5. As shown in Figure 1, in a certain city, there is a neat grid of roads between two places, $M$ and $N$. $A_{1}, A_{2}, A_{3}, A_{4}$ are four intersections located on one diagonal of the road grid. Now, two people, A and B, start from $M$ and $N$ respectively and randomly choose a shortest path along the streets, walking at the same speed towards $N$ and $M$ respectively until they reach $N$ and $M$. The probability that A and B will meet is $\qquad$
5. $\frac{41}{100}$.

If A and B walk along the shortest path, they can only meet at $A_{1}, A_{2}, A_{3}, A_{4}$. If A and B meet at $A_{2}$, A has $\mathrm{C}_{3}^{1}$ ways to walk from $M$ to $A_{2}$, and $\mathrm{C}_{3}^{1}$ ways to walk from $A_{2}$ to $N$, making a total of $\mathrm{C}_{3}^{1} \mathrm{C}_{3}^{1}$ ways; B also has $\mathrm{C}_{3}^{1} \mathrm{C}_{3}^{1}$ ways to walk from $N$ via $A_{2}$ to $M$; thus, they have $\left(\mathrm{C}_{3}^{1}\right)^{4}$ different ways to meet at $A_{2}$.

Similarly, they have $\left(\mathrm{C}_{3}^{i-1}\right)^{4}$ different ways to meet at $A_{i}(i=1,2,3,4)$. And
$$
\left(\mathrm{C}_{3}^{0}\right)^{4}+\left(\mathrm{C}_{3}^{1}\right)^{4}+\left(\mathrm{C}_{3}^{2}\right)^{4}+\left(\mathrm{C}_{3}^{3}\right)^{4}=164 .
$$

Therefore, the probability $P$ that A and B meet is $P=\frac{164}{400}=\frac{41}{100}$.