Quote from dloserFeb 06, 2011 - 10:18:25
How about p^(1+100/max(#challs,100)) or p^(1+100/(#challs+100)) or something? I think a power of 2 is a nice minimum. Going beyond it diminishes the points for low percentages so much that it almost isn't "worth" doing a site unless you know you can finish pretty much all challenges. This also avoids having to adjust for sites with few challenges (and then again if they add a bunch).
This is very much what I was trying to argue, though without the formula. It is frustrating to start a site knowing that that you are going to solve 30 challenges for a handful of points.
I'm not committed to the 1 point minimum. I was just thinking that anything would be better than 0 for half a dozen challenges. It would also be nice if the system worked without
ad hoc tinkering, which is a plus for dloser's idea.