A Dice Game

You roll a fair six-sided die and are paid its face value in dollars. If the face is a 4, 5, or 6 you roll again and are paid again; if it is a 1, 2, or 3 the game stops. Your total payoff is the sum of every face you roll. What payoff should you expect?

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#Set up the recursion

Each roll pays its face value, worth an average of $\tfrac{1+2+3+4+5+6}{6} = 3.5$ regardless of which number lands. With probability $\tfrac{1}{2}$ that number is a 4, 5, or 6 and the game restarts as an independent copy with the same expected payoff $E$ . Collecting the first roll and the possible restart,

E = 3.5 + \tfrac{1}{2}E \quad\Longrightarrow\quad \tfrac{1}{2}E = 3.5 \quad\Longrightarrow\quad E = 7. \tag{1}

#Or add up the rolls

The $k$ -th roll happens only if the previous $k-1$ all continued, which has probability $\left(\tfrac{1}{2}\right)^{k-1}$ , and contributes an expected $3.5$ when it does. Linearity sums a geometric series,

E = \sum_{k=1}^{\infty} 3.5 \left(\tfrac{1}{2}\right)^{k-1} = 3.5 \cdot \frac{1}{1 - \tfrac{1}{2}} = 3.5 \cdot 2 = 7. \tag{2}

Every roll is worth an expected 3.5, but the k-th roll only happens with probability (1/2)^(k-1). The expected contributions halve each step and sum to 7.

#Read it off

The expected payoff is $7$ dollars. The continuation rule doubles a single roll's expected $3.5$ , since on average you take two rolls before a low number ends the game.