[sql] How to count occurrences of a column value efficiently in SQL?

I have a table of students:

id | age
--------
0  | 25
1  | 25
2  | 23

I want to query for all students, and an additional column that counts how many students are of the same age:

id | age | count
----------------
0  | 25  | 2
1  | 25  | 2
2  | 23  | 1

What's the most efficient way of doing this? I fear that a sub-query will be slow, and I'm wondering if there's a better way. Is there?

Here's another solution. this one uses very simple syntax. The first example of the accepted solution did not work on older versions of Microsoft SQL (i.e 2000)

SELECT age, count(*)
FROM Students 
GROUP by age
ORDER BY age

and if data in "age" column has similar records (i.e. many people are 25 years old, many others are 32 and so on), it causes confusion in aligning right count to each student. in order to avoid it, I joined the tables on student ID as well.

SELECT S.id, S.age, C.cnt
FROM Students S 
INNER JOIN (SELECT id, age, count(age) as cnt  FROM Students GROUP BY student,age) 
C ON S.age = C.age *AND S.id = C.id*

select s.id, s.age, c.count
from students s
inner join (
    select age, count(*) as count
    from students
    group by age
) c on s.age = c.age
order by id

If you're using Oracle, then a feature called analytics will do the trick. It looks like this:

select id, age, count(*) over (partition by age) from students;

If you aren't using Oracle, then you'll need to join back to the counts:

select a.id, a.age, b.age_count
  from students a
  join (select age, count(*) as age_count
          from students
         group by age) b
    on a.age = b.age

I would do something like:

select
 A.id, A.age, B.count 
from 
 students A, 
 (select age, count(*) as count from students group by age) B
where A.age=B.age;