value = (int) (Math.random() * 6.0 + 1.0);

Posted on 2004-11-09
Last Modified: 2008-02-01

I have this homework to roll 2 dice and there is a part of the code below that i just dont get.

Can someone please help me figure out why we have to have the   +1  in this Math.Random.

value = (int) (Math.random() * 6.0 + 1.0);

The six means that there are 6 different possible number of dots, i don't know what the + 1 is for.

Question by:mdbbound
    LVL 92

    Expert Comment

    because random() returns a value greater than or equal to 0.0 and less than 1.0.
    LVL 92

    Assisted Solution

    so Math.random() * 6.0 returns a value greater than or equal to 0.0 and less than 6.0.
    which when cast to an integer becomes from 0 to 5

    LVL 6

    Expert Comment

    >>Can someone please help me figure out why we have to have the   +1  in this Math.Random.
    to make the possible values of "value" from 1 to 6, probably don't want to have 0 value.

    (Math.random() * 6.0 + 1.0
    converts it to next integer

    if value of (Math.random() * 6.0 is 0.6
    when it is added to 1 it becomes  1.6
    then finally value = (int)1.6 is 1

    the possible values of "value" will be
    1 to 6
    LVL 16

    Expert Comment

    Hi mdbbound,

    Well, I think that all the explanation above should be straight forward. So if you have codes like this.

    value = (int) (Math.random() * 6.0);

    It will return the value from 0.0 to 5.0. It will not touch 6.0 UNLESS you add it by 1.0. So in the roll game, there is no value for the dice to be zero, so the value can only between 1 to 6. Did you get it?

    So, if you need to have the random number between 10 and 200. You will have codes like

    value = (int) (Math.random() * 190 + 10);

    Hehe, now you know already right? Please don't hesitate to ask more questions


    LVL 37

    Assisted Solution

    An alternative for

            value = (int) (Math.random() * 6.0 + 1.0);


            java.util.Random r = new java.util.Random();
            value = r.nextInt(6) + 1;      // r.nextInt(6) returns an int from 0 to 5
    LVL 30

    Expert Comment

    You can think of rounding or ceiling/ flooring the value as per your requirement using the methods in the Math class.
    LVL 2

    Accepted Solution

    Let us make this simple.

    Math.random() gives you a random number between 0 and 1. [0 < Math.random() < 1]

    Math.random() * 6.0 would be between 0 and 6. [Hey, my dice does not have a 0 !!!]

    Note that I specifically mentioned 0 < Math.random() < 1, meaning Math.random() never returns me 0 or 1. It returns me something between it. In other words, with some precision loss,
    we can say
    0.00001 <= Math.random() <= 0.999999

    When I try to convert Math.random()*6.0 into an integer, the decimal part of the number is trimmed, i.e
    (int)5.6 = 5
    (int)4.1 = 4 [this is what guys call it floor of the number.]

    0.00006 <= Math.random()*6.0 <=  5.999994

    Now lets typecast to (int)

    0 <= (int)(Math.random()*6.0) <= 5 [Note my decimal part gets trimmed when I cast to integer]

    Hey, so when I write (int)(Math.random()*6.0), I get a number between 0 and 5. But what I want is a number between 1 and 6. Now you know what to do ;)
    LVL 37

    Assisted Solution

    >> Note that I specifically mentioned 0 < Math.random() < 1, meaning Math.random() never returns me 0 or 1.
    I quote from the java docs:
    public static double random()
    Returns a double value with a positive sign, greater than ***or equal to 0.0*** and less than 1.0

    Author Comment

    Hello everyone,

    Thank you so much for all the responses.  I'm just starting to learn java, i love it but it's kinda difficult.  

    I wanted to give the points to objects just as soon as i posted the question for his quick response but somehow i had a problem with my internet connection.

    So for now i will give the points to Objects.  But I still have some more questions on some parts of the code in my other lectures coz i really want to understand java.  

    zzynks and kjayaraman, Thank you for an in-depth explanation, i truly appreciate it.
    LVL 37

    Expert Comment

    Thanks for appreciating our help

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