A question about probability (conflicting solutions) There is a square of side $n$ units. Join the diagonals. Now the square is divided into 4 regions of equal area. Each of them is coloured differently. Given 2 points that can lie within any of the four regions, what is the probability that two points are in the same coloured region?


The problem:

I and my friend get two different solutions using two different lines of reasoning. It has to do with the order of the points being placed:

Solution 1 (mine): The order in which the points are placed does not matter. The number of ways 2 identical objects can be placed in 4 distinct regions is 10 (stars and bars). Out of those, there are 4 cases which satisfy the condition. Hence the probability is $2/5$.

Solution 2 (my friend's): The order of the points being placed does matter. That means there are 16 ways to place the points (4 orientations × 4 regions). Out of those, 4 cases satisfy the condition. Hence the probability is $1/4$.

Now of course, one line of reasoning must be wrong here. Which one is wrong? An explanation is appreciated.
The first solution is wrong and the second is correct. Whichever region the first point lies in, there is a $\frac14$ chance the second point will be in that same region.

The first solution assumes that the probabilities of having the two points in one given region and those of having them in two separate, given regions are the same, when in fact the second is twice the first.
The order is important. Consider the four regions and possible allocations of two points (label $1$ and $2$):
$$\begin{align}&amp;(1,2)(×)(×)(×); (1)(2)(×)(×); (1)(×)(2)(×); (1)(×)(×)(2); \\
&amp;(2)(1)(×)(×); (×)(1,2)(×)(×); (×)(1)(2)(×); (×)(1)(×)(2); \\
&amp;(2)(×)(1)(×); (×)(2)(1)(×); (×)(×)(1,2)(×); (×)(×)(1)(2); \\
&amp;(2)(×)(×)(1); (×)(2)(×)(1); (×)(×)(2)(1); (×)(×)(×)(1,2). \end{align}$$
There are four equally likely outcomes with two points in a common region, therefore:
$$P=\frac{4}{16}=\frac14.$$