Probability in coin toss Suppose that you have a fair coin. You start with $\$0$. You win $\$1$ each time you get a head and loose $\$1$ each time you get tails. Calculate the probability of getting $\$2$ without getting below $\$0$ at any time.
 A: To get exactly 2$ you must have a run consisting of succesive pairs of (1,-1) followed by (1,1).
For an arbitrary but fixed n, n even, You have 2^n different runs. 
Now, you get 2$ if your run begins with (1,1) the rest, 2^(n-2) digits, is arbitrary. Therefore you have 2^(n-2) successes out of 2^n events which yields 2^(n-2)/(2^n) = 1/2^2.
If Your run begins with (1,-1) followed by (1,1) you have 2^(n-4) successes out of 2^n events which yields 2^(n-4)/(2^n) = 1/2^4, etc. etc.
For an arbitrary but fixed n, n even, the probability to win 2$ is the sum s(n)=1/2^n + 1/(2^{n-2)+ ....+1/2^2. The Limes of s(n), n-> infinity is 1/3. 
A: $P$(getting \$$2$ without getting below \$$0$) = 
$P$(getting \$$2$ without getting below \$$0$ in $2$ tosses) +  
$P$(getting \$$2$ without getting below \$$0$ in $4$ tosses) + 
$P$(getting \$$2$ without getting below \$$0$ in $6$ tosses) + $\ldots$
= $\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \ldots$
= $\frac{1}{3}$
A: Consider $P_i$ the probability to obtain $\$2$ when you have $\$i$, then we look for $P_0$. So a recurrence can be
$$P_{i}=\frac{1}{2}P_{i-1} + \frac{1}{2}P_{i+1}$$
with $P_2=1$ and $P_{-1}=0$ because we do not allow a score below of $\$0$. Now
$$P_0=\frac{1}{2}P_{-1} + \frac{1}{2}P_{1}=\frac{1}{2}P_{1},$$
$$P_1=\frac{1}{2}P_{0} + \frac{1}{2}P_{2}=\frac{1}{2}P_{0} + \frac{1}{2}$$
therefore
$$P_0=\frac{1}{2}P_{1}=\frac{1}{2}\left(\frac{1}{2}P_{0} + \frac{1}{2}\right)$$
and $P_0=\frac{1}{3}$.
A: Consider the first two tosses of the coin. There are $4$ equally likely outcomes
$HH, HT, TH, TT$, and you have won if the outcome is $HH$ and lost if the outcome
is $TH$ or $TT$. Else, the outcome is $HT$ and you are back to Square One and
must toss the coin again to determine whether you will win $2$ before
your wealth reduces below $0$.
So consider the very first time that the outcome is not $HT$. Then,
conditioned on the event $A = \{HH, TH, TT\}$, the conditional
probability of winning is
$$P(HH \mid A) = \frac{P(HH \mid A)}{P(A)} = \frac{P(HH)}{P(A)} 
= \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}.$$

Alternatively, as I suggested in a comment on the main question,
a win occurs exactly when the successive tosses result in
$HH, HTHH, HTHTHH, \ldots$ which have probabilities 
$\frac{1}{4}, \frac{1}{16}, \frac{1}{64}, \ldots$ and thus
$$\begin{align}
P(2~\text{before}~ -1) &= P\{HH, HTHH, HTHTHH, \ldots\}\\
&= P(HH) + P(HTHH) + P(HTHTHH) + \cdots & \text{third axiom}\\
&= \sum_{n=1}^\infty \frac{1}{4^n}
= \frac{1}{4}\times \frac{1}{1 - \frac 14} =\frac 13.
\end{align}$$
