Interesting question: there is a possible interval distribution of 1

If 10 non-negative values are x1, x2 ,..., x10 meets the requirements of X1 + X2 + X3 +... + x10 = 1, are the 10 numbers evenly distributed between [0, 1? Apparently not. To illustrate this, the best way is to limit the distribution -- we can divide the [0, 1] range into several cells, it also shows that these 10 numbers cannot be evenly distributed in these intervals. For example, divide [0.25] into four sections: [0, 0.25), [0.5, 0.5), [0.75, 0.75), [, 1: if 10 numbers fall into [0, 0.25), their sum may be 1. If 10 numbers fall into [0.75, 1, obviously, their sum cannot be 1. An interesting question arises from this: Consider the distribution of 10 numbers of 4 ^ 10, and how many of them are possible to be 1?

 
 
Obviously, the sum of the right endpoints of the 10 intervals must be greater than 1. Therefore, as long as the sum of the Left endpoints of the 10 intervals does not exceed 1, the sum of the numbers selected in these intervals may be 1. [0, 0.25), [0.25, 0.5), [0.5, 0.75), [0.75, 1] are numbered 0, 1, 2, 3, since their left endpoints are 0/4, 1/4, 2/4, 3/4, respectively, the sum of the Left endpoint cannot exceed 1, which is equivalent to the sum of the numbers of the 10 intervals cannot exceed 4. The 10 non-negative integers not greater than 4 correspond to the order of the 4 small balls and the 10 partition. They have a total of 1001 cases of C (14, 4) =. But in these 1001 cases, (4, 0, 0,..., 0), (0, 4, 0,..., 0 ),......, (0, 0, 0,..., 4) These 10 cases should be excluded, because the range numbers are only 0 to 3. Therefore, only 991 of the 10 numbers of 4 ^ 10 distributions meet their sum, which may be 1.

Next, we naturally thought of the question: if we divide [0, 1] into four different sections, will the answer still be 991? It seems that this is not the case. It is easy to think that because smaller numbers are more likely to appear, more than 991 types of distributions meeting requirements will appear if the segments are more intensive. In the same way, we divide [0, 1] into four sections. How many distributions can meet the requirements?

 
 
The answer is no more than 4 ^ 10-3 ^ 10, because no matter how the interval is divided, at least 3 ^ 10 distributions do not meet the requirements. You only need to note that if a certain distribution meets the requirements, the sum of the Left endpoints of all intervals must be smaller than 1. Then, move all 10 intervals left, the sum of the Left endpoint in the original distribution is the sum of the right endpoint in the new distribution. Because it is smaller than 1, the new distribution must not meet the requirements. Now, if 10 numbers can only fall into the intervals 1, 2, and 3, the total number of distributions is 3 ^ 10. For each of these distributions, either it is not required, or the new distribution obtained after moving all the intervals left is not required. Therefore, at least 3 ^ 10 distributions do not meet the requirements, that is, a maximum of 4 ^ 10-3 ^ 10 distributions are available.
This upper bound can be reached. For example, the four intervals are [0, 0.01), [0.01, 0.02), [0.02, 0.03), [0.03, 1], the distribution that meets the requirements is exactly 4 ^ 10-3 ^ 10 = 989527.

Source: Last month, IBM ponder this