質問のカードの確率の問題をJavaで解いてみた。

　質問のカードの確率の問題をJavaで解いてみました。(^_^;
　組合せを生成して、P(a< b),P(a==b),P(a> b)を求めてみました。途中、b=2,4,6,...,20のときのa>b,a=b,a

/* 
 * ProbOfCards3.java
 * 
 */

class ProbOfCards3 {
    // 順列生成
    static boolean nextPerm(int[] p, int n, int r) {
        int i, j;
        int t;
        
        if(r <= 0 || n < r) return(false);
        for(i = r + 1; i <= n-1; i++)
            for(j = i; j >= r + 1 && p[j-1] < p[j]; j--){
                t = p[j]; p[j] = p[j-1]; p[j-1] = t;    // swap(p,j,j-1);
            }
        for(i = n - 1; i > 0 && p[i-1] >= p[i]; i--);
        if(i==0) return(false);
        for(j = n - 1; j > i && p[i-1] >= p[j]; j--);
        t = p[j]; p[j] = p[i-1]; p[i-1] = t;            // swap(p,j,i-1);
        for(j = n - 1; i < j; i++, j--){
            t = p[i]; p[i] = p[j]; p[j] = t;            // swap(p,i,j);
        }
        return(true);
    }
    // 組合せ生成
    static boolean nextComb(int[] p, int n, int r) {
        int i;
        do{
            if(!nextPerm(p,n,r)) return(false);
            for(i=0; i< r-1; i++)
                if(p[i]> p[i+1]) break;
        } while (i< r-1);
        return(true);
    }
    
    static int gcd(int a, int b){
        return( b == 0 ? a : gcd(b, a % b) );
    }
    
    public static void main(String[] args) {
        final int N = 10, R =  2;
        final int[] A = {1,2,3,4,5,6,7,8,9,10};
        int total = 0, countA = 0, countB = 0, countE = 0, g = 1;
        int[] a = new int[N];           // 組合せ生成用
        
        long tm=System.nanoTime();      // Timer Start
        System.out.println(" b:a>b,a=b,a<b");
        for(int b=2; b<=20; b+=2){
            int subCountA = 0, subCountB = 0, subCountE = 0;
            a = (int[])A.clone();
            do{
                if(a[0]+a[1]> b) { countA++; subCountA++; }
                if(a[0]+a[1]==b) { countE++; subCountE++; }
                if(a[0]+a[1]< b) { countB++; subCountB++; }
                total++;
            }while(nextComb(a,N,R));
            System.out.printf("%2d:%3d,%3d,%3d\n",b,subCountA,subCountE,subCountB);
        }
        System.out.println();
        g = (countA!=0 ? gcd(countA, total) : 1);
        System.out.println("P(a> b)= "+countA+"/"+total+"="+(countA/g)+"/"+(total/g));
        g = (countB!=0 ? gcd(countE, total) : 1);
        System.out.println("P(a==b)= "+countE+"/"+total+"="+(countE/g)+"/"+(total/g));
        g = (countB!=0 ? gcd(countB, total) : 1);
        System.out.println("P(a< b)= "+countB+"/"+total+"="+(countB/g)+"/"+(total/g));
        tm=System.nanoTime()-tm;        // Timer Stop
        System.out.printf("Runtime : %.3f [sec]\n",(double)tm/1e9);
    }
}

●実行結果

 b:a>b,a=b,a<b
 2: 45,  0,  0
 4: 43,  1,  1
 6: 39,  2,  4
 8: 33,  3,  9
10: 25,  4, 16
12: 16,  4, 25
14:  9,  3, 33
16:  4,  2, 39
18:  1,  1, 43
20:  0,  0, 45

P(a> b)= 215/450=43/90
P(a==b)= 20/450=2/45
P(a< b)= 215/450=43/90
Runtime : 0.032 [sec]