The winning probability algorithm based on PHP code can be used for scratch cards, big wheel and other lottery algorithms, _PHP tutorial

The winning probability algorithm based on PHP code can be used for scratch cards, big wheel and other lottery algorithms.

The big wheel winning probability algorithm is often encountered in our daily life. So how to implement the winning probability algorithm based on PHP code? Here is a code example to introduce the PHP winning probability algorithm. The code is simple and easy to understand, and comes with comments. The specific code is as follows:

<&#63;php
/*
 * 经典的概率算法，
 * $proArr是一个预先设置的数组，
 * 假设数组为：array(100,200,300，400)，
 * 开始是从1,1000 这个概率范围内筛选第一个数是否在他的出现概率范围之内， 
 * 如果不在，则将概率空间，也就是k的值减去刚刚的那个数字的概率空间，
 * 在本例当中就是减去100，也就是说第二个数是在1，900这个范围内筛选的。
 * 这样 筛选到最终，总会有一个数满足要求。
 * 就相当于去一个箱子里摸东西，
 * 第一个不是，第二个不是，第三个还不是，那最后一个一定是。
 * 这个算法简单，而且效率非常 高，
 * 关键是这个算法已在我们以前的项目中有应用，尤其是大数据量的项目中效率非常棒。
 */
function get_rand($proArr) { 
 $result = ''; 
 //概率数组的总概率精度 
 $proSum = array_sum($proArr); 
 //概率数组循环 
 foreach ($proArr as $key => $proCur) { 
  $randNum = mt_rand(1, $proSum); 
  if ($randNum <= $proCur) { 
   $result = $key; 
   break; 
  } else { 
   $proSum -= $proCur; 
  }   
 } 
 unset ($proArr); 
 return $result; 
} 
/*
 * 奖项数组
 * 是一个二维数组，记录了所有本次抽奖的奖项信息，
 * 其中id表示中奖等级，prize表示奖品，v表示中奖概率。
 * 注意其中的v必须为整数，你可以将对应的 奖项的v设置成0，即意味着该奖项抽中的几率是0，
 * 数组中v的总和（基数），基数越大越能体现概率的准确性。
 * 本例中v的总和为100，那么平板电脑对应的 中奖概率就是1%，
 * 如果v的总和是10000，那中奖概率就是万分之一了。
 * 
 */
$prize_arr = array( 
 '0' => array('id'=>1,'prize'=>'平板电脑','v'=>1), 
 '1' => array('id'=>2,'prize'=>'数码相机','v'=>5), 
 '2' => array('id'=>3,'prize'=>'音箱设备','v'=>10), 
 '3' => array('id'=>4,'prize'=>'4G优盘','v'=>12), 
 '4' => array('id'=>5,'prize'=>'10Q币','v'=>22), 
 '5' => array('id'=>6,'prize'=>'下次没准就能中哦','v'=>50), 
); 
/*
 * 每次前端页面的请求，PHP循环奖项设置数组，
 * 通过概率计算函数get_rand获取抽中的奖项id。
 * 将中奖奖品保存在数组$res['yes']中，
 * 而剩下的未中奖的信息保存在$res['no']中，
 * 最后输出json个数数据给前端页面。
 */
foreach ($prize_arr as $key => $val) { 
 $arr[$val['id']] = $val['v']; 
} 
$rid = get_rand($arr); //根据概率获取奖项id 
$res['yes'] = $prize_arr[$rid-1]['prize']; //中奖项 
unset($prize_arr[$rid-1]); //将中奖项从数组中剔除，剩下未中奖项 
shuffle($prize_arr); //打乱数组顺序 
for($i=0;$i<count($prize_arr);$i++){ 
 $pr[] = $prize_arr[$i]['prize']; 
} 
$res['no'] = $pr; 
print_r($res); 

Let me share with you an example code based on Java to calculate the winning probability

