedge N$, where $N = 100$.
Now the expectation will be: 
$\mathbb{E}[S_{\tau \wedge N}] = -10p + 40(1 - p)$, where $p = P(\tau\leq N)$ 
If the coin is unbiased, then $p$ can be found using reflection principle. Not sure how to find the distribution of $\tau$ for the biased coin. Any suggestions are welcome! 
Edited: I just realized, if the coin is unbiased, then $S_{t\wedge N}$ will be a martingale and we can use optimal stoping theorem to get $\mathbb{E}[S_{\tau \wedge N}] = \mathbb{E}[S_0] \mathbb{E}[\tau \wedge N] = 0$. We can apply this theorem because $\tau \wedge N$ is also a stopping time and bounded. 
 A: This is not a complete answer (since I did not find a closed form for this probability and even suspect it may not exist at all), but here are my thoughts that may give you a better insight to the problem.
First, for a fixed $k$ and a fixed sum $S$ let us count the following probability
$$f(k, S) = P(z_1 + z_2 + \dots + z_k = S)$$
For all $S > k$ and $S < -k$ this probability is equal to 0. Note that $z_k$ is either $1$ or $-1$ with probabilities 0.7 and 0.3 respectively. Thus,
$$f(k, S) = 0.7f(k - 1, S - 1) + 0.3f(k - 1, S + 1)$$
This formula allows computing $f(k, S)$ recursively.
Then, for some $m > 0$:
$$p_m := P(\tau < N, \tau \geqslant m) = P(\tau < N, \tau \geqslant m | \tau = m)P(\tau = m) + P(\tau < N, \tau \geqslant m | \tau \geqslant m + 1)P(\tau \geqslant m + 1) = P(\tau < N | \tau = m)P(\tau = m) + P(\tau < N | \tau \geqslant m + 1)P(\tau \geqslant m + 1) = f(m, -10) + P(\tau < N, \tau \geqslant m + 1)(1-f(m, -10)) = \\ = f(m, -10) + (1-f(m, -10))p_{m+1}$$
As you can see, $p_m$ also can be counted recursively (but the closed form for $p_m$ can hardly be computed). The answer to your question is $p_1$.
