The Setup

You roll a fair 6-sided die repeatedly. The rules are:

  • After each roll, you earn an amount equal to the face value shown.
  • If you roll a 4, 5, or 6, you get to roll again.
  • If you roll a 1, 2, or 3, the game ends immediately.

Question:

What is the expected total payout from this process?

Let the Expected Total Be \( E \)

We compute \( E \) using a recursive expectation. Consider the expected value of the first roll:

\[ E = \sum_{i=1}^{6} \frac{1}{6} \left[\text{earn } i + \text{continue or stop}\right] \]

Break it into two parts:

  • For rolls 1, 2, 3: the game ends after this roll
  • For rolls 4, 5, 6: you earn the value and then continue with expectation \( E \)

Compute Explicitly:

\[ E = \frac{1}{6}(1 + 2 + 3) + \frac{1}{6}[(4 + E) + (5 + E) + (6 + E)] \]

\[ E = \frac{6}{6} + \frac{1}{6}(15 + 3E) = 1 + \frac{15}{6} + \frac{3E}{6} = \frac{21}{6} + \frac{1}{2}E \]

Now solve for \( E \):

\[ E = \frac{21}{6} + \frac{1}{2}E \Rightarrow \frac{1}{2}E = \frac{21}{6} \Rightarrow E = \frac{42}{6} = \boxed{7} \]

Final Answer

\[ \boxed{7} \]

So the expected total payout under this stopping rule is 7 dollars.

Reference