Question

这是一个有趣的统计相关问题。

我有什么

平均
可能发生的范围

我的追求

在我的范围内发生一次事件的可能性

例如，平均值为10.3。 1-20。发生4的机会是什么？我需要使用10.3作为权重，因为如果没有，那么每次发生的可能性都为1/20

是否有类似这样的统计公式？

编码

public void ReboundFormula(double TeamRebound, double OTeamRebound, double OffensiveRebound, double DefensiveRebound,
        double ShotsTotalAverage, double OShotsTotalAverage, double TeamShotAveragePercent, double OTeamShotAveragePercent,
        double PlayerORebound, double PlayerDRebound, double PlayerOPercentRebound, double PlayerDPercentRebound)
{
    //Possible rebounds using amount of shots that "miss" which result in a rebound chance. Note fouls, ball going out of bounds
    //, or other event that causes a rebound to not occur on a missed shot

    predictedPossibleRebounds = (ShotsTotalAverage*(1 -(double) TeamShotAveragePercent/100)) + OShotsTotalAverage*(1 - (double) TeamShotAveragePercent/100);


    Temp = (double) TeamRebound/(TeamRebound+OTeamRebound);

    Temp2 = (double) OTeamRebound/(TeamRebound+OTeamRebound);

    System.out.println("Percent of Team 1 Rebounds: " + Temp + " | Percent of Team 2 Rebound: " + Temp2);

    System.out.println();

    //Predicted rebounds a team will grab of the amount of rebounds they will likely get
    predictedTeamRebound = (double) predictedPossibleRebounds*Temp;

    predictedOTeamRebound = (double) predictedPossibleRebounds*Temp2;

    System.out.println("Amount of Rebounds to Team: " + predictedTeamRebound + " | Amount of Rebounds to OTeam: " + predictedOTeamRebound);
    System.out.println();

    //Rebounds predicted to be grabbed by teams
    Oratio = ((double) OffensiveRebound/TeamRebound)*predictedTeamRebound;
    Dratio = ((double) DefensiveRebound/TeamRebound)*predictedTeamRebound;

    System.out.println("Amount of Offensive Rebounds: " + Oratio + " | Amount of Defensive Rebounds: " + Dratio);
    System.out.println();

    System.out.println("Predicted Rebounds: " + predictedPossibleRebounds + " | Player Offensive Percent: " + PlayerOPercentRebound + " | Player Defensive Percent: " + PlayerDPercentRebound);

    //Player data time
    Temp = (Oratio*(PlayerOPercentRebound/100)) + (Dratio*(PlayerDPercentRebound/100));

    System.out.println();
    System.out.println("Player Predicted Rebound Total: " + Temp + " | Player Rebound Total Average: " + (PlayerORebound+PlayerDRebound) );
    System.out.println();

    System.out.println( "  " + (int) Math.round((Temp) - (Temp*0.75)) + "     " + (int) Math.round((Temp) - (Temp*0.5)) + "     " + (int) Math.round((Temp) - (Temp*0.25)) + "     " + (int) Math.round((Temp) - (Temp*0.1)) + "    " + (Temp) 
            + "  " + (int) Math.round((Temp) + (Temp*0.1)) + "    " + (int) Math.round((Temp) + (Temp*0.25)) + "     " +  (int) Math.round((Temp) + (Temp*0.5)) + "     " + (int) Math.round((Temp) + (Temp*0.75)));
    System.out.println("-75%  -50%  -25%  -10%   Predicted Average  +10%  +25%   +50%   +75%");
    System.out.println();

进料数据

run.ReboundFormula(54, 48.2, 13.662, 40.338, 89.86, 87.06, 45.2, 45.7, 3.7, 6.2, 12.9, 22.8);

Answer 1

您需要指定结果的分布。仅指定平均值和范围是不够的。例如，一个不平衡的骰子有1/12的机会阅读1和3/12的阅读机会5，其他结果具有相同的可能性，则平均为3.83。但是，也可以使用具有相同平均值的其他不平衡骰子，例如，骰子有1/12的机会读取2和3/12的机会读取6。

我怀疑您想要的是binomial distribution.的二项式分布是由有限数量的离散试验（如程序的反弹次数）产生的，每个试验成功或失败，每个等重的试验。

如果您单击链接，将会看到很多公式，可能比您想要的要多，但是一个原理是公式的基础，而该原理是您主要要掌握的。那并不是那么容易，但是一旦掌握了原理，自然就会遵循主要的公式。

使用平均值获取一系列数字中一次出现的概率

1 个答案: