Handicap Calculator
Rating Range Tables
The table below shows the rating difference ranges that require specific handicap points for optimal balance. The handicap points are the same across all match formats (Best of 3, 5, or 7).
| Handicap Points |
Rating Difference Range | ||
|---|---|---|---|
| NewTTRS / RC | TTR | USATT | |
Rating Difference: The rating difference between the stronger and the weaker player.
Handicap Points: Number of points the weaker player starts with in each set (0 means no handicap needed).
Probabilistic Model
Consider a match between two players, A and B. We assume a fixed point-winning probability \( p \) for player A. While this assumption simplifies the model, it is not entirely realistic, as in real-world scenarios the point-winning probability depends on who is serving and other external factors. The probability that player B wins a point is then \( 1 - p \).
A table tennis set can be modeled as a Markov chain (see Figure 1). If the current score is \( i:j \), the next point will result in a score of either \( (i+1):j \) with probability \( p \), or \( i:(j+1) \) with probability \( 1-p \). This means the next state depends only on the current state, satisfying the Markov property.
We define \( P(i,j) \) as the probability that player A eventually wins the set given the current score \( i:j \). The recursive relation is then:
\[ P(i,j) = p \cdot P(i+1,j) + (1-p) \cdot P(i,j+1). \]The boundary conditions correspond to the winning criteria of a set. Player A wins the set by reaching 11 points with a lead of at least 2 points:
\[ P(11,j) = 1 \quad \text{for all } j \leq 9, \] \[ P(i,11) = 0 \quad \text{for all } i \leq 9. \]Deuce situation
In a deuce situation \( l : l \), where \( l \geq 10 \), the player who gains a two-point advantage wins. This scenario can be modeled again using a Markov chain (see Figure 2). The probability that player A eventually wins from deuce satisfies
\[ P(\text{deuce}) = p^2 \cdot 1 + (1-p)^2 \cdot 0 + 2 \cdot p (1-p) \cdot P(\text{deuce}). \]By rearranging, we can find a closed-form solution
\[ \begin{aligned} P(\text{deuce}) - 2p(1-p) \cdot P(\text{deuce}) &= p^2, \\ P(\text{deuce}) \left[ 1 - 2p(1-p) \right] &= p^2, \\ P(\text{deuce}) &= \frac{p^2}{1 - 2p(1-p)}. \end{aligned} \]This yields the boundary condition
\[ P(10, 10) = \frac{p^2}{p^2 + (1-p)^2}. \]Using dynamic programming we can solve for all \(P(i,j)\) quickly.
Closed-form set win probability
We can derive a closed-form solution by making use of the negative binomial distribution. To achieve victory in a table tennis set, player A must reach 11 points with a minimum 2-point advantage. This can be accomplished through two mutually exclusive scenarios:
- Direct victory: Player A reaches 11 points before player B accumulates 10 points
- Deuce resolution: Both players reach 10 points, followed by player A winning the subsequent deuce situation
Let \(W(n, k, p)\) denote the probability that player A wins exactly \(n\) points before player B wins \(k\) points, where each individual point is won by player A with probability \(p\).
Case 1: Direct victory
For player A to achieve a direct victory with final score \(11:j\) where \(j \leq 9\), player A must win exactly 11 points while player B wins exactly \(j\) points. The total number of points contested is \(11 + j\), with the constraint that player A must win the final point to secure the set.
