Expected wealth of coin toss game with stop loss

Question

You start with an initial wealth $\$X$. The coin toss game is such that where the coin is flipped $100$ times and for every heads the players get - they receive $\$2$, else they lose $\$1$. There is a stop loss at $\$0$, namely if the current wealth ever reaches $0, the game ends. What is the expected wealth?

Attempts and thoughts:

The game is a random walk, and its simple to compute the expected wealth without the stop loss. If we take $Z_i$ to be +2 with probability $1/2$ else $-1$ with probability $1/2$, then $S_k = \sum_{i=1}^{k}Z_i$, and one can show that $S_k - \frac{1}{2} t$ is a martingale, thus $\mathbb{E}[S_T] = T/2$. So, clearly the player expects to make $\$50$.

I have tried simulating this on python (taking an initial wealth to be zero):

   av = []
   for simulation in range(0,1000):
       s = 0
       l = 0
       for toss in range(0,100):
           x = np.random.randint(0,2)
           if x == 1:
               s+=2
           elif x == 0:
               s-=1

           if s <= 0:
               break
       av.append(s)
    print(np.array(av).mean())

from which I get a mean of about 16.45 or so. I am finding it difficult to work backwards and rationalise this result.

Can anyone at least point me in the right direction (just a hint/useful method please)? I'm familiar with the basic statements of optional stopping theorem (used earlier, if this can be used?).

Answer 1

Here, the stopping time $T$ is the minimum of $100$ coin flips, and the time to bankruptcy. So, on using the optional stopping theorem, we get $E[S_T]=E[T]/2$. In general, a player will get bankrupt before $100$ coin tosses, so $E[T]\neq 100$.