# What is a good method for modelling combinatorial tree (sport competition results)?

Probably newbie question here, please point me out to relevant theories/tutorials if needed :

• let say I want to evaluate the probabilities of the final rankings for a sport competition
• the competition involved 8 teams, but I can simplify the problem to 4 contestants (say A - B - C - D)
• the competition is splitted into n journeys during which every team faces another team (and only one). So with 4 contestants, I have 2 matches per journey
• at the end of a match, 6 different points attributions are possible (depending on the score)

After one journey, there are 36 different possible combinations in terms of team's points. So the model looks like a tree with a journey at each level.

Even if I simplify the situation to 2 journeys left, I can't think of a elegant way to implement this problem rather than "manually" creating the tree with all the possible combinations at each level and counting the leaves ?

I'm not familiar with this kind of problem so I'm not sure about the path forward.

I can't follow how you get 30 combinations, you have 2 matches per journey each of the 2 matches can result in 5 different attributions, so you get 5*5 = 25 different attribution per match. At each journey there are 3 possibilities to make 2 matches. So the result would be 25/3 + 25/3 + 25/3 = 25 different point attributions.
Did I got something wrong?

You want to calculate how many different point attributions there are after n journeys and the probabilities to make a match are the same for every combination at every journey?

• No you are right indeed. As for what I want to achieve : my goal is to compute, at the end of all journeys, the probability (% of situation) of each possible rankings among the 4 teams, depending on the successive score attributions. – Beinje Feb 13 at 20:22
• Actually, there are 6 different attributions per match, so a total of 36 combinations after a journey with 4 teams. I edited the post though this is not the core problem. – Beinje Feb 14 at 9:23