\[ P_{\text{direct}}(j) = \binom{10+j}{j} p^{10} (1-p)^j \cdot p. \]Summing over all possible values of \(j\) for direct victory we get
\[ P_{\text{direct}} = \sum_{j=0}^{9} \binom{10+j}{j} p^{11} (1-p)^j. \]Case 2: Victory through deuce resolution
Alternatively, player A may achieve victory by first reaching deuce score \(10:10\) and then winning two points. The probability of reaching a score of \(10:10\) is
\[ \binom{20}{10} p^{10} (1-p)^{10}. \]Combining this with our previously derived deuce win probability we get
\[ P_{\text{deuce}} = \binom{20}{10} p^{10} (1-p)^{10} \cdot \frac{p^2}{p^2 + (1-p)^2}. \]The total probability that player A wins the set from initial state \(0:0\) is the sum of the probabilities for both victory scenarios:
\[ \boxed{ P_{\text{set}} = \sum_{j=0}^{9} \binom{10+j}{j} p^{11} (1-p)^j + \binom{20}{10} p^{10} (1-p)^{10} \cdot \frac{p^2}{p^2 + (1-p)^2} }. \]Closed-form match win probability
Similarly to the point scores we can model set scores \(m : n\) again as a Markov chain (like in Figure 1) with transition probability \(P_\text{set}\). Let's assume \(s\) is the amount of sets needed to win the match (usually 2, 3 or 4). The probability that player A wins a match is
\[ \boxed{ P_{\text{match}} = \sum_{k=0}^{s-1} \binom{s-1 + k}{k} (P_{\text{set}})^s (1 - P_{\text{set}})^k }. \]Calculation of Handicap Points
NewTTRS (RC)
The NewTTRS rating system1 used in Austria and Australia (ratingscentral) calculates rating scores for each player using Bayesian statistics. It internally defines a probability-of-upset function
\[ \pi(x) := \frac{1}{1 + e^{\frac{x}{67}}}. \]The probability that player A with playing strength \(r_A\) will upset player B with playing strength \(r_B \geq r_A\) is \(\pi(r_B - r_A)\), which is equivalent to the match-win probability for player A \(P_\text{match}\). Note that NewTTRS does not differentiate between best of 3, 5 or 7 matches. Given a playing strength difference \(x\) and sets required to win \(s\) we can now calculate the point-win probability \(p\) (\(P_\text{point}\)) for the weaker player. We assume \(s\) to be \(3\), as most matches played and submitted are best of 5.
Using the \(P_\text{point}\) values and dynamic programming, we can now find the optimal amount of handicap points to balance the match best, i.e. find the \(1 \leq h \leq 10\) for which \(P_\text{match}\) is closest to \(0.5\) when \(P_\text{set}\) is redefined as \(P(h,0)\).
TTR System
Germany uses the TTR system2 which is quite similar to NewTTRS, and defines the match win probability as
\[ \pi(x) := \frac{1}{1 + 10^{\frac{x}{150}}}, \]where \(x = \text{TTR}_B - \text{TTR}_A\). We can again compute the table for the similar formula.
USATT System
The United States Association of Table Tennis (USATT) uses a discrete point exchange table rather than a continuous probability function3. Unlike other rating systems that explicitly define win probability functions, USATT's approach is based on point exchanges that implicitly encode win probabilities through an equilibrium assumption.
The USATT point exchange system awards points to the winner and deducts points from the loser based on the rating difference between players. When a higher-rated player wins as expected, they gain fewer points, while an upset victory by the lower-rated player yields more points. The system assumes that at equilibrium, the expected point change for any player should be zero over many games.
Using this equilibrium condition, we can derive the implicit win probabilities. For a given rating difference range with expected win points \(E\) and upset win points \(U\), if the win probability is \(p\), then
\[ p \cdot (-E) + (1-p) \cdot U = 0. \]Solving for \(p\) yields.
\[ p = \frac{U}{E + U}. \]By fitting a logistic curve to these derived probabilities, we estimate the implicit probability function
\[ \pi(x) \approx \frac{1}{1 + e^{\alpha x}}, \]where \(\alpha \approx 0.0173\) for the USATT system.
It is important to note that this probability function is not officially defined by USATT but rather inferred from their point exchange table through mathematical analysis. The actual USATT system uses discrete rating ranges and may not perfectly follow this continuous approximation, especially at the boundaries between rating difference ranges.
- Marcus, David J. (2001) New Table-Tennis Rating System. Journal of the Royal Statistical Society: Series D (The Statistician), 50: 191–208. doi: 10.1111/1467-9884.00271 ↩
- Beschreibung der andro-Rangliste (PDF). Retrieved from myTischtennis.de on 10.08.2025. ↩
- USA Table Tennis Rating System. Official Tournament Guide. Retrieved from usatt.simplycompete.com/info/ratings on 11.08.2025. ↩