在javascript中，如何在控制总金额的同时，向用户的余额添加随机数？

Question

I'm trying to make it to where when a user does a certain thing, they get between 2 and 100 units. But for every 1,000 requests I want it to add up to 3,500 units given collectively.

Here's the code I have for adding different amounts randomly to a user:

if (Math.floor(Math.random() * 1000) + 1 === 900) {
    //db call adding 100
}
else if (Math.floor(Math.random() * 100) + 1 === 90)  {
    //db call adding 40
}
else if (Math.floor(Math.random() * 30) + 1 === 20)  {
    //db call adding 10
}
else if (Math.floor(Math.random() * 5) + 1 === 4)  {
    //db call adding 5
}
else {
    //db call adding 2
}

If my math is correct, this should average around 4,332 units per 1,000 calls. But obviously it would vary and I don't want that. I'd also like it to add random amounts instead, as the units added in my example are arbitrary.

EDIT: Guys, Gildor is right that I simply want to have 3,500 units, and give them away within 1,000 requests. It isn't even entirely necessary that it always reaches that maximum of 3,500 either (I could have specified that). The important thing is that I'm not giving users too much, while creating a chance for them to win a bigger amount.

Here's what I have set up now, and it's working well, and will work even better with some tweaking:

Outside of call:

var remaining = 150;
var count = 0;

Inside of call:

    count += 1;
    if (count === 100) {
        remaining = 150;
        count = 0;
    }

    if (Math.floor(Math.random() * 30) + 1 === 20) {
        var addAmount = Math.floor(Math.random() * 85) + 15;
        if (addAmount <= remaining) {
            remaining -= addAmount;
            //db call adding addAmount + 2
        }
        else {
            //db call adding 2
        }
    }
    else if (Math.floor(Math.random() * 5) + 1 === 4) {
        var addAmount1 = Math.floor(Math.random() * 10) + 1;
        if (addAmount1 <= remaining) {
            remaining -= addAmount1;
            //db call adding addAmount1 + 2

        }
        else {
            //db call adding 2
        }
    }
    else {
        //db call adding 2
    } 

I guess I should have clarified, I want a "random" number with a high likelihood of being small. That's kind of part of the gimmick, where you have low probability of getting a larger amount.

Answer 1

As I've commented, 1,000 random numbers between 2 and 100 that add up to 3,500 is an average number of 3.5 which is not consistent with random choices between 2 and 100. You'd have to have nearly all 2 and 3 values in order to achieve that and, in fact couldn't have more than a couple large numbers. Nothing even close to random. So, for this to even be remotely random and feasible, you'd have to pick a total much large than 3,500. A random total of 1,000 numbers between 2 and 100 would be more like 51,000.

Furthermore, you can't dynamically generate each number in a truly random fashion and guarantee a particular total. The main way to guarantee that outcome is to pre-allocate random numbers that add up to the total that are known to achieve that and then random select each number from the pre-allocated scheme, then remove that from the choice for future selections.

You could also try to keep a running total and bias your randomness if you get skewed away form your total, but doing it that way, the last set of numbers may have to be not even close to random in order to hit your total consistently.

A scheme that could work if you reset the total to support what it should be for actual randomness (eg to 51,000) would be to preallocated an array of 500 random numbers between 2 and 100 and then add another 500 numbers that are the complements of those. This guarantees the 51 avg number. You can then select each number randomly from the pre-allocated array and then remove it form the array so it won't be selected again. I can add code to do this in a second.

function RandResults(low, high, qty) {
    var results = new Array(qty);
    var limit = qty/2;
    var avg = (low + high) / 2;
    for (var i = 0; i < limit; i++) {
        results[i] = Math.floor((Math.random() * (high - low)) + low);
        // 
        results[qty - i - 1] = (2 * avg) - results[i];
    }
    this.results = results;
}

RandResults.prototype.getRand = function() {
    if (!this.results.length) {
        throw new Error("getRand() called, but results are empty");
    }
    var randIndex = Math.floor(Math.random() * this.results.length);
    var value = this.results[randIndex];
    this.results.splice(randIndex, 1);
    return value;
}

RandResults.prototype.getRemaining = function() {
    return this.results.length;
}


var randObj = new RandResults(2, 100, 1000);

// get next single random value
if (randObj.getRemaining()) {
   var randomValue = randObj.getRand();
}

Working demo for a truly random selection of numbers that add up to 51,000 (which is what 1,000 random values between 2 and 100 should add up to): http://jsfiddle.net/jfriend00/wga26n7p/

---

If what you want is the following: 1,000 numbers that add up to 3,500 and are selected from between the range 2 to 100 (inclusive) where most numbers will be 2 or 3, but occasionally something could be up to 100, then that's a different problem. I wouldn't really use the word random to describe it because it's a highly biased selection .

