The problem is from a textbook I'm reading, but even with the hint, I'm not being able to come up with a solution.
Let $X_1$ and $X_2$ be two random variables with CDFs $F_1$ and $F_2$. We say $X_2$ stochastically dominates $X_1$, or $X_2 \succsim X_1$, if $F_1(x) \geq F_2(x)$ for any real number $x$. Find an example of $X_1$ and $X_2$ on the same probability space where $X_2 \succsim X_1$ but $P(X_1 > X_2) \geq 0.95$.
Hint: clockwork
My first thought was to compare the positions of the hour and minute hands over a period of time, but whenever the hour hand is high enough to be greater than the minute hand 95% of the time, it still stochastically dominates the minute hand.
Let $X_2$ be a discrete uniform distribution over the set $\{1, \cdots, n\}$, and define
$$X_1=X_2+1 \text{ if } X_2= 1, \cdots, n-1$$ $$X_1=1 \text{ if } X_2=n.$$
Indeed, $X_1$ is obtained by a 1-step clockwise rotation of the values of $X_2$. Then, both $X_1$ and $X_2$ have the same cdf, while
$$\mathbb P(X_1 > X_2) \geq 1-\frac{1}{n}.$$
To get the lower bound 0.95, we need to set $n \ge 20$.
To have intuition on how this ties into the clockwork hint, let us target $n=24$ (while it is nicer to consider $n=12$ as an example, but we did not becuase $\frac{11}{12}$ is less than 0.95).
Assume that we do not know the time now (so it is uniformly distributed on $]0,12]$. Suppose that the existing clock is such that you can only read the time in multiples of half an hour, i.e., $$ 12, 0.30, 1, 1.30, 2, ..., 11.30.$$ Now let $X_2$ denote the time you read now by looking at the clock and $X_1$ is the time you will read from the clock half an hour later (note that here $X_1=X_2+0.30$ for $X_2=0.30, \cdots, 11.30$, and $X_1=0.30$ for $X_2=12$).
In the above examples, $F_{X_1}(x)=F_{X_2}(x)$ for all values of $x$. To have an example such that $F_{X_1}(x)>F_{X_2}(x)$ for some $x$, consider the following setting:
$$X_1=X_2+0.5 \text{ if } X_2= 1, \cdots, n-1$$ $$X_1=0.5 \text{ if } X_2=n.$$
for which $F_{X_1}(x)>F_{X_2}(x)$ for all $x=0.5, \cdots, n-1+0.5$, whereas again we have $\mathbb P(X_1 > X_2) \geq 1-\frac{1}{n}.$