Let $F_n$ be the set of all bijective functions $f:\left\{1,2,\ldots,n\right\}\rightarrow\left\{1,2,\ldots,n\right\}$ that satisfy the conditions:

1. $f(k)\leq k+1$ for all $k\in\left\{1,2,\ldots,n\right\}$
2. $f(k)\neq k$ for all $k\in\left\{2,3,\ldots,n\right\}$

Find the probability that $f(1)\neq1$ for $f$ randomly chosen from $F_n$.
To solve the problem, we need to find the probability that \( f(1) \neq 1 \) for a randomly chosen bijective function \( f \) from the set \( F_n \). Let's denote the number of such functions by \( g(n) \). We will use the Fibonacci sequence to express \( g(n) \).

1. **Define the sets and initial conditions:**
   - Let \( g(n) \) be the number of bijective functions \( f: \{1, 2, \ldots, n\} \to \{1, 2, \ldots, n\} \) that satisfy the given conditions.
   - Let \( g(n)_1 \) be the number of such functions where \( f(1) = 1 \).
   - Let \( g(n)_2 \) be the number of such functions where \( f(1) = 2 \).

2. **Establish the recurrence relations:**
   - **Claim 1:** \( g(n)_2 = g(n-1) \)
     - If \( f(1) = 2 \), then for the remaining \( n-1 \) elements, the function \( f \) must map \( \{2, 3, \ldots, n\} \) to \( \{1, 3, 4, \ldots, n\} \) with the same properties. This is equivalent to finding a function from \( \{1, 2, \ldots, n-1\} \) to itself with the same properties.
     - Therefore, \( g(n)_2 = g(n-1) \).

   - **Claim 2:** \( g(n)_1 = g(n-1)_2 \)
     - If \( f(1) = 1 \), then for the remaining \( n-1 \) elements, the function \( f \) must map \( \{2, 3, \ldots, n\} \) to \( \{2, 3, 4, \ldots, n\} \) with the same properties, except that \( f(2) \neq 2 \).
     - This is equivalent to finding a function from \( \{1, 2, \ldots, n-1\} \) to itself with the same properties, where \( f(1) = 2 \).
     - Therefore, \( g(n)_1 = g(n-1)_2 = g(n-2) \).

3. **Combine the claims to form a recurrence relation:**
   - From the claims, we have:
     \[
     g(n) = g(n)_1 + g(n)_2 = g(n-2) + g(n-1)
     \]
   - This is the recurrence relation for the Fibonacci sequence.

4. **Initial conditions:**
   - For \( n = 1 \), there is only one function \( f \) that maps \( \{1\} \) to \( \{1\} \), and it does not satisfy \( f(1) \neq 1 \). Hence, \( g(1) = 1 \).
   - For \( n = 2 \), there is only one function \( f \) that maps \( \{1, 2\} \) to \( \{1, 2\} \) and satisfies the conditions, which is \( f(1) = 2 \) and \( f(2) = 1 \). Hence, \( g(2) = 1 \).

5. **Express \( g(n) \) in terms of the Fibonacci sequence:**
   - From the recurrence relation and initial conditions, we see that \( g(n) \) follows the Fibonacci sequence:
     \[
     g(n) = F_{n-1}
     \]
   - Therefore, \( g(n)_2 = g(n-1) = F_{n-2} \).

6. **Calculate the probability:**
   - The probability that \( f(1) \neq 1 \) is given by the ratio of the number of functions where \( f(1) = 2 \) to the total number of functions:
     \[
     \text{Probability} = \frac{g(n)_2}{g(n)} = \frac{F_{n-2}}{F_{n-1}}
     \]

The final answer is \(\boxed{\frac{F_{n-2}}{F_{n-1}}}\).