Let \(d\) be the shortest distance between the two dice, where \(0\leq d \leq 50\). In a single round, the change in distance \(\Delta d\in \{-2, -1, 0, 1, 2\}\) occurs with probabilities \(1/36\), \(8/36\), \(18/36\), \(8/36\), and \(1/36\), respectively.
The expected number of remaining turns to reach the absorbing state, \(E(d)\), satisfies the recurrence: \[E(d) = 1 + \frac{1}{36}E(d-2)+\frac{8}{36}E(d-1)+\frac{18}{36}E(d)+\frac{8}{36}E(d+1)+\frac{1}{36}E(d+2).\] Multiplying by 36 and rearranging the terms yields the linear equation: \[ E(d-2)+8E(d-1)-18E(d)+8E(d+1)+E(d+2)=-36.\] Applying the absorbing state \(E(0) = 0\). The final expected value, \(E(50)\), can then be computed by reducing the matrix via standard Gaussian elimination.