A coin is tossed $10$ times. Compute the probability that two heads will turn up in succession somewhere in the sequence of throws.
1. **Define the problem and the recursive approach:**
   We need to compute the probability that two heads will turn up in succession somewhere in a sequence of 10 coin tosses. To do this, we first count the number of sequences in which two heads are NOT in succession for the general case when a coin is tossed \( n \) times. Let \( g(n) \) be the number of such sequences.

2. **Base cases:**
   - For \( n = 1 \), the sequences are \( H \) and \( T \). Thus, \( g(1) = 2 \).
   - For \( n = 2 \), the sequences are \( HT, TH, TT \). Thus, \( g(2) = 3 \).

3. **Recursive relation:**
   For \( n \geq 2 \), any sequence of length \( n \) that does not have two heads in succession can either:
   - Begin with a \( T \) (followed by any sequence of length \( n-1 \) that does not have two heads in succession), contributing \( g(n-1) \) sequences.
   - Begin with \( HT \) (followed by any sequence of length \( n-2 \) that does not have two heads in succession), contributing \( g(n-2) \) sequences.
   
   Therefore, the recursive relation is:
   \[
   g(n) = g(n-1) + g(n-2)
   \]

4. **Recognize the Fibonacci sequence:**
   The recursive relation \( g(n) = g(n-1) + g(n-2) \) with initial conditions \( g(1) = 2 \) and \( g(2) = 3 \) is similar to the Fibonacci sequence. Specifically, \( g(n) = F_{n+2} \), where \( F_n \) is the \( n \)-th Fibonacci number.

5. **Calculate \( g(10) \):**
   We need to find \( g(10) \), which corresponds to \( F_{12} \) in the Fibonacci sequence. The first few Fibonacci numbers are:
   \[
   F_1 = 1, \quad F_2 = 1, \quad F_3 = 2, \quad F_4 = 3, \quad F_5 = 5, \quad F_6 = 8, \quad F_7 = 13, \quad F_8 = 21, \quad F_9 = 34, \quad F_{10} = 55, \quad F_{11} = 89, \quad F_{12} = 144
   \]
   Thus, \( g(10) = F_{12} = 144 \).

6. **Total number of sequences:**
   The total number of sequences of length 10 is \( 2^{10} = 1024 \).

7. **Probability calculation:**
   The number of sequences where two heads are NOT in succession is \( g(10) = 144 \). Therefore, the number of sequences where two heads ARE in succession is:
   \[
   1024 - 144 = 880
   \]
   The probability that two heads will turn up in succession is:
   \[
   \frac{880}{1024} = 1 - \frac{144}{1024} = 1 - \frac{9}{64} = \frac{55}{64}
   \]

The final answer is \(\boxed{\frac{55}{64}}\).