As we all know, the correct answer is (c). The number of rows is random (random value with binomial distribution), but the problem is with the outer part of the query - particularly with evaluation of "id = random_value" comparison, which is performed within the CASE construct.

The way the query is written is a little bit misleading (it wouldn't be a puzzler) as the "common sense" says that the inline view returns only rows with "id = random_value", so the condition in CASE has to be evaluated as true.

Truth is Oracle (10g) does not work like that - random value generated within the inline view is not used when evaluating CASE condition, but the expression "ROUND(DBMS_RANDOM.value(1,10))" is evaluated again. Due to this a new random value will be returned - probably not equal to the previous one. Thus the CASE will randomly return "EQUAL" and "NOT EQUAL".

I did not have a chance to test this on Oracle 11g, but according to the information I've received this puzzler will be processed differently - according to the common sense. So the number of rows will be random, but the expression won't be evaluated again - all the rows will contain "EQUAL".

How to "fix" this on 10g? You may use "materialize" hint, so that the inline view will be "materialized" and won't be re-evaluated in the next step - this is virtually identical to the solution 11g does automatically:

WITH t AS (SELECT /*+ materialize */ my_table.*,
                  ROUND(DBMS_RANDOM.value(1, 10)) as random_value
           FROM my_table)
SELECT (CASE WHEN (id = random_value) THEN 'ROVNOST'
             ELSE 'NEROVNOST' END) AS vysledek FROM t
WHERE id = random_value

And now some fun with probabilities ...

Matematical fun

As alteady mentioned in the previous puzzler, the number of returned rows is govened by the binomial distribution, i.e. the probability that given table of n rows (10 in our case), k rows will be returned, is

P_n,k = n! / ((n-k)! * k!) * p^k * (1-p)^n-k

where p is the probability that a particular row will be returned independently from the others (in our case p = 10^-1).

But what values will be in those returned rows? As already said, the values will be random - Oracle re-evaluates the expression and thus generates a new random values. Given a fixed number of returned rows k and the number of rows containing value "EQUAL" denote as l - logically the inequality k >= l holds. Probability that a given row contains "EQUAL" value is 1/n = 10^-1 again, so let's use the name p.

Then the number of rows with "EQUAL" value is governed by the binomial distribution - just replace n and k, resp. k and l, i.e.

P_k,l = k! / ((k-l)! * l!) * p^l * (1-p)^k-l

So the probability that from a table of n rows exactly k rows will be returned, and that l of them will contain "EQUAL" value is a product of the two probabilities

P_n,k,l = P_n,k * P_k,l = n! / ((n-k)! * (k-l)! * l!) * p^k+l * (1-p)^n-l

This for example for n = 10 gives the following table (whereas 0.00% denotes very small although not zero probabilities):