= Solution
Use positive ability parameters in the <Bradley-Terry model>, so $\mathbb P(i\text{ beats }j)=\theta_i/(\theta_i+\theta_j)$. If the course instead denotes log abilities by $\theta_i$, apply the following calculation to their exponentials; the ordering is unchanged. Up to a factor independent of the abilities, the <likelihood function> is
$$
L(\theta)=\frac{\theta_1}{\theta_1+\theta_2}
\frac{\theta_3}{\theta_1+\theta_3}
\left(\frac{\theta_2}{\theta_2+\theta_3}\right)^k.
$$
The observed wins form a directed cycle, so a finite maximum exists. The <Bradley-Terry likelihood Hessian> is negative definite on contrasts of log abilities, giving uniqueness up to common scaling. We can therefore find the <maximum-likelihood estimate> through the <Bradley-Terry score equations>.
Player 1 has one observed win in two comparisons. Its score equation is
$$
1=\frac{\theta_1}{\theta_1+\theta_2}
+\frac{\theta_1}{\theta_1+\theta_3},
$$
which simplifies to $\theta_1^2=\theta_2\theta_3$. By the model's scale invariance, set $\theta_2=1$ and write $\theta_1=r$, $\theta_3=r^2$, with $r>0$. Player 2's score equation becomes
$$
k=\frac1{1+r}+\frac{k}{1+r^2},
\qquad kr^3+(k-1)r^2=1.
$$
The left side of the polynomial equation is strictly increasing on $r>0$, starts at zero, and tends to infinity. For $k=1$, its unique solution is $r=1$. For $k>1$, its value at one is $2k-1>1$, so its solution satisfies $0<r<1$. The <Three-player Bradley-Terry comparison cycle> consequently gives
$$
\boxed{k=1:\quad\widehat\theta_1=\widehat\theta_2=\widehat\theta_3;
\qquad k>1:\quad\widehat\theta_2>\widehat\theta_1>\widehat\theta_3.}
$$
Thus there is a complete tie when each directed edge is observed once, and otherwise the decreasing ranking is \b[2, 1, 3].
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