Expose a chance for mass index

[Marcee]

There's a set of indexes 1 to 9. Each index is multiplied by 10 (i.e. index 1 is 10 per cent of the chance, 2 is 20 per cent and 9 = 90 per cent), which selected the index, the task of determining the index.

I came to this decision: Create a mass in which the number of values is equal to the index number with the index number. Example:

arr[1] = [1]
arr[2] = [2]
arr[3] = [2]
arr[4] = [3]
arr[5] = [3]
arr[6] = [3]
arr[7] = [4]
arr[8] = [4]
arr[9] = [4]
arr[10] = [4]
...
arr[..] = [7] //не считал какой именно получается номер индекса
arr[..] = [7]
arr[..] = [7]
arr[..] = [7]
arr[..] = [7]
arr[..] = [7]
...

Then make a suffle of the mass (not necessarily).

Then make $rand = rand(1,count(arr)

And the result will be arr[$rand] - the value and the volume index between 1 and 9. ♪

Question: Is it right if I do, what alternatives are easier? ♪ ♪

SUPPLEMENTING FOR NON-GOVERNING OBSERVATIONS:

In front of you, nine doors, each door has a number 1, 2, 3, 4...9. The chance you're gonna walk in the door with number 1 = 10%, the chance you're gonna get in the door with number 2 = 20 percent, the chance you're gonna get in the door with number 7 = 70 percent and so on to the door with room 9, which has a 90% chance that you're gonna get in there. Which door will you come in?

As option No. 2

// если не из 45, а из 10
$randNum = rand(1, 90)
if($randNum = 1 or $randNum = 2) return 1  // 2%
elseif($randNum >= 3 and $randNum <= 6) return 2 // 4%
elseif($randNum >= 7 and $randNum <= 12) return 3 // 6%
elseif($randNum >= 13 and $randNum <= 20) return 4 // 8%
elseif($randNum >= 21 and $randNum <= 30) return 5 // 10%
elseif($randNum >= 31 and $randNum <= 42) return 6 // 12%
elseif($randNum >= 43 and $randNum <= 56) return 7 // 14%
elseif($randNum >= 57 and $randNum <= 72) return 8 // 16%
elseif($randNum >= 73 and $randNum <= 90) return 9 // 18%
//90%

I've got it. ♪ ♪

[chanisef]

And if you don't have to count it, you just need to get the room ball out of the bag, javascript You can think of such a solution for an ambulance.

var arr = [];
var k = 0;
var b = 10;
// заполняем мешок шариками

for (var i = 1; i < b; i++) {

for (var j = 0; j < i; j++) {

arr[k] = i;

k++;

}

}
// Перемешиваем

function sRand() {

return Math.random() > 0.5 ? 1 : -1;

}

arr.sort(sRand);
// Вытаскиваем первый шарик

console.log(arr[0]);