For online games, trying to match players who have balanced chances of victory is usually a healthy goal.

The most common method relies on calculating individual rankings based on the Bradley-Terry model, which assigns a positive strength value S to each player, with the chances of victory of player P (with strength S) over player P’ (with strength S’) defined as:

$p(P>P') = \frac{S}{S+S'}.$

The most famous system based on this model is the Elo rating for Chess.

We can also encounter situations where player A usually beats player B, who usually beats player C, who usually beats player A. This nontransitive situation cannot be captured with a single strength value.

Actually, this nontransitive strength mechanism is one of the underlying design principles of many competitive multiplayer video games.

Bringing a fresh approach to the old rock paper scissor concept, the solution we experimented with relies on simply using three “strength” values for each player instead of one: $$(S_1, S_2, S_3)$$ for player $$P$$ and $$(S'_1, S'_2, S'_3)$$ for player $$P'$$. In the case of rock paper scissors, for example, these can be interpreted as the probability for each player to select rock, paper or scissors.

The probability for $$P$$ to win against $$P'$$ becomes:

$p(P>P') = \frac{ \left( \frac{S_1}{S_1 + S'_1} + \frac{S_2}{S_2 + S'_2} + \frac{S_3}{S_3 + S'_3} \right) }{3}.$

We can easily express the nontransitive situation as follows:

• Player A skills : $$(0,1,2)$$
• Player B skills : $$(2,0,1)$$
• Player C skills : $$(1,2,0)$$.

We obtain:

• A beats B with probability $$p(A>B) = \frac{5}{9}$$
• B beats C with probability $$p(B>C) = \frac{5}{9}$$
• C beats A with probability $$p(C>A) = \frac{5}{9}$$.

We were also able to confirm through simulations that these “hidden” strength values can be calculated using only the win/loss outcomes from several games between nontransitive players.

Interestingly, this sort of “nontransitive paradox” can be found in “non-transitive dice”, for example.

Patented Technology