Rating changes for last rounds are temporarily rolled back. They will be returned soon. ×

About "Standard" Score Distribution

Revision en1, by djm03178, 2020-05-14 12:13:52

I've seen phrases like this in contest announcements many times:

The score distribution is standard: 500 - 1000 - 1500 - 2000 - 2500 - 3000

However, I don't really see that this 'standard' distribution is balanced in practice or anything, so I don't get why this uniform distribution is called standard.

The score decreasing rate (i.e. the number of points lost per minute) is proportional to the initial score of the problem. For example, in a 2-hour contest, if you solve a 1000-point problem in 1:59, you get 524 points, which is still higher than solving a 500-problem in 0:00. But if you solve a 3000-point problem in 1:59, you only get 1572 points. That's much less than solving 2000 or 2500-point problem fast enough.

In most of Div. 2 contests, when I see the standings table, the general tendency of gained score of each participant is like this: A <<< B < C >= D >= E >= F. This includes my recent contest (yes, I underestimated D a lot).

I think the reason we give more points to a harder problem is to reward participants who manage to solve those problems. But in practice, a better strategy is to almost always solve easier problems first, and as a result, most people gain less points by solving harder problems. Another side effect is that when you fail on system test, the most critical one is failing on B or C, and not the harder problems. Let's exclude those who first read harder problems and decide to participate only when they're confident to solve it fast.

This is also related to Div. 1 / 2 parallel rounds. I see most of these rounds tend to use score distribution of 2C~2F = 1A~1D + 1000 when these problems are the same ones. But this means Div. 2 participants are rewarded much less for solving 2E or 2F than Div. 1 participants solving 1C or 1D, compared to how much they get for solving 2C(1A) or 2D(1B).

Of course, I didn't count getting wrong answers to get penalty, but the average number of tries to get AC won't be that much to affect this in general.

Conclusively, I'd like to argue that it will be better not to hesitate setting higher score on harder problems. Let me suggest a 'balanced' score distribution I think: 500 — 750 — 1250 — 1750 — 2500 — 3500.

History

 
 
 
 
Revisions
 
 
  Rev. Lang. By When Δ Comment
en1 English djm03178 2020-05-14 12:13:52 2335 Initial revision (published)