, 2019

Solution:
WLOG, $AB$ and $CD$ are horizontal line segments and $BC$ and $DA$ are vertical. Then observe that we can consider the reflections over vertical lines separately from those over horizontal lines, as each reflection over a vertical line moves $P_i$ horizontally to point $P_{i+1}$, and vice versa. Now consider only the reflections over horizontal segments $AB$ and $CD$. Note that it is impossible for $P_8$ to be in the same vertical location as $P$ if there are an odd number of these reflections. Then we consider the reflections in pairs: let $w$ denote reflecting twice over $AB$, let $x$ denote reflecting over $AB$ and then $CD$, let $y$ denote reflecting over $CD$ and then $AB$, and let $z$ denote reflecting twice over $CD$. Note that both $w$ and $z$ preserve the position of our point. Also note that in order to end at the same vertical location as $P$, we must have an equal number of $x$'s and $y$'s. Now we count the number of sequences of length at most 4 with this property:
- Case 1: Length 0
There is just the empty sequence here, so 1.
- Case 2: Length 1
There are just the sequences $w$ and $z$, so 2.
- Case 3: Length 2
We may either have an $x$ and a $y$ or two characters that are either $w$ or $z$. There are 2 sequences of the former type and 4 of the latter, for 6 total.
- Case 4: Length 3
There are 12 sequences with an $x$, a $y$, and either a $w$ or a $z$, and 8 sequences of only $w$'s and $z$'s, for 12 total.
- Case 5: Length 4
There are 6 sequences of $2x$'s and $2y$'s, 48 with one of each and two terms that are either $w$ or $z$, and 16 of just $w$'s and $z$'s, for a total of 70.
Now let the number of such sequences of length $k$ be $a_k$ (so $a_3 = 20$). Note that these counts work also if we consider only reflections over vertical line segments $BC$ and $AD$. Now to finish, we only need to count the number of ways to combine 2 valid sequences of total length 4. This is
$$\sum_{i=0}^{4} a_i a_{4-i} \binom{8}{2i}$$
as there are $a_i$ sequences of reflections over $AB$ and $CD$, $a_{4-i}$ sequences of reflections over $BC$ and $AD$ such that there are 8 total reflections, and $\binom{8}{2i}$ ways to choose which of the 8 reflections will be over $AB$ or $CD$. We compute that this sum is $1 \cdot 70 \cdot 1 + 2 \cdot 20 \cdot 28 + 6 \cdot 6 \cdot 70 + 20 \cdot 2 \cdot 28 + 70 \cdot 1 \cdot 1 = 4900$ total sequences of reflections that place $P_8$ at $P$. There are of course $4^8 = 65536$ total sequences of 8 reflections, each chosen uniformly at random, so our answer is $\frac{4900}{65536} = \frac{1225}{16384}$.

Solution:
Suppose that $P_0$ is the origin and the four lines are $x = \pm 0.5$ and $y = \pm 0.5$. We consider a permutation of the lattice points on the coordinate plane, where all points with even $x$-coordinates are reflected across the $y$-axis and all points with even $y$-coordinates are reflected across the $x$-axis, so that the $x$- and $y$-coordinates are both rearranged in the following order:
$$\ldots, 4, -3, 2, -1, 0, 1, -2, 3, -4, \ldots$$
It is not difficult to see that a reflection across one of the lines corresponds to changing one of the coordinates from one number to either the previous number or the next number. Therefore, after the permutation, the question is equivalent to asking for the number of lattice walks of length 8 that returns to the origin. For such a lattice walk to return to origin, there needs to be the same number of up and down moves, and the same number of left and right moves. This condition is equivalent to having four moves that are left or up (LU), and four moves that are right or up (RU). Moreover, knowing whether a move is LU and whether it is RU uniquely determines what the move is, so it suffices to designate four LU moves and four RU moves, giving $\binom{8}{4}^2 = 4900$ possible walks. Hence the probability is $\frac{4900}{4^8} = \frac{1225}{16384}$.