What is the minimal common number probability distribution? There is a subset of natural numbers $\mathcal N=\{1,\ldots,N\}$ and there are four people who want to pick a number from that set $\mathcal N$. 
How do we calculate the four independent probability distributions  $p_1,p_2,p_3,p_4$ on $\mathcal N$ (one for each person) that gives the minimum probability of the event $A = \{\text{any two or more of the persons picking the same number}\}$? Is it when all $p_1,p_2,p_3,p_4$ are i.i.d uniform?
Edit: The persons have no communication between them. They each pick the probability distribution individually. And there is no central authority giving them the distribution. After they pic the distributions they tell it to a central authority and that authority will calculate the probability of the event $A$.
 A: To calculate the probability of two specific people choosing the same number, given their chosen distributions, we simply add up the probabilities of each pair matching, like this...
Supposing that $p_k(n)$ is the probability that person $k$ chooses $n$ (so $p_k$ is the distribution for person $k$), then the probability that they match is
$$
P_{j,k} = \sum_{i=1}^N p_j(i) p_k(i)
$$
These can then be extended, to determine the probabilities for three or four, and thus the probability that at least two match can be found by inclusion/exclusion.
However, if each person is unable to have any knowledge of any other person's distribution, then it can easily be seen that the expected probabilities (as reported by the authority) are $P_{j,k} = \frac1n$, where $n$ is the size of the subset being used. Similarly, $P_{i,j,k}=\frac1{n^2}$ and $P_{i,j,k,l} = \frac1{n^3}$. The expected probability that at least two match can then be found by inclusion/exclusion, as noted above.
A: There is a set of all possible distributions $D$. Each person has it's own probability distribution on this set (so for each distribution, there is a probability a player will choose it). Because all players are identical, these distributions are too. The distribution of the numbers a player will choose can now be calculated (by taking a weighted average of the probabilities in the distributions in $D$). Because $D$ and the distributions over $D$ are identical, so is the final distribution for each player. Thus we conclude that all players have identical distributions. Now, it is not hard to prove the identical distribution with probability $\frac 1n$ for each number is optimal.
EDIT
Suppose the probabilities for the numbers are $p_1$, $p_2$, $\dots$, $p_n$, with $\sum_{i=1}^np_i=1$. If $p_1<p_2$ (or $p_i<p_j$ for some $i$ and $j$) the probability that two (chosen) players 'collide' on $1$ or $2$ is $p_1^2+p_2^2$. (This is not the probability that two out of the five players collide, but this is sufficient to show that $\frac 1n$ is optimal.) We know that $p_1^2+p_2^2\geq 2\left(\frac {p_1+p_2}2\right)^2$ (because of the quadratic-arithmetic mean on $p_1$ and $p_2$). Thus, it is better to replace $p_1$ and $p_2$ by their average. Now, if follows (intuitively) that it is optimal to have all $p_i$ equal, and thus equal to $\frac 1n$.
