Suppose I have 10 dollars and I'm able to make fair 50/50 bets like flipping a coin. Now suppose each bet is for 1 dollar. What is the probability that if I keep making bets until I hit 0 dollars that the highest amount of money reached is k dollars?
This may help: In the past I solved a similar problem I was wondering about of the form what is the probability that I will hit k dollars before I hit 0 dollars by drawing it out and seeing that the possibilities formed an infinite tree and that an equation could be formed by creating an equation which expresses the probability sought in terms of itself.
What I'm asking above is not just that I would reach k dollars but that k dollars would be the highest level if I kept betting until I hit 0.
Suppose player A and B start with $a$ dollars and $b$ dollars respectively. They play a fair game, betting $1$ dollar each time. It is a standard result that goes back to Huygens that the probability A will be ruined is $\frac{b}{a+b}$ and the probability B will be ruined is $\frac{a}{a+b}$.
We apply this to our game. Suppose that you start with $10$ dollars. Then the probability you will reach a total of $10+t$ dollars is $\frac{10}{10+t}$. Given that you have reached $10+t$, the probability that you will not reach $10+t+1$ is the probability of ruin if you have $10+t$ and your opponent has $1$, which is $\frac{1}{11+t}$.
Thus the probability your maximum will be exactly $10+t$ is $$\frac{10}{10+t}\cdot \frac{1}{11+t}.$$ In terms of $k$, our probability is $\dfrac{10}{k(k+1)}$ for $k\ge 10$.