#include<bits/stdc++.h>
using namespace std;
typedef long double ld;
typedef long long ll;
//#define int ll
#define endl '\n'
#define fi first
#define se second
#define pii pair<int, int>
#define rep(i,a,n,k) for(int i=a;i<=n;i+=k)
#define per(i,a,n,k) for(int i=a;i>=n;i-=k)
/*·········································*/
const int N = 1e5 + 10;
int n, m;
ld dp[N][3];
ll p[N], q[N];

ld vec(ld x, ld y)
{
	if(x <= y) return (x + 1) / 2.0 / (y + 1);
	return (2 * x - y) / 2.0 / ld(n);
}

ld deu(ld x, ld y)
{
	return min(x, y) / ld(n) / ld(m + 1);
}

ld fai(ld x, ld y)
{
	return 1 - vec(x, y) - deu(x, y);
}

void solve()
{
	cin >> n;
	rep(i,0,n-1,1) cin >> p[i];
	rep(i,0,n-1,1) cin >> q[i];
	per(i,n-1,0,1) rep(j,1,min(p[i],(ll)n-i),1)
	{
		dp[i][1] = max(dp[i][1], 1 - dp[i + j][0]);
		dp[i][2] = max(dp[i][2], dp[i + j][0]);
		if(j != p[i])
			dp[i][0] = max({dp[i][0], dp[i][1], dp[i + j][1] * vec(p[i] - j, q[i] + q[i + j]) + (1 - dp[i + j][0]) * deu(p[i] - j, q[i] + q[i + j]) + (1 - dp[i + j][2]) * fai(p[i] - j, q[i] + q[i + j])});
		else dp[i][0] = max(dp[i][0], dp[i][1]);
	}
	cout << fixed << setprecision(10) << dp[0][0];
}

signed main()
{
	ios::sync_with_stdio(0);
	cin.tie(0),cout.tie(0);
	int t = 1;
	//cin>>t;
	while(t--) solve();
	return 0;
}