Question

Probability distribution using MySQL RAND() function

Asked by: bobbit31

I am trying to right a FUNCTION in MySQL that will return a random value.  Here is the tricky part:

31 has 70% chance of being returned
32 has 20% chance of being returned
33 has 10% chance of being returned

I am basically gifting an item but its luxury value is random..

Hope someone could figure this out using RAND()

Here is the only function in MySQL that would help but still doesnt give me a chance to set the probability:
 
 The effect of using a non-constant argument is undefined. As of MySQL 5.0.13, non-constant arguments are disallowed.
 
To obtain a random integer R in the range i <= R < j, use the expression FLOOR(i + RAND() * (j  i)). For example, to obtain a random integer in the range the range 7 <= R < 12, you could use the following statement:
 
SELECT FLOOR(7 + (RAND() * 5));
                                  
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Asked On
2009-03-05 at 00:50:21ID24200651
Tags

mysql RAND() probability

Topics

Microsoft Office Suite

,

MySQL Server

,

Databases Miscellaneous

Participating Experts
1
Points
500
Comments
1

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Answers

 

by: StraySodPosted on 2009-03-05 at 02:19:23ID: 23804004

maybe you could do it like this:

choose a random number between 1 and 100 included:
if the value is in range 1..70 return 31
if it's 71..90 return 32
for >90 return 33

CREATE FUNCTION rand_probability()
  RETURNS INT
BEGIN
  declare number int(5);
 
  SELECT FLOOR(1 + (RAND() * 99)) INTO number;
 
  if number < 71 then return 31;
  elseif number < 91 then return 32;
  else return 33;
  end if;
 
END

                                              
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