### Applying the Birthday Paradox to World Cup Group Scores

First, I should apologize. The seventh MLP puzzle I was promising hasn't been finished, and it won't be finished for some time yet. But I do have something to show today.

Alexandre Muñiz posted some combinatorial musings on the World Cup recently, and I had a stab at the main posed problem. Figured out some shortcuts, but it's quite combinatorially tricky.

The question is this: given that the soccer teams in each of several groups are evenly matched, and there is a certain likelihood of each game resulting in a tie, how likely is it that the groups will all have different scores?

Well, that's about it for the main problem. I don't know the answer. But to make sure my methods work, I tried them in a spreadsheet on three groups of three teams each. The probability of no repeats when ties are 1/3 probability is about .53315 or exactly 1166/2187. This reaches a maximum of about .533468755 when ties have a probability of about .34308.

Alexandre Muñiz posted some combinatorial musings on the World Cup recently, and I had a stab at the main posed problem. Figured out some shortcuts, but it's quite combinatorially tricky.

The question is this: given that the soccer teams in each of several groups are evenly matched, and there is a certain likelihood of each game resulting in a tie, how likely is it that the groups will all have different scores?

**( Collapse )**Well, that's about it for the main problem. I don't know the answer. But to make sure my methods work, I tried them in a spreadsheet on three groups of three teams each. The probability of no repeats when ties are 1/3 probability is about .53315 or exactly 1166/2187. This reaches a maximum of about .533468755 when ties have a probability of about .34308.