蒙特卡罗模拟来模拟选举

Question

I have been tormented by this problem for the last week or so and cant seem to find a solution..

My question is, given a presidential election, I am given odds for the 6 candidates, I have to find the probability of a certain three candidates finishing in the last 3 places in any order using a monte carlo simulation.

My first approach was to run a simulation x amount of times with 1000000 + votes.. the problem with this was that I was never finding any cases where these candidates finished last because with so many votes and with the random number generator in java, i eliminated the possibility of luck and therefore the placings just stayed with the probability..

I have been told a better way to do it is to stage an election, and then if the random number falls into the probability of one candidate, i must give him first place, and then with whatever probability is left over i must decide second place and then third place etc etc.. I know the answer has to be 0.7.. does anyone know how I can approach this? many thanks.

Here is the code I have so far, this is not giving me the answer I am supposed to have but I cant seem to find a fault in it..

import java.util.*;

public class Lab4 {

public static void main(String[] args) {

    Random rd = new Random();

    int numElections = 50000000;
    int count = 0;

    ArrayList<String> ar2 = new ArrayList<String>();

    for(int i = 1; i <= numElections; i++)
    {
        double mh = (double)25/26 / 1.0951899766899766;     
        double pc = (double)1/10 / 1.0951899766899766;
        double sg = (double)1/66 / 1.0951899766899766;
        double lnr = (double)1/80 / 1.0951899766899766;
        double jf = (double)1/250 / 1.0951899766899766;
        double gd = (double)1/500 / 1.0951899766899766;

        double sumOdds = (mh + pc + sg + lnr + jf + gd);

        for(int j = 1; j <= 6; j++)
        {
            double randomValue = rd.nextDouble() * sumOdds;

            if(randomValue > 0 && randomValue <= mh)
            {
                ar2.add("mh");
                sumOdds -= mh;
                mh = 0;
            }
            else if(randomValue > 0 && randomValue <=  mh + pc)
            {
                ar2.add("pc");
                sumOdds -= pc;
                pc = 0;
            }
            else if(randomValue > 0 && randomValue <= mh + pc + sg)
            {
                ar2.add("sg");
                sumOdds-= sg;
                sg = 0;
            }
            else if(randomValue > 0 && randomValue <= mh + pc + sg + lnr)
            {
                ar2.add("lnr");
                sumOdds-= lnr;
                lnr = 0;
            }
            else if(randomValue > 0 && randomValue <= mh + pc + sg + lnr + jf)
            {
                ar2.add("jf");
                sumOdds-= jf;
                jf = 0;
            }
            else if(randomValue > 0 && randomValue <= mh + pc + sg + lnr + jf + gd)
            {
                ar2.add("gd");
                sumOdds-= gd;
                gd = 0;
            }   
        }

        if(ar2.get(3)  == "pc"|| ar2.get(4) == "pc" || ar2.get(5) == "pc")
        {
            if(ar2.get(3)  == "gd"|| ar2.get(4) == "gd" || ar2.get(5) == "gd")
            {
                if(ar2.get(3)  == "sg"|| ar2.get(4) == "sg" || ar2.get(5) == "sg")
                {
                    count++;
                }
            }
        }
        ar2.clear();

    }   

    System.out.println((double)count/numElections);


}   

}

Answer 1

The odds for a given candidate are either a percentage of the total vote, or X:Y - like 10:1. You can convert the latter into the former by Y/X = 0.1.

Your simulation idea is great. The question is, how do you choose based on the random number generator .

Let's say we had an array of Candidates, each of which had a normalised chance of winning. A percentage would be divided by 100 - so 80% = 0.8, and the odds format X:Y would be converted as shown above.

We know that the array of candidates' chance column adds up to 1.0.

Using https://docs.oracle.com/javase/8/docs/api/java/util/Random.html#nextFloat-- we can generate a number between 0. and 1.0.

Keeping a running total of the total probability so far, count up the list to find the candidate that the nextFloat has hit.

Something like this

Random random = new Random();
Candidate[] candidates = ...;


// do this for each vote
float total = 0.0f;
int index = 0;
float randomVote = random.nextFloat();
while (total<randomVote && index<candidates.length) {
   total += candidates[index].getProbability();
   if (total<randomVote) { 
       // only increment to the next candidate if this one didn't win the vote
       index++;
   }
}

// at this point, the index is the candidate who won the vote