In a building with $n$ floors and two staircases leading from the first to the last floor, there is a door on each intermediate landing between any two floors (you can go from the staircase to the floor even if the door is locked). The caretaker decided that too many open doors is bad, and locked exactly half of the doors, choosing them randomly. What is the probability that one can climb from the first floor to the last, passing only through open doors?
Doors leading from one floor to the next will be called the doors of floor $k(k=1,2, \ldots, n-1)$.

If both doors of a floor are closed (on both staircases), it is impossible to get from floor $k$ to floor $k+1$, and consequently, it is impossible to get from the first to the last floor. If at least one door is open on each floor, passage is possible. Therefore, the event $A=\{$ passage is possible $\}$ occurs if and only if there are no two doors simultaneously closed on the same level. However, exactly half of the doors are closed. Thus, when the event $A$ occurs, if a door on the left staircase is open, the corresponding door on the right staircase is closed. There are a total of $2^{n-1}$ such combinations.

The total number of ways to close half of the $2(n-1)$ doors is $C_{2 n-2}^{n-1}$. Therefore, $\mathrm{P}(A)=\frac{2^{n-1}}{C_{2 n-2}^{n-1}}$.

## Answer

$\frac{2^{n-1}}{C_{2 n-2}^{n-1}}$

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