Two Thirds of a Dollar, and the Branch That Pays Nothing

The pile really does average exactly one dollar, which is why almost everyone answers one dollar and why the trap is a correct calculation of the wrong quantity. The play is worth two thirds, because the roll that ends the game pays on two of its three faces and that roll is independent of how big the pile grew.

Lambdia
August 13, 2026 · 9 min read

Roll a fair die. A 1, 2 or 3 puts a dollar on the table and you roll again. A 4 or a 5 ends the game and you keep everything on the table. A 6 ends the game and you keep nothing. What is one play worth?

Almost everyone answers one dollar, and the interesting part is that they have done the hard piece of the calculation correctly. The pile really does average exactly one dollar. The answer is $2/3$ of a dollar, and the missing third is the branch nobody writes down.

Two different dollars

There are two random quantities here and they are easy to confuse. Let $N$ be the size of the pile in dollars, meaning the number of rolls that landed low. Let $X$ be what you are actually handed at the end. Then

X = \begin{cases} N & \text{if the stopping roll is a 4 or a 5},\\[2pt] 0 & \text{if the stopping roll is a 6}.\end{cases}

(1)

The question asks for $\mathbb{E}[X]$ . The reflex computes $\mathbb{E}[N]$ . They are different numbers, and the gap is a third.

Fig. 1 — Three faces run the loop and three end it. Of the three that end it, two pay.

The pile, which really does average a dollar

Each roll continues the game with probability $3/6 = 1/2$ and ends it with the same probability, so $N$ counts the low rolls before the first high one. That gives

\Pr(N = k) = \left(\tfrac{1}{2}\right)^{k+1}, \qquad k = 0, 1, 2, \dots

(2)

Half of all plays never put anything on the table at all, because the very first roll is high. The fastest way to the average is the loop itself. Whatever happens, one roll is made. With probability $1/2$ the game is over with the pile as it stands. With probability $1/2$ you have gained a dollar and you are facing an identical game, so

\mathbb{E}[N] = \tfrac{1}{2}\cdot 0 + \tfrac{1}{2}\bigl(1 + \mathbb{E}[N]\bigr) \quad\Longrightarrow\quad \mathbb{E}[N] = 1

(3)

Summing the series in (2) directly gives the same 1, and the number of rolls averages 2. Nothing here is wrong, and it is worth saying so plainly, because the trap in this problem is a correct calculation of the wrong quantity.

The roll that ends it

Condition on the game ending. The stopping roll is a 4, a 5 or a 6, and those three faces are equally likely, so the chance of being paid is

\Pr(\text{paid}) = \frac{2/6}{3/6} = \frac{2}{3}

(4)

Now the step that makes the whole thing collapse to one multiplication. Which face stops the game is decided by a single roll, and that roll is independent of everything before it. So the identity of the stopping face carries no information about how big the pile grew:

\mathbb{E}[X] = \Pr(\text{paid}) \cdot \mathbb{E}[N] = \frac{2}{3} \cdot 1 = \frac{2}{3}

(5)

Fig. 2 — The bust does not fall harder on big piles than on small ones. That is why the factor two thirds can be pulled out in front of the average.

Why the factorisation is legitimate

Equation (5) needs $\mathbf{1}\{\text{paid}\}$ and $N$ to be independent, and they are. $N$ is determined by the rolls before the stop, and whether the stopping roll pays is determined by that roll alone, conditional on it being a stopper. Without independence the answer would need $\mathbb{E}[N \mid \text{paid}]$ instead, which is a genuinely harder object.

The same answer the long way

If the factorisation feels like a sleight of hand, sum over the size of the pile. A pile of $k$ dollars happens with probability $(1/2)^{k+1}$ , and it is paid with probability $2/3$ , so

\mathbb{E}[X] = \sum_{k \ge 1} k \left(\tfrac{1}{2}\right)^{k+1} \cdot \frac{2}{3} = \frac{2}{3}

(6)

Two routes, one number. This is a habit worth keeping in problems with a bust branch, because the two routes fail differently: the factorisation breaks if the stop is correlated with the pile, and the series breaks if the tail does not converge.

Changing the rules

Write $L$ for the number of faces that continue the game and pay a dollar, and $m$ for how many of the remaining $6 - L$ faces pay out. The loop argument gives $\mathbb{E}[N] = L/(6-L)$ , and the same factorisation gives

\mathbb{E}[X] = \frac{m}{6-L} \cdot \frac{L}{6-L}

(7)

Our game is $L = 3$ , $m = 2$ , giving $(2/3)(1) = 2/3$ . Widening the low band is worth more than it looks: with $L = 4$ , so 1 through 4 paying and rolling again, the pile averages $4/2 = 2$ dollars, and if both remaining faces pay then the play is worth the whole 2. Set $m = 0$ and the play is worth nothing however long the loop runs, which is the sanity check that equation (7) has the right shape.

Nothing above needs the die to be a die. With continuation probability $q$ and a chance $r$ of being paid at the stop, the play is worth $r \cdot q/(1-q)$ . As $q \to 1$ the pile grows without bound and so does the value, which is the mathematically respectable way of saying that a game which almost never ends is worth a lot.

Where the reasoning stops working

Two assumptions carry equation (5), and both can fail in problems that look almost identical.

The first is termination. The number of rolls is geometric with mean 2, and the chance of never stopping is zero, so no probability mass is hiding at infinity. Change the rules so the game continues on five faces out of six and the pile still has a finite average, 5 dollars, but the variance grows sharply and a simulation will look badly behaved long before the algebra does.

The second is independence, and it is the fragile one. Suppose the rule were that a 6 wipes you out only if the pile has reached three dollars or more. Now the bust probability depends on $N$ , the two thirds cannot come out front, and the honest computation is a sum over $k$ with a different weight in each term. Any problem where the payout condition mentions the accumulated total should be treated as guilty until proven independent.

Sources and further reading

The distribution in equation (2) — Geometric distribution
The loop in equation (3), stated properly — Law of total expectation
The quantity being computed — Expected value

The answer was checked three ways before publication: exact rational arithmetic on the factorised route, the closed form of the series in equation (6), and ten million seeded plays of the literal procedure, which returned a mean payoff of 0.6673, a mean pile of 1.0001 and a mean of 2.0001 rolls. The simulation also confirmed the independence step numerically: the average pile among plays that paid came back at 1.001, the same as the average pile overall.

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