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Chinmay Ravi Jha's avatar

*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%?

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Sherlock's avatar

Nice thought,

but nope answer shall not be 50% or around in every case...

It shall depend on n = the no. of teams taking part in the round-robin tournament...

You can try it out at n=3, wherein there will be 3 matches in the league stage, thus making total no. of ways in which tournament can occur = 2^3 = 8 ways

And if you try the no. of ways in which CoP will be formed manually then you shall get 2, making the probability = 0.25 = 25%

Similarly, we can also evaluate for n = 4,5 ; you can try finding out on these as well, and infact at n=5, the probability equals 17/32 (exceeding 50%)...

We shall talk more about this & how we came up with results at n=3,4,5 in the Issue#2 of the Circle Of Parity; till then keep trying more & sharing with us, stay tuned & engaged with us in our journey :)

Thank you

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Chinmay Ravi Jha's avatar

based on the number of hamiltonian cycles in a complete undirected graph and complete directed graph.

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