The Dynamic Dice Game

A casino lets you roll a die as many times as you like. Each roll showing $1$ to $5$ adds that many dollars to your bank, but a $6$ wipes out everything you have banked and ends the game. After any roll you may instead stop and keep your bank. How much should a risk-neutral player pay to play?

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#When to stop

Say your bank holds $m$ . Rolling once more lands $1$ through $5$ with probability $\tfrac{1}{6}$ each, adding that to the bank, and a $6$ with probability $\tfrac{1}{6}$ that drops it to zero. The expected bank right after one more roll is

\frac{1}{6}\sum_{f=1}^{5}(m + f) + \frac{1}{6}\cdot 0 = \frac{5m + 15}{6}. \tag{1}

Rolling beats stopping exactly when $\tfrac{5m+15}{6} > m$ , that is when $m < 15$ . A larger bank only makes another roll less attractive, so the stopping region is closed upward and this one-step rule is already optimal. Roll while the bank is under $15$ and stop once it reaches $15$ .

Rolling again is worth an expected (5m+15)/6, the stop value is the bank m itself, and the two cross at 15. Below 15 another roll pays; above 15 it does not, so you roll until the bank reaches 15.

#The value of the game

Price the game as the expected payoff from an empty bank under that rule. Writing $V(m)$ for the value at bank $m$ , we have $V(m) = m$ for $m \ge 15$ and

V(m) = \frac{1}{6}\sum_{f=1}^{5} V(m + f), \qquad m < 15. \tag{2}

Working down from $15$ to $0$ unwinds the recursion to

V(0) \approx 6.15. \tag{3}

A risk-neutral player should pay about $6.15. The threshold sits well above that price because the rare $6$ punishes greed, since every roll past a bank of $15$ risks more than it expects to add.