Percentage chance of saying something?

可紊 提交于 2019-12-02 15:31:33
Ernest Friedman-Hill

Yes, Math.random() is an excellent way to accomplish this. What you want to do is compute a single random number, and then make decisions based on that:

var d = Math.random();
if (d < 0.5)
    // 50% chance of being here
else if (d < 0.7)
    // 20% chance of being here
else
    // 30% chance of being here

That way you don't miss any possibilities.

sarnold

For cases like this it is usually best to generate one random number and select the case based on that single number, like so:

int foo = Math.random() * 100;
if (foo < 80) // 0-79
    sendMessage("hi");
else if (foo < 85) // 80-84
    sendMessage("bye");
else // 85-99
    sendMessage("test");

I made a percentage chance function by creating a pool and using the fisher yates shuffle algorithm for a completely random chance. The snippet below tests the chance randomness 20 times.

var arrayShuffle = function(array) {
   for ( var i = 0, length = array.length, swap = 0, temp = ''; i < length; i++ ) {
      swap        = Math.floor(Math.random() * (i + 1));
      temp        = array[swap];
      array[swap] = array[i];
      array[i]    = temp;
   }
   return array;
};

var percentageChance = function(values, chances) {
   for ( var i = 0, pool = []; i < chances.length; i++ ) {
      for ( var i2 = 0; i2 < chances[i]; i2++ ) {
         pool.push(i);
      }
   }
   return values[arrayShuffle(pool)['0']];
};

for ( var i = 0; i < 20; i++ ) {
   console.log(percentageChance(['hi', 'test', 'bye'], [80, 15, 5]));
}

Here is a very simple approximate solution to the problem. Sort an array of true/false values randomly and then pick the first item.

This should give a 1 in 3 chance of being true..

var a = [true, false, false]
a.sort(function(){ return Math.random() >= 0.5 ? 1 : -1 })[0]
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