Ok. I've woken up more now, and I have a way of phrasing my nebulous-sounding concerns better (I hope!), and I've got some math. What timchen is looking at is exactly what to do to figure out the probability of any one player reaching a certain threshold of points. Incidentally, here's the breakout of that, from 15 points up:

4p chance Points 3p chance

1/256 ~= 00.39% 23-24 1/81 ~= 01.23%

5/256 ~= 01.95% 22 1/81 ~= 01.23%

5/256 ~= 01.95% 21 5/81 ~= 06.17%

15/256 ~= 05.86% 19-20 5/81 ~= 06.17%

35/256 ~= 13.67% 17-18 15/81 ~= 18.52%

66/256 ~= 25.78% 16 15/81 ~= 18.52%

66/256 ~= 25.78% 15 31/81 ~= 38.27%

As we can see, 19-20 is by far the closest, and the rest have pretty substantial gaps.

However, the important thing comes here: what will the threshold actually be? To figure out what you'd expect this to be, you'd need to know what the maximum score from any given pod will be, and then how many pods there are. And it's going to be different for 3p pods and 4p pods, so you actually need to know how many of each type you have.

Now, if we make a few assumptions, we can actually break this down. For now, let's assume that every game is independent of every other (bad because of player skill and seating order), and that every person is equally likely to get every place in every game.

Note that we can't just use what timchen started and I expanded on above, because the different players in a pod are NOT independant of each other. i.e., if one guy gets 1st 4 times, we know that nobody else got first, so those possibilities get sliced out, if somebody gets 1st twice, nobody else can get 1st more than twice, if two guys get first twice, nobody else can get first, etc.

Anyway, in a 3p pod, there are 6 distinct orderings that any game can finish in. 4 games makes 1296 orderings, but we get one factor of 6 back because at the end, if we switch all the As for Bs, or all the Bs for Cs, etc., this doesn't make a substantive difference (i.e. ABC ABC ABC ABC == BCA BCA BCA BCA == BAC BAC BAC BAC == CAB CAB CAB CAB, etc.).

So we have 216 possibilities. Now, I'm not going to sit and count them all out right now, because I don't have time. But just take the top case, 24 points. To get 24 points, someone had to score perfect. The other guys could have (B gets 2nd 4 times, C gets 3rd 4 times; B gets 2nd thrice; B gets 2nd twice, etc.).

...aaaaand I realize my counting brain has gone offline here. Someone else who is skilled in the mathematical arts want to progress this?