Solution:

Strategy for the Detective: Pick a card $A$ and compare against all others except one. If he ever gets a "Yes", that pair works; else the remaining card is consecutive with $A$. This process takes at most $50$ queries.

Strategy for the Magician: We show that it is not always possible to obtain a "Yes" in $50$ turns, hence showing that $49$ turns are not enough to figure out a consecutive pair. It is enough to conjure a labelling of cards for which denying all $50$ inquiries is valid.

Replace $52$ by any $N>3$. Think of the cards as vertices of a complete graph $K_{N}$. Delete all edges joining vertices which correspond to pairs of cards the Detective inquired about. We will show that deleting any $N-2$ edges of $K_{N}$ still leaves a graph that admits a path containing all the vertices. Labelling all cards along this path as $1$ to $N$ would finish. Several proofs of this claim are possible. We present three of them.

## Proof 1

For any two vertices $a$ and $b$, since $\operatorname{deg} a + \operatorname{deg} b \geq 2(N-1)-(N-2)=N$, they share a common neighbour. Hence the graph is connected.

Pick the longest path $\mathcal{P}: u=u_{0} \rightarrow u_{1} \rightarrow \cdots \rightarrow u_{k}=v$. All neighbours of $u$ and $v$ must remain within the path, else we could get a longer path. Let $u$ have $x$ neighbours $\{u_{i_{1}}, u_{i_{2}}, \ldots, u_{i_{x}}\}$ with $1=i_{1}<i_{2}<\cdots<i_{x} \leq k$. Let $v$ have $y$ neighbours $\{u_{j_{1}}, \ldots, u_{j_{y}}\}$. Since $x+y \geq n$, we see that $i_{s}=j_{r}+1$ for some $r$ and $s$. Thus there exists $i$ such that $u \rightarrow u_{i+1}$ and $u_{i} \rightarrow v$ are edges. Thus the path is a cycle
$$\mathcal{C}=u_{i+1} \rightarrow u_{0} \rightarrow u_{1} \cdots \rightarrow u_{i} \rightarrow v \rightarrow u_{k-1} \cdots \rightarrow u_{i+1}$$
Suppose a vertex $w$ is not in the path $\mathcal{P}$. By connectedness, we have a path $\mathcal{P}'$ from $w$ to some vertex of $\mathcal{P}$. Continue along this path via the cycle $\mathcal{C}$ to obtain a path longer than $\mathcal{P}$; contradiction! Thus the graph has a path of length $N-1$, as desired.

## Proof 2

Pick the longest cycle $\mathcal{C}=v_{1} \rightarrow \cdots \rightarrow v_{k} \rightarrow v_{1}$. Note that any vertex $w$ not in the cycle can be incident to no more than $\frac{k}{2}$ of the vertices in it; else there exists $i$ such that $w v_{i}$ and $w v_{i+1}$ (indices mod $k$) are edges, so we can put $w$ in to get a longer cycle. Thus our graph is missing at least $\frac{1}{2} k(N-k)$ edges. So $2(N-2) \geq k(N-k)$. Clearly $k>2$ so we see that $k \in\{N-2, N-1\}$.

Case 1. $k=N-1$. Pick the leftover $w$ outside $\mathcal{C}$. Not all edges from $w$ to the cycle are missing (since only $N-2$ are missing in total), so follow an edge from $w$ to $\mathcal{C}$ and continue along $\mathcal{C}$ to get a path of length $N-1$.

Case 2. $k=N-2$. Pick the leftover $a, b$ outside $\mathcal{C}$. It is clear that both of them have edges to the cycle and $ab$ is also an edge (since $k(N-k)=2(N-2)$ in this case). So starting at $a$, going to $b$, to some vertex of $\mathcal{C}$ and following along $\mathcal{C}$ gives us a path of length $N-1$.

The proof is complete.

## Proof 3

The idea is to prove the stronger claim by induction on $N \geq 3$: a graph on $N$ vertices with $\binom{N-1}{2}+2$ edges has a cycle of length $N$. Deleting the extra edge will give a path of length $N-1$ through all the vertices.

The base case $N=3$ is trivial. Suppose it holds for all $k \leq N$, we prove it for $N+1$. Since $\frac{2\left(2+\binom{N}{2}\right)}{N+1}>N-2$ we see that some vertex $v$ has degree either $N-1$ or $N$.

Case 1. If degree of $v$ is $N-1$. Then we have an edge $e=uv$ missing among all the edges through $v$. Delete $v$ along with all the edges through it in the graph. The induced graph has a cycle of length $N$. Pick two consecutive vertices that are not $u$, and append $v$ between them.

Case 2. If degree of $v$ is $N$. Delete $v$ along with all its edges. Add an arbitrarily chosen extra edge to the graph so obtained. By induction hypothesis, this resulting graph has a cycle of length $N$. If removing the extra edge does not disrupt the cycle, append $v$ anywhere between two consecutive vertices. If it does break the cycle, use $v$ to connect the vertices it joined.

The induction is complete.