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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 isThe 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 iswhich 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 becomesThe 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 givesThus there is a complete tie when each directed edge is observed once, and otherwise the decreasing ranking is 2, 1, 3.

For one win of player 1 over 2, one win of 3 over 1, and $k\ge 1$ wins of 2 over 3, the Bradley-Terry score equations force abilities proportional to $(r,1,r^2)$, where $r>0$ satisfies the displayed equation. Its left side is strictly increasing on $r>0$. For $k=1$, $r=1$ and all three estimates tie. For $k>1$, $r<1$, giving the ranking $2>1>3$.