Elo Ratings & Bradley–Terry
Elo Ratings & Bradley–Terry Models
The Elo system, invented by physicist Arpad Elo for chess, represents each competitor by a single rating $R$. The expected score between players $A$ and $B$ is $E_A = 1/(1 + 10^{(R_B - R_A)/400})$, a logistic function of the rating difference. A 400-point gap implies $\sim 10\times$ odds in favor of the higher rating; 200 points $\sim 3.2\times$ odds; 100 points $\sim 1.78\times$.
After each game, ratings update by $R'_A = R_A + K(S_A - E_A)$, where $S_A \in \{0, 0.5, 1\}$ is the actual result and $K$ is the K-factor controlling responsiveness. Larger $K$ means faster adaptation but noisier ratings. FIDE uses $K=40$ for new players, $K=10$ for masters. FIFA uses K variable by tournament importance; 538's NFL Elo uses $K \approx 20$ with margin-of-victory adjustment.
Elo is a special case of the Bradley–Terry model: $P(A \text{ beats } B) = \pi_A/(\pi_A + \pi_B)$, where $\pi$ is a positive strength parameter. Taking $R = 400 \log_{10} \pi$ recovers Elo. Bradley–Terry generalizes to multiple competitors, ties, and home-field advantage.
$$E_A = \frac{1}{1 + 10^{(R_B - R_A)/400}}, \qquad R'_A = R_A + K(S_A - E_A)$$
Zero-sum: $R'_A + R'_B = R_A + R_B$. A draw transfers points from higher to lower rating.
$P(H \text{ beats } A) = \pi_H h /(\pi_H h + \pi_A)$, where $h > 1$ is a multiplicative home bonus. In Elo units this is an additive shift — typical NFL home advantage is $\approx 55$ Elo points; NBA $\approx 100$; MLB $\approx 24$.
Player A is rated 1800, B is rated 2000. Compute A's expected score.
$E_A = 1/(1 + 10^{200/400}) = 1/(1+10^{0.5}) = 1/(1+3.162) = 1/4.162 = 0.240$.
A expected to score 24% — either a 24% win-or-draw probability or average fractional points.
A (1800) beats B (2000) with $K = 32$. Compute new ratings.
$E_A = 0.240$, $S_A = 1$. $\Delta = 32(1 - 0.240) = 24.3$. New: $R'_A = 1824.3$, $R'_B = 1975.7$.
Two NFL teams rated 1500 and 1520 play at Team A's home (home advantage 55 Elo). Compute Team A's win probability.
Effective rating gap: A home = 1555 vs B = 1520. Diff = +35 for A. $E_A = 1/(1+10^{-35/400}) = 1/(1+0.820) = 0.550$.
Practice Problems
Show Answer Key
1. $1/(1+10^{-1}) = 0.909$ (≈91%).
2. No change — both had $E=0.5$, $S=0.5$, so $\Delta = 0$.
3. To stabilize established ratings; low K reduces noise once a rating is considered mature.
4. Let $\pi_A = 10^{R_A/400}$. Then $P(A) = \pi_A/(\pi_A + \pi_B) = 1/(1 + 10^{(R_B-R_A)/400})$.
5. $1/(1+10^{-120/400}) = 1/(1+0.501) = 0.666$.
6. $\pi$ ratio $= 10^{50/400} = 10^{0.125} \approx 1.33$ — 33% stronger by BT.