The Circle of Parity
What do you think is common between this IPL and last year's ODI World Cup?
19th November, 2023
The hearts of millions of cricket fans across India sank as their beloved team, riding high on a remarkable streak of 10 consecutive victories, suffered a loss to Australia in the all important finals of the ICC Men’s Cricket World Cup, 2023. With all factors seemingly aligned in our favor, and fostering hopes to continue the tradition of host countries winning the tournament, witnessing our star players teary eyed was undeniably heart wrenching.
But do you know, it led to the occurrence of an interesting Mathematical phenomenon. The results of the tournament became such that each team in the tournament could be mapped in this flow of a team losing to one and winning to another.
Often referred to as the “Circle of Parity” or the “Parity Problem”, this Circle is more frequently observed in Seasonal Leagues like NFL and UEFA Champions League which follow a round robin / double round robin format. In a round robin format every team plays every other team exactly once (and exactly twice in a double round robin).
Defining formally, Circle Of Parity is basically a circular arrangement of teams in which the teams arranged in the circle have won and lost from each other thus completing a circle. Suppose there are ‘k’ teams; T₁, T₂, T₃, ……, Tₖ ; if a CoP is possible between these teams the circle would start from T₁: T₁ beats T₂, T₂ beats T₃, T₃ beats T₄ , and so on up till Tₖ beats T₁, and a circle would thus complete.
So, coming back to the Cricket World Cup, the tournament structure in its entirety was not exactly Round Robin. Yes, the league stages were so, followed by the two semifinals and the heart wrenching finals. But it led to the creation of an Interesting Problem Statement in our minds.
If an n-team tournament is organized in a round robin format where every team plays every other team exactly once. What is the probability of the formation of at least one CoP ? In other words, out of the ⁿC₂ (n choose 2) matches being played, which can generate a total of 2^(ⁿC₂) combinations of unique tournament results. What is the probability that a CoP will be formed, assuming that for each match, either of the teams have an equal probability to win (also called a Competitive Parity), and assuming that each match does yield a result (we get a result for each match for sure where one team wins and the other losses) ?
To present in a different way, if we represent these teams by mere points on a sheet, then this actually is a Graph Theory problem where we want a complete cycle among all the points (vertices). And such a cycle which traverses all the vertices exactly once indeed has a special name, it is called the Hamiltonian Cycle. There can be a case where there are multiple Hamiltonian cycles in a graph, and a case where there is no Hamiltonian cycle. Also note that the Network we will obtain on plotting all the results of a tournament for our case will be a Directed Network in a complete Graph- since we are dealing with a Round Robin tournament where every team plays every other team once resulting in each vertex being connected to every other vertex, the Graph will be complete and the result of a particular match decides in what direction the edge connecting two nodes is, the Network will be directed. Hence, the probability of tracing a Circle of Parity in an n-team tournament is equivalent to looking at the formation of at least one Hamiltonian cycle in a complete graph in a directed network.
Let’s take a random result for a 5-team tournament, to put things in perspective. Representing the teams as vertices numbered 1, 2, 3, 4 and 5 respectively and the matches between them by the directed edges with the arrow pointing towards the winning team, let’s say the matches occurred in such a way that we could represent it by the above graph. A Hamiltonian cycle (1→3→5→2→4→1) traced by the yellow edges does get formed in this network. Hence, we do get a CoP in this particular 5-team tournament.
Do take a moment to ponder on this and share your approach with us in the comment section on Substack. And Stay Tuned for our upcoming series of articles where we delve into different methodologies to reach the said probability expression and how we use certain algorithms and simulations to substantiate our hypothesis.
Keywords:
ICC on X , Complete Graph, Directed Graph, Hamiltonian Path, Circle of Parity
*Out of the ⁿC₂ (n choose 2) matches being played, which can generate a total of 2^(ⁿC₂) combinations of unique tournament results. What is the probability that a CoP will be formed?*
Is the answer 50%?