Problem Statement

You are given a fair 6-sided die and you roll it. You can either:

  • Stop and receive the observed value (in dollars), or
  • Roll again. But if the sum of both rolls is at least 7, you pay the value of your first roll.
    If the sum is less than 7, you receive the second roll’s value.

Assuming optimal play, what is your expected payout?

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Solution

Let’s denote the value of the first roll as \(x \in \{1, 2, 3, 4, 5, 6\}\).
For each \(x\), we’ll compute:

  • The expected payout if we stop: \(= x\)
  • The expected payout if we roll again, based on the outcome of the second roll.

We will then compare the two and decide whether to stop or roll again.

Step 1: Compute Expected Payout on Re-Roll

Suppose your first roll is \(x\). Now, depending on the second roll \(y\), we have:

  • If \(x + y \geq 7\), you pay \(x\) → Payout = \(-x\)
  • If \(x + y < 7\), you receive \(y\) → Payout = \(y\)

Let’s go case by case.

🔸 Case 1: \(x = 1\)

Here are the second roll outcomes:

\(y\) \(x + y\) Result Payout
1 2 \(< 7\) 1
2 3 \(< 7\) 2
3 4 \(< 7\) 3
4 5 \(< 7\) 4
5 6 \(< 7\) 5
6 7 \(\geq 7\) -1

Expected payout if re-rolling:

\[\mathbb{E}_{x=1} = \frac{1 + 2 + 3 + 4 + 5 + (-1)}{6} = \frac{14}{6} = \frac{7}{3} > 1\]

So, rolling again is better than stopping (which gives only 1).

🔸 Cases \(x = 2\) to \(x = 6\)

We’ll briefly summarize results:

\(x = 2\):

  • Bad (lose \(x\)) if \(y \geq 5\) (i.e. 2 values)
  • Good if \(y \leq 4\)

Expected payout:

\[\mathbb{E}_{x=2} = \frac{1 + 2 + 3 + 4 -2 -2}{6} = \frac{6}{6} = 1 < 2\]

\(x = 3\):

\[\mathbb{E}_{x=3} = \frac{1 + 2 + 3 -3 -3 -3}{6} = \frac{-3}{6} = -0.5 < 3\]

\(x = 4\):

Only 2 values (1 and 2) will give positive outcome

\[\mathbb{E}_{x=4} = \frac{1 + 2 -4 -4 -4 -4}{6} = \frac{-13}{6} \ll 4\]

\(x = 5, 6\): even worse.

Step 2: Apply Optimal Strategy

From the above:

  • If \(x = 1\): roll again
  • If \(x \geq 2\): stop

Step 3: Compute Final Expected Value

\[\mathbb{E} = \sum_{x=1}^{6} \mathbb{P}(x) \cdot \text{Expected Payout for } x\]

Using optimal decisions:

  • \(x = 1\): expected = \(\frac{7}{3}\)
  • \(x = 2\) to \(6\): expected = just \(x\)
\[\mathbb{E} = \frac{1}{6} \left( \frac{7}{3} + 2 + 3 + 4 + 5 + 6 \right) = \frac{1}{6} \left( \frac{7}{3} + 20 \right) = \frac{1}{6} \cdot \frac{67}{3} = \frac{67}{18}\]

Thus the expected payoff assuming optimal play is \(\boxed{\frac{67}{18}}\)

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