最大化N个容器中一种类型的概率

给定N个容器，其中N个容器由A个副本组成，N个容器由B个副本组成。我们需要在N个容器中排列数字，以使随机选择一个容器并抽取一个A副本的概率最大。

例子：

Input : N = 1
Output : 0.5,

Input : N = 2
Output : 0.667

我们需要在N个容器中寻找编号为A的N个副本和编号为B的N个副本的最佳安排。

Pi = (Favorable Outcomes) / (Total Outcomes)

where Pi denotes the probability of an event i.

因此，为了最大化这一点，我们要么使分母(即总结果)最小化，要么使分子(即总有利结果总数)最大化，并使另一个保持恒定。

考虑一种可能的安排，在第一个(N-1)个容器中我们只放置数字A的副本，这样第一个(N-1)个容器中包含(N-1)个数字A的副本，而没有数字B的副本。然后将遗留的数字A副本和数字N的N副本放在最后一个容器中。

Probability (copy of A from any of the first (N-1) containers) = P_{n-1 containers} = 1
Probability (copy of A from N^th container) = P_N = ¹⁄_(N+1)
P_max = P_{N-1 containers} * (N – 1) + P_N
∴ P_max = N / (N + 1)

C++

// CPP program to find maximum probability of
// getting a copy 'A' from N
#include 
using namespace std;
 
// Returns the Maximum probability for Drawing
// 1 copy of number A from N containers with N
// copies each of numbers A and B
double calculateProbability(int N)
{
    // Pmax = N/(N+1)
    double probability = (double)N / (N + 1);
    return probability;
}
 
int main()
{
    int N;
    double probabilityMax;
 
    // 1. N = 1
    N = 1;
    probabilityMax = calculateProbability(N);
    cout << "Maximum Probability for N = "
         << N << " is, " << setprecision(4)
        << fixed << probabilityMax << endl;
 
    // 2. N = 2
    N = 2;
    probabilityMax = calculateProbability(N);
    cout << "Maximum Probability for N = "
         << N << " is, " << setprecision(4)
         << fixed << probabilityMax << endl;
 
    // 3. N = 10
    N = 10;
    probabilityMax = calculateProbability(N);
    cout << "Maximum Probability for N = "
         << N << " is, " << setprecision(4)
         << fixed << probabilityMax << endl;
 
    return 0;
}

Java

// Java program to find maximum probability of
// getting a copy 'A' from N
class GFG {
     
    // Returns the Maximum probability for Drawing
    // 1 copy of number A from N containers with N
    // copies each of numbers A and B
    static double calculateProbability(int N)
    {
         
        // Pmax = N/(N+1)
        double probability = (double)N / (N + 1);
         
        return probability;
    }
     
    // Driver code
    public static void main(String[] args)
    {
         
        int N;
        double probabilityMax;
 
        // 1. N = 1
        N = 1;
        probabilityMax = calculateProbability(N);
         
        System.out.println("Maximum Probability for"
                  + " N = " + N + " is, " + Math.round(
                  probabilityMax * 10000.0) / 10000.0);
 
        // 2. N = 2
        N = 2;
        probabilityMax = calculateProbability(N);
        System.out.println("Maximum Probability for N = "
                          + N + " is, " + Math.round(
                     probabilityMax * 10000.0) / 10000.0);
 
        // 3. N = 10
        N = 10;
        probabilityMax = calculateProbability(N);
        System.out.println("Maximum Probability for N = "
                            + N + " is, " + Math.round(
                     probabilityMax * 10000.0) / 10000.0);
    }
}
 
// This code is contributed by Anant Agarwal.

Python3

# Python3 program to find maximum
# probability of getting a copy 'A' from N
 
# Returns the Maximum probability for
# Drawing 1 copy of number A from N
# containers with N copies each of
# numbers A and B
def calculateProbability(N):
 
    # Pmax = N / (N+1)
    probability = N / (N + 1)
    return probability
 
# Driver code
 
# 1. N = 1
N = 1
probabilityMax = calculateProbability(N)
print("Maximum Probability for N = ",
       N , "is, %.4f" %probabilityMax)
 
# 2. N = 2
N = 2
probabilityMax = calculateProbability(N);
print("Maximum Probability for N =",
       N, "is, %.4f" %probabilityMax)
 
# 3. N = 10
N = 10
probabilityMax = calculateProbability(N);
print("Maximum Probability for N =",
       N,"is, %.4f" %probabilityMax)
 
# This code is contributed by Anant Agarwal.

C#

// C# program to find maximum probability of
// getting a copy 'A' from N
using System;
 
class GFG {
     
    // Returns the Maximum probability for Drawing
    // 1 copy of number A from N containers with N
    // copies each of numbers A and B
    static double calculateProbability(int N)
    {
         
        // Pmax = N/(N+1)
        double probability = (double)N / (N + 1);
        return probability;
    }
     
    //Driver code
    public static void Main ()
    {
        int N;
        double probabilityMax;
      
        // 1. N = 1
        N = 1;
        probabilityMax = calculateProbability(N);
        Console.WriteLine("Maximum Probability for N = "
             + N + " is, " +
        Math.Round(probabilityMax * 10000.0) / 10000.0);
      
        // 2. N = 2
        N = 2;
        probabilityMax = calculateProbability(N);
        Console.WriteLine("Maximum Probability for N = "
             + N + " is, " +
        Math.Round(probabilityMax * 10000.0) / 10000.0);
      
        // 3. N = 10
        N = 10;
        probabilityMax = calculateProbability(N);
        Console.WriteLine("Maximum Probability for N = "
             + N + " is, " +
        Math.Round(probabilityMax * 10000.0) / 10000.0);
      
    }
}
 
// This code is contributed by Anant Agarwal.

Javascript

 
// JavaScript program to find maximum probability of
// getting a copy 'A' from N
 
// Returns the Maximum probability for Drawing
// 1 copy of number A from N containers with N
// copies each of numbers A and B
function calculateProbability(N)
{
     
    // Pmax = N/(N+1)
    let probability = N / (N + 1);
       
    return probability;
}
 
// Driver Code
let N;
let probabilityMax;
 
// 1. N = 1
N = 1;
probabilityMax = calculateProbability(N);
   
document.write("Maximum Probability for" +
               " N = " + N + " is, " + Math.round(
               probabilityMax * 10000.0) / 10000.0 + "<br/>");
 
// 2. N = 2
N = 2;
probabilityMax = calculateProbability(N);
document.write("Maximum Probability for N = " + N +
               " is, " + Math.round(
               probabilityMax * 10000.0) / 10000.0 + "<br/>");
 
// 3. N = 10
N = 10;
probabilityMax = calculateProbability(N);
document.write("Maximum Probability for N = " +
               N + " is, " + Math.round(
               probabilityMax * 10000.0) / 10000.0 + "<br/>");
                
// This code is contributed by avijitmondal1998
 

输出：

Maximum Probability for N = 1 is, 0.5000
Maximum Probability for N = 2 is, 0.6667
Maximum Probability for N = 10 is, 0.9091

上面程序的时间复杂度是O(1)。