Q&A for pro webmasters
select COUNT(*), qpc.QuestionerPosts as OutOf, MAX(a.CreationDate) as LastOccurance, q.OwnerUserId as [User Link], qu.UpVotes as AskerUp, qu.Reputation as AskerReputation, au.DisplayName, a.OwnerUserId as [User Link], (cast(COUNT(*) as decimal)/cast(qpc.QuestionerPosts as decimal)) as PercentSuspicious, au.Reputation as AnswererReputation, apc.AnswererPosts, qpc.QuestionerPosts as AskerPosts from Posts q inner join Posts a on q.AcceptedAnswerId=a.Id inner join Users qu on qu.Id = q.OwnerUserId inner join Users au on au.Id = a.OwnerUserid and 1=1 inner join (select COUNT(*) as AnswererPosts, OwnerUserId from Posts GROUP BY OwnerUserId) apc on apc.OwnerUserId=au.Id inner join (select COUNT(*) as QuestionerPosts, OwnerUserId from Posts GROUP BY OwnerUserId) qpc on qpc.OwnerUserId=qu.Id where qu.Id != au.Id and au.Reputation>15 and qu.Reputation<100 group by au.DisplayName, qu.UpVotes, q.OwnerUserId, a.OwnerUserId, au.Reputation, qu.Reputation, apc.AnswererPosts, qpc.QuestionerPosts having COUNT(*)>2 AND (cast(COUNT(*) as decimal)/cast(qpc.QuestionerPosts as decimal))>0.6 and au.Reputation<=5000 and MAX(a.CreationDate) >= DATEADD(m, -6, current_timestamp) and MAX(a.CreationDate) <= DATEADD(m, -3, current_timestamp) order by LastOccurance desc