Toss a fair coin a number of times. [closed]

Toss a fair coin, number of times. You always bet on heads and gain one dollar if it is a head and lose one dollar if it is a tail. You stop whenever you gain 3 dollars or you lose 2 dollars. Then which of the following is the correct one?

1. you end up winning with $2\over 3$ probability
2. you end up winning with $1\over 3$ probability
3. you end up winning with $3\over 4$ probability
4. you end up winning with $3\over 5$ probability
5. None of the above is correct.

If we don't know the number of trials, how can we do it?

For elegance, note that we can start from $2$ dollars instead of nothing, and win if you get to $5$ dollars and lose if you get to $0$ dollars.

Let $p_n$ be the probability that you will win if you start with $n$ dollars.

Then $p_0 = 0$ and $p_5 = 1$.

Also $p_n = \frac{1}{2} p_{n-1} + \frac{1}{2} p_{n+1}$ for any integer $n$ (strictly) between $0$ and $5$.

Thus $p_{n+1}-p_n = p_n-p_{n-1}$ for any integer $n$ (strictly) between $0$ and $5$.

Thus $p_5-p_0 = 5 ( p_1-p_0 )$ and hence $p_1-p_0 = \frac{1}{5}$.

Thus $p_n = \frac{n}{5}$ for any integer $n$ from $0$ to $5$.

I assume that you stop when your net gain becomes +3 or -2.

On average, you don't win or lose money in one toss. So your average gain in the whole game must be $0$ as well. (I don't really know how to prove it strictly, but if you just need to give the answer in a test, this consideration is enough.)

If you win with probability $p$ and lose with probability $1-p$, then your average gain is $3p-2(1-p)$. So you have $3p-2(1-p) = 0$, which means that $p = \frac25$.