as stated in MDN reference about Math.random()

Returns a floating-point, pseudo-random number in the range [0, 1) that is, from 0 (inclusive) up to but not including 1 (exclusive), which you can then scale to your desired range.

Since Math.random can return 0, then Math.ceil(Math.random()*10) could also return 0 and that value is out of your [1..10] range.

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About your second question, see Most efficient way to create a zero filled JavaScript array?

It's great your showing your code and ideas-- good stuff! I think you meant to have a single key/value pair with an array. That makes this work. Good luck with your JavaScript journey! ```javascript const theBigExercise = { exercise: ['Bench Press', 'Shoulder Press', 'Deadlift' , 'Squats'] } function mainExercise() { let randomMain = Math.floor(Math.random() * theBigExercise.exercise.length); console.log(theBigExercise.exercise[randomMain]); return randomMain; } ```

Math.floor() is the floor function. Which means it returns the integer part of a number. In your example Math.floor(5.94) = 5

The numbers from Math.random() are in the range 0 (inclusive) to 1 (exclusive). That means you might get a 0, but you'll never get 1.

Thus that formula starts off with multiplying by an integer, which in your example will give a value from 0 (inclusive) up to 4 (exclusive). If you want random integers like 0, 1, 2, and 3, you have to use .floor() to get rid of any fractional part entirely. In other words, you might get 3.999, and you don't want a 4, so .floor() throws away the .999 and you're left with 3.

Now this presumes that what you want is what most people want, a random integer between a minimum value and a maximum value. Once you understand what Math.random() returns, you're free to generate integers however you want.

That i understood. I even got the answer using that method and using the following code: function randomWholeNum() { // Only change code below this line var x = Math.random()*10; var y = Math.floor(x); return y; } What I want to understand is why the first method doesn’t work.

What would make it easier?

I'll explain this formula:

Math.floor(Math.random() * (myMax - myMin + 1) + myMin);

Say we want a random number from 5-15 (including both 5 and 15 as possible results). Well, we're going to have to work with Math.random(), which only produces values from 0 through approximately 0.99999999999999, so we need to do two tricks to be able to work with this.

The first trick is recognizing that Math.random()'s lowest possible return value is 0, and 0 times anything is 0, so we need to start our range at 0 and adjust to account for this at the end. Instead of calculating 5-15 from the beginning, we recognize that there are 11 values in 5-15 (5, 6, 7, 8, 9, 10, 11, 12, 13, 14, and 15) and count up that many from 0 (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10) to use 0-10 as our range instead. This is what the myMax - myMin part of the formula does. It defines our new max as 10. Then, at the end of the calculations, we'll just add 5 back to whatever result we get to make the possible result range change from 0-10 to 5-15. This is what the + myMin part of the formula does.

The second trick is recognizing that multiplying Math.random() by our new max range of 10 can only give us a result as high as about 9.999999999999 because Math.random() only goes as high as about 0.99999999999 (never actually 1). When we Math.floor() that later to make it an integer, it brings the result down to 9, so we need to add 1 there to make the maximum possible value 10 instead of 9. That's what the + 1 part of the formula does.

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Let's finish this off by walking through an example.

Math.random() can be 0 at lowest and approximately 0.99999999999999 at highest (never 1). Let's just look at what happens with those two cases to see how the range works out.

If we run the case where we've called randomRange(5, 15) and Math.random() gives us 0, here's what we end up with:

Math.floor(Math.random() * (myMax - myMin + 1) + myMin);

=

Math.floor(0 * (15 - 5 + 1) + 5);

=

Math.floor(0 * 11 + 5);

=

Math.floor(0 + 5);

=

Math.floor(5);

=

5

So the lowest value possible is 5. If we run the case where we've called randomRange(5, 15) and Math.random() gives us 0.99999999999999, here's what we end up with:

Math.floor(Math.random() * (myMax - myMin + 1) + myMin);

=

Math.floor(0.99999999999999 * (15 - 5 + 1) + 5);

=

Math.floor(0.99999999999999 * 11 + 5);

=

Math.floor(10.9999999999999 + 5);

=

Math.floor(15.9999999999999);

=

15

So the highest value possible is 15.