The number $15$ is written on a blackboard. A move consists of erasing the number $x$ and replacing it with $x+y$ where $y$ is a randomly chosen number between $1$ and $5$ (inclusive). The game ends when the number on the blackboard exceeds $51$. Which number is most likely to be on the blackboard at the end of the game?

$\text{(A) }52\qquad\text{(B) }53\qquad\text{(C) }54\qquad\text{(D) }55\qquad\text{(E) }56$
1. We start with the number \(15\) on the blackboard.
2. Each move consists of erasing the current number \(x\) and replacing it with \(x + y\), where \(y\) is a randomly chosen number between \(1\) and \(5\) (inclusive).
3. The expected value of \(y\) can be calculated as follows:
   \[
   \mathbb{E}(y) = \frac{1 + 2 + 3 + 4 + 5}{5} = \frac{15}{5} = 3
   \]
4. We need to determine the number of moves required for the number on the blackboard to exceed \(51\). Let \(n\) be the number of moves.
5. After \(n\) moves, the expected value of the number on the blackboard is:
   \[
   15 + 3n
   \]
6. We need this value to exceed \(51\):
   \[
   15 + 3n > 51
   \]
7. Solving for \(n\):
   \[
   3n > 51 - 15
   \]
   \[
   3n > 36
   \]
   \[
   n > 12
   \]
   Therefore, \(n\) must be at least \(13\) moves.
8. On the \(13\)th move, the number on the blackboard will be:
   \[
   15 + 3 \times 13 = 15 + 39 = 54
   \]
9. Since the number \(54\) exceeds \(51\), the game will end.

The final answer is \(\boxed{54}\).