Can you help me understand this exam answer related to “breaking discrete logarithm over P256”?
I hope I interpreted this correctly, as both the question here and the pdf are not completely clear to me. As the question seems to do, I will everywhere assume that the chosen walks are completely random.

The idea is that we are going to run the ${\tt Dlog}$ function, and we want to know the expected value of the number of times we have to iterate its loop. Suppose we start a loop, and we have chosen points $P_0,\ldots,P_{\ell-1}$. Three things can happen:


We have seen the point $P_{\ell-1}$ before.
We have not seen the point $P_{\ell-1}$ before, and the point is distinguished.
We have not seen the point $P_{\ell-1}$ before, and it is not distinguished.


This is an exhaustive list of options. Now let's suppose that if we are in 3, that we choose $P_{\ell}$ and continue the loop. Then after a while we are going to end up either in 1 or 2. This shows that every loop either ends up on a point which we have seen before (let's call this event $X$), or ends up on a point which we have not seen before, and the point is distinguished (let's call this event $Y$).

Estimating some probabilities:

Let's call $\mathcal{P}(X)=x$ and $\mathcal{P}(Y)=y$. Since these two options are exhaustive, we have $$x+y=1.$$
Since by assumption we visited at least $k\frac{n}{t}$ points, we have 
$$x\geq \frac{1}{n}\cdot k\frac{n}{t}=k\frac{1}{t}.$$
Moreover, suppose we visited $v$ points, then
$$y=\frac{n-v}{n}\frac{1}{t}\leq\frac{1}{t}.$$
Combining the last two equalities, we conclude that $x\geq k\frac{1}{t}\geq ky$. Writing $x=1-y$, from this it follows that $1-y\geq ky$, and therefore that $y\leq\frac{1}{k+1}$. Thus $-y\geq\frac{-1}{k+1}$. From this we we have that
$$x=1-y\geq1-\frac{1}{k+1}=\frac{k}{k+1}.$$
Finally we conclude that $$\frac{1}{x}\leq \frac{k+1}{k}=1+\frac{1}{k}.$$

The final argument:

So why did we do all of this. Well, we have defined a "successful" event, namely $X$, and an "unsuccessful" event, namely $Y$. How many times do we have to repeat before we end up in $X$? This is a standard geometric distribution with probability $x$. It has expected value
$$\sum_{i=1}^{\infty}\left(i\cdot xy^{i-1}\right),$$
which is known to equal $\frac{1}{x}$. As $\frac{1}{x}\leq 1+\frac{1}{k}$, we are done.