Нахождение самой длинной серии побед

Это должно работать, Fiddle здесь: http://sqlfiddle.com/#!4/31f95/27

SELECT   team_id, MAX(seq_length) AS longest_sequence
      FROM (SELECT   team_id, COUNT(*) AS seq_length
                 FROM (SELECT team_id, won_ind,match_id, SUM(new_group) OVER(ORDER BY match_id) AS group_no
                         FROM (SELECT   team_id, won_ind, match_id,
                                        DECODE(LAG(won_ind) OVER(ORDER BY match_id), won_ind, 0, 1) AS new_group
                                   FROM matches
                               ORDER BY team_id))
                WHERE won_ind = 'Y'
             GROUP BY team_id, group_no)
   GROUP BY team_id
   ORDER BY 2 DESC, 1;

Используя вариант ответа, который я разместил здесь

select
    team_id,
    max(wins)
  from
    (
     select
            a.team_id,
            a.match_id amatch,
            b.match_id bmatch,
    (select count(distinct match_id) 
       from matches matches_inner
      where a.team_id = matches_inner.team_id
        and matches_inner.match_id between a.match_id and b.match_id) wins
      from
            matches a
            join matches b on a.team_id = b.team_id 
                      and b.match_id > a.match_id
     where
    not exists 
    (select 'x'
       from matches matches_inner
      where a.team_id = matches_inner.team_id
        and matches_inner.match_id between a.match_id and b.match_id
        and matches_inner.won_ind = 'N')

group by team_id

У меня была аналогичная задача на Teradata, я модифицировал ее для работы на Oracle:

SELECT
   team_id,
   MAX(cnt)
FROM
 (
   SELECT
      team_id,
      COUNT(*) AS cnt
   FROM 
    (
      SELECT
        team_id, 
        match_id,
        won_ind,
        SUM(CASE WHEN won_ind <> 'Y' THEN 1 END) 
        OVER (PARTITION BY team_id 
              ORDER BY match_id 
              ROWS UNBOUNDED PRECEDING) AS dummy
      FROM matches
    ) dt
   WHERE won_ind = 'Y'
   GROUP BY team_id, dummy
 ) dt
GROUP BY team_id;