Is there another challenge I can try to practice this concept?

Let's take a look at the following formula:

Math.floor(Math.random() * (6 - 1 + 1)) + 1; 

If we break it down one by one, we see that the outermost operation is Math.floor(x) + 1. The floor function works by rounding down a float regardless of the decimal places. For example 1.0000001 rounds down to 1 and 1.9999999 also rounds down to 1. Floor() basically truncates everything to the right of the decimal place if you want to think of it that way.

Then, what ever x is, it is simply rounded down and then 1 is added to the truncated number. I will explain why the 1 is added after I explain the inner funciton (i.e., Math.random() * (6 - 1 + 1))

Math.random() generates a random number between 0 and 1, including 0 but not including 1. So it can go from 0 up to 0.9999999999...

If we simply apply the floor function to just the generated random number, the result will always be 0 because Math.random() only generates numbers greater than or equal to 0 and less than 1. This is why we multiply random numbers with larger numbers to scale the result up so it's not always zero.

Let's look at ten numbers generated by Math.random() * 10:

 0: 0.00757 * 10 = 0.0757
 1: 0.70006 * 10 = 7.0006
 2: 0.14752 * 10 = 1.4752
 3: 0.90371 * 10 = 9.0371
 4: 0.23131 * 10 = 2.3131
 5: 0.78659 * 10 = 7.8659
 6: 0.23670 * 10 = 2.3670
 7: 0.81528 * 10 = 8.1528
 8: 0.51122 * 10 = 5.1122
 9: 0.43452 * 10 = 4.3452

Notice how each value generated by the random function will round down to zero but when multiplied by 10, the numbers vary between 0 and 9 (see below):

 0: Math.floor(0.0757) = 0
 1: Math.floor(7.0006) = 7
 2: Math.floor(1.4752) = 1
 3: Math.floor(9.0371) = 9
 4: Math.floor(2.3131) = 2
 5: Math.floor(7.8659) = 7
 6: Math.floor(2.3670) = 2
 7: Math.floor(8.1528) = 8
 8: Math.floor(5.1122) = 5
 9: Math.floor(4.3452) = 4

So now we see that by multplying by 10 we can get a random number between 0 and 10 (including 0 but not 10). Now if we add 1 to whatever is generated we will always have a number between 1 and 10 (both 1 and 10 included)!!

You can see here that 10 sets the upper limit on the random function (remember the excercise?) so basically what ever you multiply with Math.random() becomes its upper limit. Adding 1 to the product is what makes the upper limit inclusive or exclusive.

Now the tricky part is understanding the latter part of Math.floor(Math.random() * (6 - 1 + 1)) + 1. Think of it as Math.floor(Math.random() * (upper - lower + 1)) + lower;

Say we want a value between 6 and 10. The lower limit is 6 and upper limit is going to be 10. The random function will be multiplied by 5 (see below):

upper - lower + 1  :  (10 - 6) + 1 = 5

This means that the left term will always return 0, 1, 2, 3, or 4. Remember 5 will not be included since Math.random() only returns 0 to 0.99999999999...

Now add the lower limit to 0, 1, 2, 3, or 4:

lower = 6

0 + lower = 6
1 + lower = 7
2 + lower = 8
3 + lower = 9
4 + lower = 10

So the whole formula will always give 6, 7, 8, 9, or 10! It's just a little math trick that's complicating the formula but hopefully it makes sense now.

When it comes to trying to parse out what a formula does or how it works, the best thing I've learned (as I mentioned in my previous post!!) is to handwrite your code and manually calculate your expected results just like I did here. This will flush out all the values you should expect and eventually, you should see a number pattern that solves your problem.

Happy coding!