Chance of obtaining fish element Probability

Question

I'm trying to get the percentage of getting some fish. X is the unique number here, it's between 0-99, but in my example code it's 70, for test purposes.

But currently it returns really wrong results, so I'm doing something wrong here.

Current output:

[-0.17142857142857143, 0.04285714285714286, 0.2571428571428571, 0.2714285714285714]

Which means

• -17% to get fish1,
• 4.2% for fish2,
• 25.7% fish3,
• 27% fish4.

The correct math is as follows:

If X number is 82+, it will get fish 1. If X number is 67+, it's fish 2 (unless 82+ etc). If x number is 52+ it's fish 3. if it's 51 or less, it's fish 4.

So, in theory, with a x of 85, its a 3/85 chance of obtaining it.

Then I need the probability by the current X of getting that specific fish. What am I doing wrong here?

function fishesPercentage(X) {
  var NumberSelected = RandomNumber(0,X-1);
  NumberSelected = 70;
  console.log("RANDOOM NUMBER IS : ",NumberSelected);
  let Fish1 = NumberSelected - 82;
  let Fish2 = NumberSelected - 67;
  let Fish3 = NumberSelected - 52;
  let Fish4 = NumberSelected - 51;

  return [
    Fish1/ NumberSelected,
    Fish2/ NumberSelected,
    Fish3/ NumberSelected,
    Fish4/ NumberSelected
  ];
}

Answer 1

You want the probabilities of obtaining fishes 1 to 4 for a given number X, and a series of ranges. The total number of fishes is:

Fish 4) 51 or X (if X is lower)

Fish 3) 68-51 o X-51 (if X is lower than 68)

etc.

If any of these X-N is negative, that means that there are no fishes of that type.

This could be achieved with any of the two functions below:

function fishProb(x) {
    let fish4 = Math.max(Math.min(x, 51),0)
    let fish3 = Math.max(Math.min(x-51, 68-51),0)
    let fish2 = Math.max(Math.min(x-68, 82-68),0)
    let fish1 = Math.max(Math.min(x-82, 99-82),0)
    return([fish1, fish2, fish3, fish4].map(f => f/x))
}

function fishProb2(x) {
    var allFishes = x
    var fishType = [51, 68-51, 82-68, 99-82]
    var fishes = []
    for (const currentFishes of fishType) {
        fishes.push( Math.max(Math.min(allFishes, currentFishes), 0) )
        allFishes -= currentFishes
    }
   return(fishes.map(f => f/x).reverse())
}

Both functions return the same thing. The second one I think expresses the problem in a simpler way.

If you have X=70 then you have 51 Fish4, from the remaining 70-51 you have (68-51) of Fish3, and the other 3 remaining are Fish2. No Fish1 are in the lot.

//X = 70
[0, 0.02857142857142857, 0.24285714285714285, 0.7285714285714285]

//X = 85
[0.03529411764705882, 0.16470588235294117, 0.2, 0.6]

//X = 20. In this case you only have 20 Fish4 and none of the others.
[0, 0, 0, 1]