Statistical aid...please help me!!!

[Paparazzo]

Hello, I'm a newbie in this forum, after some months of observation I decided to take part of this very well informed comunity. I'd like to post a question which I'm not really able to solve. Basically is a cross probability calculation problem: Given the AVERAGE outcome fo a home team (at home) and the visiting (away) how can I match the probabilities in order to obtain the correct ones for the given cross? I'll give an example to clarify: Say that the leader plays against the least, probabilites as follows: 1st at home H:77% D:16,5% A:6,5% 20th away H:68% D:23,5% A:9,5% Moreover the overall average (all teams) is: H:47% D:27% A:26% Now since both the leader wins more than average, and the least looses more than average I should expect something like that: H:91% D:7,5% A:1,5% ... But how could be the right formula to derive these results??? Thanks in advance for any help!!!

[Pawn_pt]

Re: Statistical aid...please help me!!! Hi, I use something like this: Home %W x 2/3 + Away %L x 1/3 = New Home %W Home %L x 2/3 + Away %W x 1/3 = New Away %W New Home %W / (New Home %W + New Away %W) = Adjusted Home %W Considering your example: New Home %W: 77% x 2/3 + 68% x 1/3 = 74% New Away %W: 6,5% x 2/3 + 9.5% x 1/3 = 7,5% Adjusted Home %W: 74% / (74% + 7.5%) = 91% Which is exactly your result :p