Elo Calculator

Enter each player's current Elo and a score to see how much Elo the bot would award.

A port of FoosballEloRankingSystem from Quantco/elo-slackbot. The numbers here match the bot to floating-point precision.

Notation

$r_{i}$: rating of player $i$
$R_{A}$: reduced rating of team $A$
$p_{A}$: points scored by team $A$
$g_{i}$: matches played by player $i$
$D$: rating gap that means “ten times as strong”, 500
$k$: gain coefficient, 10
$f$: participation bonus, 0.5
$m$: placement matches, 5

Step 1 · Team rating

R_{A} = D \log_{10} (\frac{1}{| A |} \sum_{i \in A} 10^{\frac{r_{i}}{D}})

A logarithmic mean, not an arithmetic one. One strong player pulls the team rating above the average, which is why a ringer cannot be hidden behind a weak partner.

Step 2 · Expected point share

E_{A} = \frac{1}{1 + 10^{\frac{R_{B} - R_{A}}{D}}}

A logistic curve on the rating gap. At $R_{A} - R_{B} = D$ team A is expected to take $\frac{10}{11}$ of the points.

Step 3 · Actual point share

S_{A} = \frac{p_{A}}{p_{A} + p_{B}}

The point ratio counts, not the win. 10 : 9 barely moves anything, 10 : 0 moves a lot.

Step 4 · Raw change

Δ = k (p_{A} + p_{B}) (S_{A} - E_{A})

Scaled by total points, so a long game carries more weight than a short one.

Step 5 · Per player

δ_{i} = f + c_{i} \frac{σ_{i} Δ}{| T_{i} |}

$σ_{i} = + 1$ for team A and $- 1$ for team B; $| T_{i} |$ is the size of that player's team. Everyone collects the participation bonus $f$ , win or lose.

Step 6 · Placement damping

c_{i} = {\begin{cases} 1 & if g_{i} < m \\ min_{g_{j} < m} \frac{g_{j}}{m} & otherwise \end{cases}

Players still inside their placement period take the full change; everyone else is damped by the least established player in the match. That is what stops people from farming Elo off newcomers. A 2 v 2 game counts as half a match per player, so $g_{i}$ grows by $1 / | T_{i} |$ .

Constants

Why combining is never exact

The participation bonus is paid per entry, not per game played. Three games hand everyone $3 f = 1.5$ Elo for free, a single entry only $f$ . That gap of $(n - 1) f$ hits every player equally and no score can make up for it, so it is reported separately in the result.