Talk:Review Algorithm/1.4.4/@comment-188.107.198.251-20130506200821/@comment-188.107.198.251-20130506221910

So, essentially, you get q = 2/3*q1 + 1/3*q2, p = 2/3*p1 + 1/3*p2 and the number of experts is a float value?

By the way:
 * If you have exactly one game rated >= 9, b = 2
 * If you have exactly two games rated >= 9, b is
 * Whatever is greater between (Top Score - 2nd Score) and (0.1 * (Top Score + 2*Y))
 * But it has to be less than 0.2 * Top Score or it just becomes 0.2 * Top Score
 * If you have three or more games rated >= 9, b is
 * Whatever is greater between (Top Score - 2nd Score) and (0.1 * (Top Score + (Value of b for previous game)*Y))
 * But it has to be less than 0.2 * Top Score or it just becomes 0.2 * Top Score

If I understand it correctly, option 2 is actually a special case of option 3 (because previous b = 2) and not a genuinely independent case. So this could be simplified somewhat.