When multiple entries have the same score, the next rank should not be consecutive. The next rank should be incremented by number of scores that share the same rank.

To display scores like that requires two rank variables

• rank variable to display
• rank variable to calculate

Here is a more stable version of ranking with ties:

SET @rnk=0; SET @rank=0; SET @curscore=0;
SELECT score,ID,rank FROM
(
    SELECT AA.*,BB.ID,
    (@rnk:[email protected]+1) rnk,
    (@rank:=IF(@curscore=score,@rank,@rnk)) rank,
    (@curscore:=score) newscore
    FROM
    (
        SELECT * FROM
        (SELECT COUNT(1) scorecount,score
        FROM scores GROUP BY score
    ) AAA
    ORDER BY score DESC
) AA LEFT JOIN scores BB USING (score)) A;

Let's try this out with sample data. First Here is the sample data:

use test
DROP TABLE IF EXISTS scores;
CREATE TABLE scores
(
    id int not null auto_increment,
    score int not null,
    primary key (id),
    key score (score)
);
INSERT INTO scores (score) VALUES
(50),(40),(75),(80),(55),
(40),(30),(80),(70),(45),
(40),(30),(65),(70),(45),
(55),(45),(83),(85),(60);

Let's load the sample data

mysql> DROP TABLE IF EXISTS scores;
Query OK, 0 rows affected (0.15 sec)

mysql> CREATE TABLE scores
    -> (
    ->     id int not null auto_increment,
    ->     score int not null,
    ->     primary key (id),
    ->     key score (score)
    -> );
Query OK, 0 rows affected (0.16 sec)

mysql> INSERT INTO scores (score) VALUES
    -> (50),(40),(75),(80),(55),
    -> (40),(30),(80),(70),(45),
    -> (40),(30),(65),(70),(45),
    -> (55),(45),(83),(85),(60);
Query OK, 20 rows affected (0.04 sec)
Records: 20  Duplicates: 0  Warnings: 0

Next, let initialize the user variables:

mysql> SET @rnk=0; SET @rank=0; SET @curscore=0;
Query OK, 0 rows affected (0.01 sec)

Query OK, 0 rows affected (0.00 sec)

Query OK, 0 rows affected (0.00 sec)

Now, here is the output of the query:

mysql> SELECT score,ID,rank FROM
    -> (
    ->     SELECT AA.*,BB.ID,
    ->     (@rnk:[email protected]+1) rnk,
    ->     (@rank:=IF(@curscore=score,@rank,@rnk)) rank,
    ->     (@curscore:=score) newscore
    ->     FROM
    ->     (
    ->         SELECT * FROM
    ->         (SELECT COUNT(1) scorecount,score
    ->         FROM scores GROUP BY score
    ->     ) AAA
    ->     ORDER BY score DESC
    -> ) AA LEFT JOIN scores BB USING (score)) A;
+-------+------+------+
| score | ID   | rank |
+-------+------+------+
|    85 |   19 |    1 |
|    83 |   18 |    2 |
|    80 |    4 |    3 |
|    80 |    8 |    3 |
|    75 |    3 |    5 |
|    70 |    9 |    6 |
|    70 |   14 |    6 |
|    65 |   13 |    8 |
|    60 |   20 |    9 |
|    55 |    5 |   10 |
|    55 |   16 |   10 |
|    50 |    1 |   12 |
|    45 |   10 |   13 |
|    45 |   15 |   13 |
|    45 |   17 |   13 |
|    40 |    2 |   16 |
|    40 |    6 |   16 |
|    40 |   11 |   16 |
|    30 |    7 |   19 |
|    30 |   12 |   19 |
+-------+------+------+
20 rows in set (0.18 sec)

Please note how multiple IDs that share the same score have the same rank. Also note that rank is not consecutive.

Give it a Try !!!

SELECT 
    id, 
    Name,
    1+(SELECT count(*) from table_name a WHERE a.Score > b.Score) as RNK,
    Score
FROM table_name b;

One option would be to use USER variables:

SET @i=0;
SELECT id, name, score, @i:[email protected]+1 AS rank 
 FROM ranking 
 ORDER BY score DESC;

The accepted answer has a potential problem. If there are two or more identical scores, there will be gaps in the ranking. In this modified example:

 id name  score rank
 1  Ida   100   2
 2  Boo    58   5
 3  Lala   99   3
 4  Bash  102   1
 5  Assem  99   3

The score of 58 has rank 5, and there is no rank 4.

If you want to make sure there are no gaps in the rankings, use DISTINCT in the GROUP_CONCAT to build a list of distinct scores:

SELECT id, name, score, FIND_IN_SET( score, (
SELECT GROUP_CONCAT( DISTINCT score
ORDER BY score DESC ) FROM scores)
) AS rank
FROM scores

Result:

id name  score rank
1  Ida   100   2
2  Boo    58   4
3  Lala   99   3   
4  Bash  102   1
5  Assem  99   3

This also works for getting a single user's rank:

SELECT id, name, score, FIND_IN_SET( score, (    
SELECT GROUP_CONCAT(DISTINCT score
ORDER BY score DESC ) 
FROM scores )
) AS rank
FROM scores
WHERE name =  'Boo'

Result:

id name score rank
 2  Boo   58    4

Here's the best answer:

SELECT 1 + (SELECT count( * ) FROM highscores a WHERE a.score > b.score ) AS rank FROM
highscores b WHERE Name = 'Assem' ORDER BY rank LIMIT 1 ;

This query will return:

3

This solution gives the DENSE_RANK in case of ties:

SELECT *,
IF (@score=s.Score, @rank:[email protected], @rank:[email protected]+1) rank,
@score:=s.Score score
FROM scores s,
(SELECT @score:=0, @rank:=0) r
ORDER BY points DESC

Wouldn't the following work (assuming your table is called Scores)?

SELECT COUNT(id) AS rank FROM Scores 
WHERE score <= (SELECT score FROM Scores WHERE Name = "Assem")

I have this, which gives the same results as the one with variables. It works with ties and it may be faster:

SELECT COUNT(*)+1 as rank
FROM 
(SELECT score FROM scores ORDER BY score) AS sc
WHERE score <
(SELECT score FROM scores WHERE Name="Assem")

I didn't test it, but I'm using one that works perfect, which I adapted to this with the variables you were using here.