output_description: Print the probability of having more heads than tails. The output is
  considered correct when the absolute error is not greater than 10^{-9}.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
      p_1 p_2 \ldots p_N
attempt: # -*- coding: utf-8 -*-
  
  #############
  # Libraries #
  #############
  
  import sys
  
  input = sys.stdin.readline
  
  import math
  
  # from math import gcd
  import bisect
  import heapq
  from collections import defaultdict
  from collections import deque
  from collections import Counter
  from functools import lru_cache
  
  #############
  # Constants #
  #############
  
  MOD = 10**9 + 7
  INF = float("inf")
  AZ = "abcdefghijklmnopqrstuvwxyz"
  
  #############
  # Functions #
  #############
  
  
  ######INPUT######
  def I():
      return int(input().strip())
  
  
  def S():
      return input().strip()
  
  
  def IL():
      return list(map(int, input().split()))
  
  
  def SL():
      return list(map(str, input().split()))
  
  
  def ILs(n):
      return list(int(input()) for _ in range(n))
  
  
  def SLs(n):
      return list(input().strip() for _ in range(n))
  
  
  def ILL(n):
      return [list(map(int, input().split())) for _ in range(n)]
  
  
  def SLL(n):
      return [list(map(str, input().split())) for _ in range(n)]
  
  
  ######OUTPUT######
  def P(arg):
      print(arg)
      return
  
  
  def Y():
      print("Yes")
      return
  
  
  def N():
      print("No")
      return
  
  
  def E():
      exit()
  
  
  def PE(arg):
      print(arg)
      exit()
  
  
  def YE():
      print("Yes")
      exit()
  
  
  def NE():
      print("No")
      exit()
  
  
  #####Shorten#####
  def DD(arg):
      return defaultdict(arg)
  
  
  #####Inverse#####
  def inv(n):
      return pow(n, MOD - 2, MOD)
  
  
  ######Combination######
  kaijo_memo = []
  
  
  def kaijo(n):
      if len(kaijo_memo) > n:
          return kaijo_memo[n]
      if len(kaijo_memo) == 0:
          kaijo_memo.append(1)
      while len(kaijo_memo) <= n:
          kaijo_memo.append(kaijo_memo[-1] * len(kaijo_memo) % MOD)
      return kaijo_memo[n]
  
  
  gyaku_kaijo_memo = []
  
  
  def gyaku_kaijo(n):
      if len(gyaku_kaijo_memo) > n:
          return gyaku_kaijo_memo[n]
      if len(gyaku_kaijo_memo) == 0:
          gyaku_kaijo_memo.append(1)
      while len(gyaku_kaijo_memo) <= n:
          gyaku_kaijo_memo.append(
              gyaku_kaijo_memo[-1] * pow(len(gyaku_kaijo_memo), MOD - 2, MOD) % MOD
          )
      return gyaku_kaijo_memo[n]
  
  
  def nCr(n, r):
      if n == r:
          return 1
      if n < r or r < 0:
          return 0
      ret = 1
      ret = ret * kaijo(n) % MOD
      ret = ret * gyaku_kaijo(r) % MOD
      ret = ret * gyaku_kaijo(n - r) % MOD
      return ret
  
  
  ######Factorization######
  def factorization(n):
      arr = []
      temp = n
      for i in range(2, int(-(-(n**0.5) // 1)) + 1):
          if temp % i == 0:
              cnt = 0
              while temp % i == 0:
                  cnt += 1
                  temp //= i
              arr.append([i, cnt])
      if temp != 1:
          arr.append([temp, 1])
      if arr == []:
          arr.append([n, 1])
      return arr
  
  
  #####MakeDivisors######
  def make_divisors(n):
      divisors = []
      for i in range(1, int(n**0.5) + 1):
          if n % i == 0:
              divisors.append(i)
              if i != n // i:
                  divisors.append(n // i)
      return divisors
  
  
  #####MakePrimes######
  def make_primes(N):
      max = int(math.sqrt(N))
      seachList = [i for i in range(2, N + 1)]
      primeNum = []
      while seachList[0] <= max:
          primeNum.append(seachList[0])
          tmp = seachList[0]
          seachList = [i for i in seachList if i % tmp != 0]
      primeNum.extend(seachList)
      return primeNum
  
  
  #####GCD#####
  def gcd(a, b):
      while b:
          a, b = b, a % b
      return a
  
  
  #####LCM#####
  def lcm(a, b):
      return a * b // gcd(a, b)
  
  
  #####BitCount#####
  def count_bit(n):
      count = 0
      while n:
          n &= n - 1
          count += 1
      return count
  
  
  #####ChangeBase#####
  def base_10_to_n(X, n):
      if X // n:
          return base_10_to_n(X // n, n) + [X % n]
      return [X % n]
  
  
  def base_n_to_10(X, n):
      return sum(int(str(X)[-i - 1]) * n**i for i in range(len(str(X))))
  
  
  def base_10_to_n_without_0(X, n):
      X -= 1
      if X // n:
          return base_10_to_n_without_0(X // n, n) + [X % n]
      return [X % n]
  
  
  #####IntLog#####
  def int_log(n, a):
      count = 0
      while n >= a:
          n //= a
          count += 1
      return count
  
  
  #############
  # Main Code #
  #############
  
  N = I()
  P = [0] + list(map(float, input().split()))
  
  dp = [[0 for j in range(N + 1)] for i in range(N + 1)]
  dp[0][0] = 1
  
  
  for i in range(1, N + 1):
      for j in range(i + 1):
          if j:
              dp[i][j] = dp[i - 1][j] * (1 - P[i]) + dp[i - 1][j - 1] * P[i]
          else:
              dp[i][j] = dp[i - 1][j] * (1 - P[i])
  print(sum(dp[-1][-(-N // 2) :]))
  
problem_description: Statement
  
  Let N be a positive odd number.
  
  There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N),
  when Coin i is tossed, it comes up heads with probability p_i and tails with
  probability 1 - p_i.
  
  Taro has tossed all the N coins. Find the probability of having more heads
  than tails.
samples: [{"input": "3\n    0.30 0.60 0.80", "output": "0.612\n    \n\nThe probability of each case where we have more heads than tails is as\nfollows:\n\n  * The probability of having (Coin 1, Coin 2, Coin 3) = (Head, Head, Head) is 0.3 \u00d7 0.6 \u00d7 0.8 = 0.144;\n  * The probability of having (Coin 1, Coin 2, Coin 3) = (Tail, Head, Head) is 0.7 \u00d7 0.6 \u00d7 0.8 = 0.336;\n  * The probability of having (Coin 1, Coin 2, Coin 3) = (Head, Tail, Head) is 0.3 \u00d7 0.4 \u00d7 0.8 = 0.096;\n  * The probability of having (Coin 1, Coin 2, Coin 3) = (Head, Head, Tail) is 0.3 \u00d7 0.6 \u00d7 0.2 = 0.036.\n\nThus, the probability of having more heads than tails is 0.144 + 0.336 + 0.096\n+ 0.036 = 0.612.\n\n* * *"}, {"input": "1\n    0.50", "output": "0.5\n    \n\nOutputs such as `0.500`, `0.500000001` and `0.499999999` are also considered\ncorrect.\n\n* * *"}, {"input": "5\n    0.42 0.01 0.42 0.99 0.42", "output": "0.3821815872"}]