#include using namespace std; #define REP(i,n) for(int i=0;i<(int)(n);++i) #define FOR(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();++i) #define ALL(c) (c).begin(), (c).end() #define valid(y,x,h,w) (0<=y&&y pii; templatebool chmax(T &a, const T &b) { if (abool chmin(T &a, const T &b) { if (bbasic_ostream& operator<<(basic_ostream&os, const C& c){os<<'[';for(auto i=begin(c);i!=end(c);++i)os<<(i==begin(c)?"":" ")<<*i;return os<<']';} templateostream&operator<<(ostream &o,const pair&t){return o<<'('<void output(ostream&,const Tp&){} templatevoid output(ostream &o,const Tp&t){if(N)o<<',';o<(t);output(o,t);} templateostream&operator<<(ostream&o,const tuple&t){o<<'(';output<0,tuple,Ts...>(o,t);return o<<')';} templatevoid output(T t,char z=10){if(t<0)t=-t,putchar(45);int c[20]; int k=0;while(t)c[k++]=t%10,t/=10;for(k||(c[k++]=0);k;)putchar(c[--k]^48);putchar(z);} templatevoid outputs(T t){output(t);} templatevoid outputs(S a,T...t){output(a,32);outputs(t...);} templatevoid output(T *a,int n){REP(i,n)cout<void output(T *a,int n,int m){REP(i,n)output(a[i],m);} templatebool input(T &t){int n=1,c;for(t=0;!isdigit(c=getchar())&&~c&&c-45;); if(!~c)return 0;for(c-45&&(n=0,t=c^48);isdigit(c=getchar());)t=10*t+c-48;t=n?-t:t;return 1;} templatebool input(S&a,T&...t){input(a);return input(t...);} templatebool inputs(T *a, int n) { REP(i,n) if(!input(a[i])) return 0; return 1;} double dp[101][101][101]; vector p; int n; double solve(int i, int j, int k) { if (dp[i][j][k] >= 0) return dp[i][j][k]; int S = 0; if (i>0) S |= 1<<0; if (j>0) S |= 1<<1; if (k>0) S |= 1<<2; if (S == 0) return 0; double p[3] = {(double)i / n, (double)j / n, (double)k / n }; double pp = 0; REP(l,3) { if (S>>l&1) pp += p[l]; } // outputs(i,j,k); double res = 0; REP(l,3) { if (S>>l&1) { int ii = i, jj = j, kk = k; if (l == 0) { ii--; } else if (l == 1) { jj--; ii++; } else { kk--; jj++; } res += (solve(ii,jj,kk) + 1.0 / pp) * (p[l]/pp); } } // outputs(i,j,k); // cout << res << " " << pp << endl; return dp[i][j][k] = res; } int main() { while(cin >> n) { int A[100]; int B[4] = {}; REP(i,n) { cin >> A[i]; chmin(A[i],3); B[3-A[i]]++; } double denom = B[0]+B[1]+B[2]+B[3]; p = vector({B[1] / denom, B[2] / denom, B[3] / denom}); memset(dp,-1,sizeof(dp)); double ans = solve(B[1],B[2],B[3]); printf("%.10f\n", ans); } }