#include<iostream>
#include<algorithm>
#include<cstring>

using namespace std;
const int N=14;
const double INF=1e11;

int A,B,C,D;
double f[N][N][N][N][5][5];

double dp(int a,int b,int c,int d,int x,int y)   // 求解 f[a][b][c][d][x][y]
{
    double &v = f[a][b][c][d][x][y];
    if(v >= 0) return v;
    int as = a + (x == 0) + (y == 0);
    int bs = b + (x == 1) + (y == 1);
    int cs = c + (x == 2) + (y == 2);
    int ds = d + (x == 3) + (y == 3);
    if(as >= A && bs >= B && cs >= C && ds >= D) return v = 0;     // 说明此状态为终点，按定义来 f = 0 ！！！
    int sum = as + bs + cs + ds;
    if(sum >= 54) return v = INF;
    sum = 54 - sum;
    v = 0;          // v = 0 或 v = 1 都是可以的，只是含义和下面的计算过程不同，自己明白即可 ！
    if(a < 13) v += (13.0 - a) / sum * (dp(a + 1,b,c,d,x,y) + 1);
    if(b < 13) v += (13.0 - b) / sum * (dp(a,b + 1,c,d,x,y) + 1);
    if(c < 13) v += (13.0 - c) / sum * (dp(a,b,c + 1,d,x,y) + 1);
    if(d < 13) v += (13.0 - d) / sum * (dp(a,b,c,d + 1,x,y) + 1);
    if(x == 4)
    {
        double t = INF;
        for(int i = 0;i < 4;i ++) t = min(t,1.0/sum * (dp(a,b,c,d,i,y) + 1));   // 如果题目变成大小鬼任意放到一叠牌堆，写法又不一样，请斟酌体会！
        v += t;
    }
    if(y == 4)
    {
        double t = INF;
        for(int i = 0;i < 4;i ++) t = min(t,1.0/sum * (dp(a,b,c,d,x,i) + 1));
        v += t; 
    }

    return v;
}



int main()
{
    cin >> A >> B >> C >> D;
    memset(f,-1,sizeof f);                  // ！！！！
    double t = dp(0,0,0,0,4,4);
    if(t > INF / 2) t = -1;
    printf("%.3lf\n",t);

    return 0;
}