抛硬币正面期望_如果抛硬币，正面的数量多于反面的可能性

抛硬币正面期望

Problem statement:

问题陈述：

Let N be a positive odd number. There are N coins, numbered 1, 2 , ... , N. For each i (1≤i≤N), when Coin i is tossed, it comes up heads with probability pi and tails with probability 1-pi.

令N为正奇数。有N个硬币，编号为1，2，...，N 。对于每个i ( 1≤i≤N )，当投入硬币i时，它的出现概率为pi出现在正面，出现的概率为1-pi出现在尾部。

Find the probability of having more heads than tails if coins are tossed.

找到抛硬币后正面多于尾的概率。

Input:

输入：

The first line contains N, number of coins. The second line contains the probability of coming head in the coins.

第一行包含N个硬币数量。第二行包含进入硬币的概率。

Output:

输出：

Output the probability of having more number of heads than tails if coins are tossed.

如果抛硬币，则输出正面比背面多的概率。

Example with explanation:

带有说明的示例：

    Input: 
N=3
[0.30, 0.60, 0.80]
Output: 
0.612
The probability of having (Coin1,Coin2,Coin3) = (Head,Head,Head) 
is 0.3×0.6×0.8 = 0.144;
The probability of having (Coin1,Coin2,Coin3) = (Tail,Head,Head) 
is 0.7×0.6×0.8 = 0.336;
The probability of having (Coin1,Coin2,Coin3) = (Head,Tail,Head)
is 0.3×0.4×0.8 = 0.096;
The probability of having (Coin1,Coin2,Coin3) = (Head,Head,Tail)
is 0.3×0.6×0.2 = 0.036
Thus, the probability of having more heads than tails is 
0.144 + 0.336 + 0.096 + 0.036 = 0.612

Solution Approach

解决方法

Dynamic Programming

动态编程

Since the problem can be solved with recursion with time complexity of O(2^n) which is not accepted as there are other better approach to solve the problem with time complexity of O(n*n);

由于可以用O(2 ^ n)的时间复杂度递归来解决问题，这是不被接受的，因为还有其他更好的方法可以解决O(n * n)的时间复杂度问题；

Let's assume prob[i][j] to be the probability of getting j heads with first i coins. To get j heads at the i^th position, there are two possibilities:

假设prob [i] [j]是第一个i个硬币获得j个头的概率。为了使j个头位于第i ^个位置，有两种可能性：

If the number of heads till (i - 1) coins is equal to j then a tail comes at i^th. If the number of heads till (i - 1) coins is equal to (j - 1) then a head comes at i^th position.

如果直到(i-1)个硬币的正面数等于j，则^第 i ^个背面为^尾数 。如果直到(i-1)个硬币的正面数等于(j-1)，则正面在^第 i ^个位置。

The probability of occurring jth head at ith coin toss depends on the number of heads in (i-1)^th coins, if (i-1) coin have (j-1) head then jth coin occurs at i^th coin(p[i]), and if (i-1) coins have already j coins then at jth toss tails come(1-p[i]).

第i个硬币抛掷第j个硬币的概率取决于第(i-1) ^个硬币的硬币的数目，如果(i-1)个硬币具有(j-1) 个硬币的硬币，则第j个硬币出现在^第 i ^个硬币( p [ i] )，如果(i-1)个硬币已经有j个硬币，则第j个抛尾出现( 1-p [i] )。

Pseudo Code:

伪代码：

// p[] is the probability array,
// n is the size of array.
solve(p[],n):
// declare prob[n+1][n+1] array of having required head.
prob[n+1][n+1]  
// no head out of 1 coins
prob[1][0]=1-p[0] 
// one head out of one coins
prob[1][1]=p[0]	  
for(i=2;i<=n;i++)
// probability of having no heads.
prob[i][0]=prob[i-1][0]*(1-p[i])  
for(i=2;i<=n;i++)
for(j=1;j<=i;j++)
prob[i][j]=prob[i-1][j-1]*p[i-1]+prob[i-1][j]*(1-p[i-1])
// probability of having jth head can take 
// place in (i-1) coins or at (i)th coin.
sum=0
for(i=n/2+n%2;i<=n;i++)
//probability of heads>tails.
sum+=prob[n][i]   
return sum

Time complexity: In worst case O(n*n)

时间复杂度 ：在最坏的情况下O(n * n)

C++ Implementation:

C ++实现：

#include <bits/stdc++.h>
using namespace std;
typedef double ll;
int main()
{
cout << "Enter number of test cases: ";
int t;
cin >> t;
while (t--) {
cout << "Enter the number of coins: ";
int n;
cin >> n;
ll p[n];
cout << "Enter the probabilities of head: ";
for (int i = 0; i < n; i++)
cin >> p[i];
ll prob[n + 1][n + 1];
// probability of one heads with one coins
prob[1][1] = p[0];
// probability of no heads with one coins
prob[1][0] = 1 - p[0];
for (int i = 2; i <= n; i++)
// probability of no heads with i coins.
prob[i][0] = prob[i - 1][0] * (1 - p[i - 1]);
for (int i = 2; i <= n; i++) {
for (int j = 1; j <= i; j++)
// probability of head at ith coin it it occurs at(i-1)
// then just multiply with tail probability otherwise
// multiply with head probability.
prob[i][j] = prob[i - 1][j - 1] * p[i - 1] + prob[i - 1][j] * (1 - p[i - 1]);
}
ll sum = 0;
for (int i = n / 2 + n % 2; i <= n; i++)
// take those probability which have more heads than tails.
sum += prob[n][i];
cout << "The probability of heads more than the tails is: ";
cout << setprecision(10) << sum << "\n";
}
return 0;
}

Output

输出量

Enter number of test cases: 3
Enter the number of coins: 5
Enter the probabilities of head: 0.42 0.01 0.42 0.99 0.42
The probability of heads more than the tails is: 0.3821815872
Enter the number of coins: 3
Enter the probabilities of head: 0.30 0.60 0.80
The probability of heads more than the tails is: 0.612
Enter the number of coins: 1
Enter the probabilities of head: 0.50
The probability of heads more than the tails is: 0.5