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


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.

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

>>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
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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


zzynxSoftware engineerCommented:
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
Mayank SAssociate Director - Product EngineeringCommented:
You can think of rounding or ceiling/ flooring the value as per your requirement using the methods in the Math class.
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 ;)

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zzynxSoftware engineerCommented:
>> 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
mdbboundAuthor Commented:
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.
zzynxSoftware engineerCommented:
Thanks for appreciating our help
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