从一系列数字产生概率

Question

If there's a number series as so: 11.5, 6.7, 3.2, and 5.11

How does one convert these numbers into probabilities so that the sum of the resultant probabilities is 1?

Thank you.

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What if the number series also includes negative numbers: -1.2, -100.34, 3.67, and 2.1 ?

Yes, they are weights associated with 4 possible classes for an instance.

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ok. Here's my solution. Feel free to suggest improvements if any.

1) shift the range of numbers to be between 1 to any n(I choose 100) by method .

2) use the solution listed in answers.

The reason for choosing the lower bound as 1 in step 1) is to not loose the min value after applying the range conversion formula.

Solved Example:

Say there 2 instances that can take on 2 possible class values with weights as shown below.
Instance1: -11.0  -2.0
Instance2: 4.0    52.0

old_max = 52.0, old_min = -11.0, new_max = 100, and new_min = 1

After applying step1), the weights are now in range 1 to 100.
Instance1: 1       15.1
Instance2: 24.5    100    

On applying step2), the following probabilities are obtained.
Instance1: 0.0708   0.937
Instance2: 0.19     0.803

Answer 1

You can divide each number by the sum:

double arr[] = {11.5, 6.7, 3.2, 5.11};
double total = 11.5 + 6.7 + 3.2 + 5.11;
for (int i = 0; i < arr.length; i++) {
    System.out.println(arr[i] / total);
}

This works because the sum divided by the sum is 1.

Answer 2

A weighted sum?

x1= 11.5, x2 = 6.7, x3 = 3.2, x4= 5.11
sum = x1+x2+x3+x4
p(x1) = x1/sum
p(x4) = x4/sum
....
p(total) = p(x1) + p(x2) + p(x3) + p(x4) = 1