def exp(p,n):
    if n == 1: return p
    x = exp(p,n//2)**2
    if n % 2 == 1: x *= p
    return x

N = 10**10
M = 4000
ans = 0
for i in range(N//2):
    flips = (i - 0 + 1) * (N-1 - i + 1) * 2 - 1
    rest = 0.5 * exp(1-2*(flips/(N*N)),M)
    ans += 0.5 + rest
    if rest <= 1e-12:
        ans += 0.5 * ((N//2 - 1) - i) # For j > i, the contribution is almost exactly 0.5
        ans *= 2 # Factor in N//2 <= j <= N-1
        break
print(ans)