When doing mobile projects, there is a demand, so let’s do a lottery! The calculation of winnings is quite disgusting. Users have to change the winning probability of each award, and there is a limit on the number of awards per day. The probabilities of prizes one, two, three, four, five and six are unreasonable. How can one calculate whether a user has won or not? After thinking hard, I can use the nextInt(int x) method of the Random class to generate a random number within a range. The winning range will be the winning range. The winning range is generated dynamically. The source code is posted for reference only!

package Mzone;
import java.util.ArrayList;
import java.util.Random;
public class Mzone {
 /**
 * CopyRright(c)2009-04: 
 * Project: 
 * Module ID: 
 * Comments: 概率计算
 * JDK version used: <JDK1.4>
 * Author：ch
 * Create Date：2009-04-20
 * Modified By： 
 * Modified Date: 
 * Why & What is modified 
 * Version: 1.0
 */ 
 static Random r = new Random();
 public static void main(String[] args) {
  //各个奖项的中奖概率的分母
  Integer _5m = new Integer(5);
  Integer _500m = new Integer(30);
  Integer _ipod = new Integer(500);
  Integer _phone = new Integer(1000);
  Integer _notebook = new Integer(1500);
  Integer _jay = new Integer(50);
  ArrayList list = new ArrayList();
  if(_5m.intValue()!=0)
   list.add(_5m);
  if(_500m.intValue()!=0)
   list.add(_500m);
  if(_ipod.intValue()!=0)
   list.add(_ipod);
  if(_phone.intValue()!=0)
   list.add(_phone);
  if(_notebook.intValue()!=0)
   list.add(_notebook);
  if(_jay.intValue()!=0)
   list.add(_jay);
  //计算最小公倍数
  int common = getN(list);
  System.out.println("最小公倍数:"+common);
  int a = 0;int b = 0;int c = 0;int d = 0;int e = 0;int f = 0;int g = 0;
  int first = 0;int second = 0;int third = 0;int four = 0;int fifth = 0;int sixth = 0;
  if(_5m.intValue()!=0){
   first = common/_5m.intValue();
  }
  if(_500m.intValue()!=0){
   second = first + (common/_500m.intValue());
  }else second = first;
  if(_ipod.intValue()!=0){
   third = second + (common/_ipod.intValue());
  }else third = second;
  if(_phone.intValue()!=0){
   four = third + (common/_phone.intValue());
  }else four = third;
  if(_notebook.intValue()!=0){
   fifth = four + (common/_notebook.intValue());
  }else fifth = four;
  if(_jay.intValue()!=0){
   sixth = fifth + (common/_jay.intValue());
  }else sixth = fifth;
  int times = 30000;//循环次数
  for(int i = 0;i < times; i++){
   int ri = getRandom(common);//产生随机数
   if(ri >= 0 && ri < first){
    a++;
   }else if(ri >= first && ri < second){
    b++;
   }else if(ri >= second && ri < third){
    c++;
   }else if(ri >= third && ri < four){
    d++;
   }else if(ri >= four && ri < fifth){
    e++;
   }else if(ri >= fifth && ri < sixth){
    f++;
   }else{
    g++;
   }
  }
  System.out.println("5m值:" + a + " 500m值:" + b + " ipodMP3:" + c + " 手机:" + d + " 笔记本电脑:" + e + " 演唱会门票:" + f + " 谢谢参与:" + g);
 }
 /**
  * 求最大公约数
 */
 public static int gcd(int m, int n){
  while (true){
   if ((m = m % n) == 0)
   return n;
   if ((n = n % m) == 0)
   return m;
  }
 }
 /**
 * 求最小公倍数
 */
 public static int gys(int z, int y){
  int t = 0;
  int c = 0;
  c = gcd(z,y);
  t = z * y / c;
  return t;
 }
 /**
  * 求几个数的最小公倍数
 */
 public static int getN(ArrayList list){
  int t = 1;
  for(int i = 0;i<list.size();i++){
   Integer temp = (Integer)list.get(i);
   t = gys(t,temp.intValue());
  }
  return t; 
 }
 /**
  * 产生随机数
 */
 public static int getRandom(int y){
  int result = r.nextInt(y);
  return result;
 }
}

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