Math.floor( Math.random () * (max - min + 1)) + min) Explanation

Hi Chelsea,

Maybe it will help to see how the formula can be derived starting only with Math.random() and Math.floor(). That way you'll know that it's correct and more importantly why it's correct.

I'll try to provide a conceptual understanding of what's going on with the formula.

We know that Math.random() returns a value in the range [0, 1). 0 is included but 1 is excluded. You can think of it as the range 0 to 0.999999...

Now let's pass that result to the Math.floor() function.

Math.floor(Math.random());

Since floor will truncate the decimal and give you back the whole number part we're always going to get 0 here. Make sure you understand that or the rest isn't going to make any sense. Ask questions if you need to.

We can only get 1 integer out of this. It's not so much important that the integer is 0 but that we can only get 1 integer out.

If we are going to be able to get more numbers out of this then we need to make that range bigger. When you multiply something by 2 it becomes twice as big as it was before. It scales up by a factor of 2. Let's see what happens when we multiply Math.random() by 2

Math.floor(Math.random() * 2);

That will give us a new range of 0 to 1.999999... which is twice as big as the range we started with.

What happens when those numbers are passed into Math.floor? All the numbers generated from 0 to 0.9999... will be truncated to 0 and all the numbers from 1 to 1.9999... will be truncated to 1

Now we're able to get 2 different integers out of this. If we multiply by 2 we can get 2 numbers back. It stands to reason that if we multiply by 6 then we will be able to get 6 numbers out.

That will give us a range that's 6 times as big, 0 to 5.99999.....

I won't write it all out but after passing through the floor function you would get

0 to .99999... -> 0
1 to 1.99999... -> 1
...
5 to 5.99999... -> 5

In general, whatever you multiply Math.random() by is how many integers you'll be able to generate.

Now we can start deriving the formula and I'll use a specific example to help.

Let's say we want to generate numbers from 5 to 10 inclusive. We need to know how many numbers are there.

Setting up the variables -

var max = 10;
var min = 5;

If we list them out, 5, 6, 7, 8, 9, 10 and count them we see that there are 6 total numbers. We know from before that we're going to have to multiply by 6 in order to get 6 numbers out.

How can we come up with the 6 using our max and min variables?? If I do max - min I get 5 which is 1 short. max - min gives you the distance from 5 to 10. You always have to add 1 to that if you want the total amount of numbers.

That gives us the expression max - min + 1

Putting that into the formula,

Math.floor(Math.random() * (max - min + 1));

It's important that max - min + 1 is enclosed in parentheses so that all of that happens before the multiplication.

At this point the formula can generate the correct amount of numbers but they always start at 0 because the range from Math.random starts from 0.

0, 1, 2, 3, 4,  5 // What we have
5, 6, 7, 8, 9, 10 // What we want

Notice that if we add 5 to all the numbers in the first row, we'll get the second row. 5 is what our min value is in the example.

So if we add the min value onto the end of our formula, it will shift all the numbers over to the ones we want.

Math.floor(Math.random() * (max - min + 1)) + min;

You can think of it as a 2 step operation, you scale up the range, and then you shift it.

I meant for this to be shorter but I hope it helps you out.