Oh no! While playing Mario Party, Theo has landed inside the Bowser Zone. If his next roll is between $1$ and $5$ inclusive, Bowser will shoot his ``Zero Flame" that sets a player's coin and star counts to zero. Fortunately, Theo has a double dice block, which lets him roll two fair $10$-sided dice labeled $1$-$10$ and take the sum of the rolls as his "roll". If he uses his double dice block, what is the probability he escapes the Bowser zone without losing his coins and stars?

[i]Proposed by Connor Gordon[/i]
To solve this problem, we need to determine the probability that the sum of two rolls of a fair 10-sided die is not between 1 and 5 inclusive. 

1. **Calculate the total number of possible outcomes:**
   Each die has 10 faces, so when rolling two dice, the total number of possible outcomes is:
   \[
   10 \times 10 = 100
   \]

2. **Identify the outcomes where the sum is between 1 and 5 inclusive:**
   We need to find the pairs \((a, b)\) such that \(1 \leq a + b \leq 5\). Let's list these pairs:
   - For \(a + b = 2\): \((1, 1)\)
   - For \(a + b = 3\): \((1, 2), (2, 1)\)
   - For \(a + b = 4\): \((1, 3), (2, 2), (3, 1)\)
   - For \(a + b = 5\): \((1, 4), (2, 3), (3, 2), (4, 1)\)

   Summarizing, the pairs are:
   \[
   (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)
   \]
   There are 10 such pairs.

3. **Calculate the probability of the sum being between 1 and 5 inclusive:**
   The number of favorable outcomes is 10, and the total number of possible outcomes is 100. Therefore, the probability is:
   \[
   P(\text{sum between 1 and 5}) = \frac{10}{100} = \frac{1}{10}
   \]

4. **Calculate the probability of escaping the Bowser zone:**
   The probability of escaping the Bowser zone is the complement of the probability of landing in the Bowser zone:
   \[
   P(\text{escape}) = 1 - P(\text{sum between 1 and 5}) = 1 - \frac{1}{10} = \frac{9}{10}
   \]

The final answer is \(\boxed{\frac{9}{10}}\).