Assuming optimal play, I think Bndrew cannot win more than

50% of the time.

Why?

Let $a$ and $b$ the two secret numbers such that $a \leq b$, and $x$ the number given by Andrew ($x=a$ or $x=b$).

If we exclude strategies that involve saying "largest" or "smallest" at random (which cannot exceed 50% win rate), Bndrew's choice can only rely on the value of $x$. That means any strategy will consist of choosing a set $S \subseteq [0,1]$ such that Andrew will say "largest" if $x \in S$, and say "smallest" otherwise.

What's the best possible choice for $S$? It would not make sense for the upper bound of $S$ to be lower than $1$, because it would mean that there exists $a \leq b$ such that Bndrew will say "highest" for $a$ but "lowest" for $b$. Hence, we can assume that $S=[s, 1]$ for some $s \in [0,1]$.

Now, what's the best possible choice for $s$? Well, to ensure a win no matter the choice of Andrew, we would like to maximise the probability of having $a \leq s \leq b$. This happens when $s=\frac{1}{2}$.

Therefore, the best strategy for Bndrew is to say "largest" if $x>\frac{1}{2}$ and "smallest" if $x \leq \frac{1}{2}$. There is a $\frac{1}{2}$ probability of having $a \leq \frac{1}{2} \leq b$ which ensures a win for Bndrew. If $a \leq b \leq \frac{1}{2}$, Andrew can prevent Bndrew from winning by giving $x=b$ (which happens with probability $\frac{1}{4}$) and if $\frac{1}{2} \leq a \leq b$, Andrew can prevent Bndrew from winning by giving $x=a$ (which also happens with probability $\frac{1}{4}$).

Assuming optimal play, I think Bndrew cannot win more than

50% of the time.

Why?

Let $a$ and $b$ the two secret numbers such that $a \leq b$, and $x$ the number given by Andrew ($x=a$ or $x=b$).

If we exclude strategies that involve saying "largest" or "smallest" at random (which cannot exceed 50% win rate), Bndrew's choice can only rely on the value of $x$. That means any strategy will consist of choosing a set $S \subseteq [0,1]$ such that Andrew will say "largest" if $x \in S$, and say "smallest" otherwise.

What's the best possible choice for $S$? It would not make sense for $S$ to not be an interval with an upper bound lower than $1$, because it would mean that there exists $a \leq b$ such that Bndrew will say "highest" for $a$ but "lowest" for $b$. Hence, we can assume that $S=[s, 1]$ for some $s \in [0,1]$.

Now, what's the best possible choice for $s$? Well, to ensure a win no matter the choice of Andrew, we would like to maximise the probability of having $a \leq s \leq b$. This happens when $s=\frac{1}{2}$.

Therefore, the best strategy for Bndrew is to say "largest" if $x \geq \frac{1}{2}$ and "smallest" if $x < \frac{1}{2}$. There is a $\frac{1}{2}$ probability of having $a \leq \frac{1}{2} \leq b$ which ensures a win for Bndrew. If $a \leq b \leq \frac{1}{2}$, Andrew can prevent Bndrew from winning by giving $x=b$ (which happens with probability $\frac{1}{4}$) and if $\frac{1}{2} \leq a \leq b$, Andrew can prevent Bndrew from winning by giving $x=a$ (which also happens with probability $\frac{1}{4}$).