I was recently looking at a debate tournament and the question arose about how to allocate adjudicators at any given tournament.

The first question is to ask how many adjudicators you have. Some tournaments run a ‘n-1’ rule, usually in BP format, whilst ‘n=1’ is more common in 3on3 tournaments or their variants. In both instances, n represents the number of teams at the tournament. ‘n-1’ makes things far more complicated because you have no real way of knowing for certain how many adjudicators will attend even if you have the number of teams. (You could have 12 teams from 12 institutions and thus no adjudicators, or 12 teams from 3 institutions giving you 9 adjudicators) It also doesn’t help that you have four teams to a debate. This isn’t too big of an issue, but does nonetheless make things more complicated. I will thus concentrate for the time being on 3on3 debates.

If you assume a strict ‘n=1’ rule and that all debates have an odd number of adjudicators (thus avoiding the awkward situation of a perfectly split panel without a deciding ballot). GIven n teams, you would then have n adjudicators and only n/2 debates. You would then nominally have 2 adjudicators per debate. But the preference for odd panels would thus give rise to the question, how many panels of 3 can you afford to have while still providing adjudicators for all the debates?

If there are 12 teams, you can have 6 panels of 3 and 6 panels of 1

If there are 13 teams, you can have 6 panels of 3, 7 panels of 1 and will still have 1 extra adjudicator

If there are 14 teams, you can have 7 panels of 3 and 7 panels of 1

If there are 15 teams, you can have 7 panels of 3, 8 panels of 1 and will still have 1 extra adjudicator

I could continue, but you may already see a pattern emerging. A strict n=1 rule means half, or just under half the debates can get panels of 3 while the rest have panels of 1.

But this is under a strict ‘n=1’ rule. What if more or fewer adjudicators are available? To cut a long explanation short, it turns out you can easily find how many debates get panels of 3 and thus by deduction how many get panels of 1. The formula is simply:

**number of 3 person panels = (number of adjudicators minus number of teams) divide by 2**

Unfortunately, this only works with panels of 1 and 3. But it is a start, and frankly I think very few tournaments can afford to have panels of more than 3 at tournaments anyways.