Option 5: none of the above. We can of course exclude 1, 3, and 4 out of hand because otherwise gambling would work more often than it failed, but to exclude 2 we must get the actual value.
Starting from $0$, your odds of ending up (after any number of rolls) at $+2$ or $-2$ are even.
If it's $-2$, then you lose $= 0.5$ chance of losing so far, and the game ends so we can ignore this branch from now on.
If you are at $+2$ then your odds of ending up at $+3/+1$ (after any number of rolls) are even.
If it's $+3$, you win $= 0.5*0.5 = 0.25$ chance of winning so far.
If $+1$, then odds of $+3/-1$ are even.
If $+3$, you win $= 0.25 + (0.5*0.5*0.5) = 0.375$ chance of winning so far.
If $-1$, odds of $-2/0$ are even.
If $-2$, you lose $= 0.5 + (0.5*0.5*0.5*0.5) = 0.5625$ chance of losing so far.
If $0$, then we're back at zero dollars.
So we have a:
- $0.5625$ chance of losing;
- $0.375$ chance of winning;
- $0.625$ chance of having to "roll again", which will always be split into the same proportions.
So the proportion of wins:losses is hence $0.5625:0.375$.
Lose: $0.5625/(0.5625+0.375) = 0.6$
Win: $0.375/(0.5625+0.375) = 0.4$
I am a math noob (which likely shows), but the odds of winning being $2/5$ and of losing being $3/5$ just feels intuitively right, too.