(Partial Answer)
Let's consider a discrete version of the problem, where Andrew is secretly shown three distinct numbers from $\{1,2,3,...,N\}$.
We'll show that for $N=6$, Andrew has a strategy that limits Bndrew's probability of winning to $0.35$. (This value tends towards $\frac{1}{3}$ as $N$ increases.)
Andrew's strategy
As an example, let's consider $N=6$. There are $\binom{N}{3}=20$ scenarios. Each row/scenario in the table below shows which numbers Andrew is shown, and the probability that he chooses that number. (For example, the first row corresponds to the scenario where Andrew is shown $\{1,2,3\}$. Andrew always picks $3$ in this scenario.)
$$
\begin{array} {cc} 1&2&3&4&5&6 \\ \hline 0&0&1&&&&\\0&0&&1&&&\\0&0&&&1&&\\0&1&&&&0&\\0&&0&1&&&\\0&&0&&1&&\\0&&1&&&0&\\0&&&1&0&&\\0&&&1&&0&\\0&&&&1&0&\\&0&0&1&&&\\&0&1&&0&&\\&0&1&&&0&\\&0&&1&0&&\\&0&&1&&0&\\&0&&&1&0&\\&&0&1&0&&\\&&0&1&&0&\\&&0&&1&0&\\&&&0&1&0&\\ \end{array}
$$
Bndrew's strategy
We can assume that Bndrew knows Andrew's strategy. Let's calculate the probability that Bndrew wins. To simplify the numbers, let's calculate the Bndrew's expected number of wins if we play each scenario once (and later divide by $20$ to get the probability).
$2$ is only shown once, and Andrew will pick an opaque box. Expected wins: $0.5$
$3$ is shown four times, and $3$ is the highest number only once. Picking $3$ would lead to $1$ win, but picking an opaque box will give $0.5$ expected wins $3$ times (so Bndrew will pick an opaque box). Expected wins: $1.5$
$4$ is shown nine times. One-third of the time it's the highest. Bndrew's strategy is irrelevant here. Expected wins: $3$
$5$ is shown six times. One-third of the time it's the highest. Bndrew's strategy is irrelevant here. Expected wins: $2$
Bndrew expects $7$ wins out of $20$ attempts. So, for $N=6$, Bndrew wins with probability $\frac{7}{20} = 0.35$.
Probability of winning
As $N\rightarrow\infty$, the probability of Bndrew winning approaches $\frac{1}{3}$.
Example: an optimal solution for N=20
I'll explain an optimal strategy for Andrew for N=20. It limits Bndrew's probability of winning to $\frac{1}{3}+\frac{1}{N(N-1)(N-2)}=0.333479...$.
It turns out that Andrew never needs to select the lowest number (of his 3 options).
Here is one optimal strategy for Andrew:
![enter image description here](https://cdn.statically.io/img/i.sstatic.net/9me9T5KN.png)
Let's say Andrew has a choice of $\{15,10,7\}$. Andrew ignores the $7$ and looks up $10,15$ on the chart, which has a value of $4$.
The entry at $10,15$ actually represents $9$ scenarios: $\{15,10,1\}$, $\{15,10,2\}$, ..., $\{15,10,9\}$, and the entry indicates that in $4$ of the $9$ scenarios, Andrew should pick the biggest number offered. So, Andrew picks $15$ with a probability of $\frac{4}{9}$ (otherwise, he picks $10$).
The chart was constructed such that for each number, if you consider all the scenarios (triplets) which pick that number, the biggest number is picked in exactly one-third of those scenarios. EXCEPTION: In the $\{1,2,20\}$ scenario, Andrew picks $2$, and that is the only scenario in which he picks $2$, which gives Bndrew a probability of $\frac{1}{2}$ of winning.
Example: Let's find the scenarios where Andrew picks $3$, according to the chart: $\{1,2,3\}$, $\{1,3,20\}$, $\{2,3,20\}$. $3$ is the biggest one-third of the time.
I found the chart's values in an ad-hoc manner, but it should be possible to smooth the values out. (It's straight-forward to modify the chart and have it remain valid).