#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for(int i = 0; i < n; i++)
#define rep2(i, x, n) for(int i = x; i <= n; i++)
#define rep3(i, x, n) for(int i = x; i >= n; i--)
#define each(e, v) for(auto &e: v)
#define pb push_back
#define eb emplace_back
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define sz(x) (int)x.size()
using ll = long long;
using pii = pair<int, int>;
using pil = pair<int, ll>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
const int MOD = 1000000007;
//const int MOD = 998244353;
const int inf = (1<<30)-1;
const ll INF = (1LL<<60)-1;
template<typename T> bool chmax(T &x, const T &y) {return (x < y)? (x = y, true) : false;};
template<typename T> bool chmin(T &x, const T &y) {return (x > y)? (x = y, true) : false;};

struct io_setup{
    io_setup(){
        ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout << fixed << setprecision(15);
    }
} io_setup;

int main(){
    int N; cin >> N;

    vector<int> A(N), R(N);
    rep(i, N) cin >> A[i];
    rep(i, N) cin >> R[i];

    vector<bool> used(1001, false);
    rep(i, N) used[A[i]] = true;

    vector<vector<double>> P(N, vector<double>(1001, 0.0));
    vector<vector<double>> dp(N+1, vector<double>(N+1, 0.0));

    rep(i, 1001){
        if(!used[i]) continue;

        rep(j, N+1){
            rep(k, N+1){
                dp[j][k] = 0.0;
            }
        }

        dp[0][0] = 1.0;

        rep(j, N){
            rep(k, N){
                if(A[j] < i){
                    dp[j+1][k] += dp[j][k]*0.5;
                    dp[j+1][k+1] += dp[j][k]*0.5;
                }
                else{
                    if(k == 0){
                        P[j][i] += dp[j][k]*0.5;
                        dp[j+1][k] += dp[j][k];
                    }
                    else{
                        dp[j+1][k] += dp[j][k]*0.5;
                        dp[j+1][k-1] += dp[j][k]*0.5;
                    }
                }
            }
        }
    }

    double ans = 0;
    int pre = 0;

    rep(i, 1001){
        if(used[i]){
            rep(j, N){
                ans += P[j][i]*R[j]*(i-pre);
            }
            pre = i;
        }
    }

    cout << ans << '\n';
}