Problem Statement

On average, how many times does a fair 6-sided die need to be rolled to obtain two consecutive rolls that differ by exactly 1?

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Solution

We start by analyzing the process after the first roll, which sets the “current” number.

Let:

  • \(E_i\) be the expected number of additional rolls needed starting from number \(i\) on the die.

Due to symmetry of the die:

  • \[E_1 = E_6\]
  • \[E_2 = E_5\]
  • \[E_3 = E_4\]

So we only need to compute \(E_1, E_2, E_3\).
We’ll derive a system of equations using expected value analysis.

Step 1: Transition Probabilities

From number \(i\), the next roll (value \(j\)) can be:

  • If \(abs(i - j) = 1\), the condition is met → Done in 1 roll
  • Otherwise, we spend 1 roll, and now have a new state with expected cost \(E_j\)

Let’s take the case of \(E_1\):

Possible next rolls:

  • Roll a 1 → stay at 1 → cost: \(1 + E_1\)
  • Roll a 2 → diff = 1 → done → cost: \(1\)
  • Roll a 3, 4 → cost: \(1 + E_3\)
  • Roll a 5 → cost: \(1 + E_2\)
  • Roll a 6 → cost: \(1 + E_1\)

So: \(E_1 = \frac{1}{6} \left( (1 + E_1) + (1) + (1 + E_3) + (1 + E_3) + (1 + E_2) + (1 + E_1) \right)\)

Simplify: \(E_1 = \frac{1}{6} (2 + 2E_1 + 1 + 2 + 2E_3 + 1 + E_2) = \frac{6 + 2E_1 + E_2 + 2E_3}{6}\)

Multiply both sides by 6: \(6E_1 = 6 + 2E_1 + E_2 + 2E_3 \Rightarrow 4E_1 - E_2 - 2E_3 = 6 \tag{1}\)

Step 2: Similar Equations for \(E_2\) and \(E_3\)

For \(E_2\) (symmetry with 5):

From 2, adjacent values: 1 and 3
→ done if roll is 1 or 3

\[E_2 = \frac{1}{6} \left( (1) + (1 + E_2) + (1) + (1 + E_3) + (1 + E_2) + (1 + E_1) \right)\]

Simplify:

\[E_2 = \frac{1 + 1 + E_2 + 1 + 1 + E_3 + 1 + E_2 + 1 + E_1}{6} = \frac{6 + E_1 + 2E_2 + E_3}{6} \Rightarrow 6E_2 = 6 + E_1 + 2E_2 + E_3 \Rightarrow 4E_2 - E_1 - E_3 = 6 \tag{2}\]

For \(E_3\) (symmetry with 4):

Adjacent values are 2 and 4
→ done if roll is 2 or 4

\[E_3 = \frac{1}{6} \left( (1 + E_1) + (1) + (1 + E_3) + (1) + (1 + E_2) + (1 + E_1) \right)\]

Simplify: \(E_3 = \frac{1 + E_1 + 1 + 1 + E_3 + 1 + 1 + E_2 + 1 + E_1}{6} = \frac{6 + 2E_1 + E_2 + E_3}{6} \Rightarrow 6E_3 = 6 + 2E_1 + E_2 + E_3 \Rightarrow 5E_3 - 2E_1 - E_2 = 6 \tag{3}\)

Step 3: Solve the System

We solve equations (1), (2), and (3):

  1. \[4E_1 - E_2 - 2E_3 = 6\]
  2. \[-E_1 + 4E_2 - E_3 = 6\]
  3. \[-2E_1 - E_2 + 5E_3 = 6\]

Solving gives:

  • \[E_1 = \frac{70}{17}\]
  • \[E_2 = \frac{58}{17}\]
  • \[E_3 = \frac{60}{17}\]

Step 4: Final Expected Value

Initial roll is uniform over \(1\) to \(6\), so:

\[\text{Expected rolls} = 1 + \frac{1}{6}(2E_1 + 2E_2 + 2E_3) = 1 + \frac{2}{6}(E_1 + E_2 + E_3)\]

Add values: \(E_1 + E_2 + E_3 = \frac{70 + 58 + 60}{17} = \frac{188}{17}\)

So:

\[\text{Final} = 1 + \frac{2}{6} \cdot \frac{188}{17} = 1 + \frac{188}{51} = \frac{188 + 51}{51} = \frac{239}{51}\]

The expected number of rolls is \(\boxed{\frac{239}{51}}\)

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