替代版本，不使用任何专有字符串处理：

select distinct t1.person_id, t2.person_id
from tbl_hobby t1
join tbl_hobby t2
  on t1.person_id < t2.person_id
where 2 = all (select count(*)
               from tbl_hobby 
               where person_id in (t1.person_id, t2.person_id)
               group by hobby_id);

也许效率较低，但携带方便！

每个人加起来，得到他们的爱好串。然后汇总每个爱好列表，找出属于多个人的爱好。

select hobbies, group_concat(person_id order by person_id) as persons
from
(
  select person_id, group_concat(hobby_id order by hobby_id) as hobbies
  from tbl_hobby
  group by person_id
) persons
group by hobbies
having count(*) > 1
order by hobbies;

这给出了每个爱好的人的列表。这是输出解决方案的最简单方法，否则我们必须构建所有可能的对。
更新：如果需要对，则必须查询表两次：

select p1.person_id as person 1, p2.person_id as person2
from
(
  select person_id, group_concat(hobby_id order by hobby_id) as hobbies
  from tbl_hobby
  group by person_id
) p1
join
(
  select person_id, group_concat(hobby_id order by hobby_id) as hobbies
  from tbl_hobby
  group by person_id
) p2 on p2.person_id > p1.person_id and p2.hobbies = p1.hobbies
order by person1, person2;