Here's a way to do that. It generates 1,000 random numbers between 2 and 100, keeping track of the total. Then, afterwards it corrects the random numbers to hit the right total by randomly selected values and decrementing them until the total is down to 3,500. You can see it work here: http://jsfiddle.net/jfriend00/m4ouonj4/

The main part of the code is this:

function RandResults(low, high, qty, total) {
    var results = new Array(qty);
    var runningTotal = 0, correction, index, trial;
    for (var i = 0; i < qty; i++) {
        runningTotal += results[i] = Math.floor((Math.random() * (high - low)) + low);
    }
    // now, correct to hit the total
    if (runningTotal > total) {
        correction = -1;
    } else if (runningTotal < total) {
        correction = 1;
    }
    // loop until we've hit the total
    // randomly select a value to apply the correction to
    while (runningTotal !== total) {
        index = Math.floor(Math.random() * qty);
        trial = results[index] + correction;
        if (trial >= low && trial <= high) {
            results[index] = trial;
            runningTotal += correction;
        }
    }

    this.results = results;
}

This meets an objective of a biased total of 3,500 and all numbers between 2 and 100, though the probability of a 2 in this scheme is very high and the probably of a 100 in this scheme is almost non-existent.

---

And, here's a weighted random generator that adds up to a precise total. This uses a cubic weighting scheme to favor the lower numbers (the probably of a number goes down with the cube of the number) and then after the random numbers are generated, a correction algorithm applies random corrections to the numbers to make the total come out exactly as specified. The code for a working demo is here: http://jsfiddle.net/jfriend00/g6mds8rr/

function RandResults(low, high, numPicks, total) {
    var avg = total / numPicks;
    var i, j;
    // calculate probabilities for each value
    // by trial and error, we found that a cubic weighting
    // gives an approximately correct sub-total that can then
    // be corrected to the exact total
    var numBuckets = high - low + 1;
    var item;
    var probabilities = [];
    for (i = 0; i < numBuckets; i++) {
        item = low + i;
        probabilities[i] = avg / (item * item * item);
    }

    // now using those probabilities, create a steps array
    var sum = 0;
    var steps = probabilities.map(function(item) {
        sum += item;
        return sum;
    });

    // now generate a random number and find what
    // index it belongs to in the steps array
    // and use that as our pick
    var runningTotal = 0, rand;
    var picks = [], pick, stepsLen = steps.length;
    for (i = 0; i < numPicks; i++) {
        rand = Math.random() * sum;
        for (j = 0; j < stepsLen; j++) {
            if (steps[j] >= rand) {
                pick = j + low;
                picks.push(pick);
                runningTotal += pick;
                break;
            }
        }
    }

    var correction;
    // now run our correction algorithm to hit the total exactly
    if (runningTotal > total) {
        correction = -1;
    } else if (runningTotal < total) {
        correction = 1;
    }    

    // loop until we've hit the total
    // randomly select a value to apply the correction to
    while (runningTotal !== total) {
        index = Math.floor(Math.random() * numPicks);
        trial = picks[index] + correction;
        if (trial >= low && trial <= high) {
            picks[index] = trial;
            runningTotal += correction;
        }
    }
    this.results = picks;    
}

RandResults.prototype.getRand = function() {
    if (!this.results.length) {
        throw new Error("getRand() called, but results are empty");
    }
    return this.results.pop();
}

RandResults.prototype.getAllRand = function() {
    if (!this.results.length) {
        throw new Error("getAllRand() called, but results are empty");
    }
    var r = this.results;
    this.results = [];
    return r;
}


RandResults.prototype.getRemaining = function() {
    return this.results.length;
}

Answer 2

As some comments pointed out... the numbers in the question does not quite make sense, but conceptually there are two approaches: calculate dynamically just in time or ahead of time.

To calculate just in time:

You can maintain a remaining variable which tracks how many of 3500 left. Each time when you randomly give some units, subtract the number from remaining until it goes to 0.

In addition, to make sure each time at least 2 units are given, you can start with remaining = 1500 and give random + 2 units each time.

To prevent cases that after 1000 gives there are still balances left, you may need to add some logic to give units more aggressively towards the last few times. However it will result in not-so-random results.

To calculate ahead of time:

Generate a random list with 1000 values in [2, 100] and sums up to 3500. Then shuffle the list. Each time you want to give some units, pick the next item in the array. After 1000 gives, generate another list in the same way. This way you get much better randomized results.

Be aware that both approaches requires some kind of shared state that needs to be handled carefully in a multi-threaded environment.

Hope the ideas help.