#include <bits/stdc++.h>
#define rep(i, a) for (int i = 0; i < (a); i++)
#define rep2(i, a, b) for (int i = (a); i < (b); i++)
#define repr(i, a) for (int i = (a) - 1; i >= 0; i--)
#define repr2(i, a, b) for (int i = (b) - 1; i >= (a); i--)
using namespace std;
typedef long long ll;
const ll inf = 1e9;
const ll mod = 1e9 + 7;

int N;
double memo[101][101][101]; // num 0, num 1, num 2

double E(int n0, int n1, int n2) {
	if (n0 == 0 && n1 == 0 && n2 == 0) return 0;
	if (memo[n0][n1][n2] >= 0) return memo[n0][n1][n2];
	int n3 = N - n0 - n1 - n2;
	double res = (double)n3 / N;
	if (n0 > 0) res += (E(n0 - 1, n1 + 1, n2) + 1) * ((double)n0 / N);
	if (n1 > 0) res += (E(n0, n1 - 1, n2 + 1) + 1) * ((double)n1 / N);
	if (n2 > 0) res += (E(n0, n1, n2 - 1) + 1) * ((double)n2 / N);
	res /= 1 - (double)n3 / N;

	return memo[n0][n1][n2] = res;
}

int main() {
	cin >> N;
	vector<int> A(N), num(4);
	rep (i, N) {
		cin >> A[i];
		A[i] = min(3, A[i]);
		num[A[i]]++;
	}
	fill_n((double *)memo, sizeof(memo) / sizeof(memo[0][0][0]), -1);

	double ans = E(num[0], num[1], num[2]);
	printf("%.20f\n", ans);

    return 0;
}