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For a connected tree of comparisons with every observed win fraction strictly between zero and one, the Bradley-Terry model can fit every empirical edge probability exactly. Fix one positive strength to remove scaling ambiguity, then propagate strength ratios $p/(1-p)$ along tree edges. Edge log-ratios are independent real coordinates and each binomial log-likelihood term is strictly concave, proving existence and uniqueness of the normalized estimate. Cycles would impose extra compatibility relations.

For observed neighbor win fractions $p_0,\dots ,p_{n-1}\in (0,1)$ on the path graph $0,1,\dots ,n$, the normalized Bradley-Terry model estimate with ${\hat{\theta }}_n=1$ is ${\hat{\theta }}_i={\prod }_{k=i}^{n-1}{p}_k/(1-p_k)$. This is backward propagation of empirical odds along the path. Every edge proportion is fitted exactly, and strict concavity in edge log-ratios proves that this is the unique global maximum.

Orient the edges of a comparison tree and set ${\eta }_{ij}=\log {\theta }_i-\log {\theta }_j$. After anchoring one vertex strength, these edge quantities are unconstrained coordinates. The Bradley-Terry model probability is the logistic function of ${\eta }_{ij}$, and its binomial log-likelihood contribution is $m[p{\eta }_{ij}-\log (1+e^{{\eta }_{ij}})]$. Differentiating gives the fitted odds $p/(1-p)$.

The Bradley-Terry model assigns comparison probability $q_i={\theta }_i/({\theta }_i+{\theta }_{i+1})$ to each observed edge. Up to factors independent of the parameters, its likelihood function isThe comparison graph is a path graph, hence a tree. After fixing ${\theta }_n=1$, its edge ratios are unconstrained positive coordinates: every collection $r_i={\theta }_i/{\theta }_{i+1}>0$ determines uniquely ${\theta }_i={\prod }_{k=i}^{n-1}{r}_k$.

For a convenient strict-concavity calculation, use the edge log-ratios in a Bradley-Terry comparison tree ${\eta }_i=\log r_i$. The log-likelihood separates asEach derivative is $m(p_i-q_i)$, where $q_i$ is the logistic function of ${\eta }_i$, and each second derivative is $-m{q}_i(1-q_i)<0$. Because $0<p_i<1$, there is a unique finite global maximum atConverting back gives the Bradley-Terry maximum-likelihood estimate on a path:Equivalently, recurse backwards using ${\hat{\theta }}_i={\hat{\theta }}_{i+1}{p}_i/(1-p_i)$. All estimates are finite and positive. The sample size $m$ changes the curvature of the likelihood but not this maximizer; absence of comparison cycles is what permits every empirical edge proportion to be fitted simultaneously.

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.