Last Roll of a Cumulative Sum Game
You roll a standard fair 6-sided die repeatedly, accumulating the sum of all rolls. You stop the moment the running total reaches or exceeds 1000. Which face value is most likely to be your final roll?
Hints
- Think about what the running sum must look like just before the final roll. For the last roll to be $k$, the pre-terminal sum has to fall in a specific range -- how wide is that range?
- For a large threshold, renewal theory says the running sum visits any particular value $s$ with probability about $1/\mu = 1/3.5 = 2/7$ -- one landing per $3.5$ integers on average. It is this visit probability that is flat (not the distribution of the pre-terminal sum itself). Use it to compare the 'capture windows' for each face value.
- The window of pre-terminal sums that triggers a final roll of $k$ has exactly $k$ elements ($\{1000 - k, \ldots, 999\}$). The total number of elements across all windows is $1 + 2 + \cdots + 6 = 21$, giving $P(\text{last roll} = k) = k/21$.
Worked Solution
How to Think About It: Ask what must be true just before the last roll. If the final roll is $k$, the running sum immediately before it sat somewhere in $\{1000-k, 1000-k+1, \ldots, 999\}$ -- a window of exactly $k$ values. So bigger faces get wider "capture windows," and the question reduces to weighing those windows. The one renewal-theory fact you need: far from the start, the running sum hits any particular integer $s$ with the same probability $1/\mu = 1/3.5 = 2/7$ (the average step is $3.5$, so the walk lands on about one integer in every $3.5$). Watch the trap, though: it is the *visit* probability that is flat. The pre-terminal sum itself is not uniform on $\{994, \ldots, 999\}$ -- sums near $999$ are likelier launch points precisely because more faces terminate from them.
Quick Estimate: The die averages $3.5$ per roll, so on the way to $1000$ the running sum lands on roughly $1000 / 3.5 \approx 286$ of the $1000$ integers -- about $2$ of every $7$. So for each $s$ near $1000$, $P(\text{sum ever hits } s) \approx 2/7$. The last roll is $k$ exactly when the walk stops on one of the $k$ values in $\{1000-k, \ldots, 999\}$ and then rolls $k$ (probability $1/6$):
$$P(\text{last roll} = k) \approx k \cdot \frac{2}{7} \cdot \frac{1}{6} = \frac{k}{21}$$
Sanity check: $\sum_{k=1}^{6} k/21 = 21/21 = 1$. So face $6$ comes in at $6/21 = 2/7 \approx 28.6\%$, six times as likely as face $1$ at $1/21 \approx 4.8\%$.
Formal Solution: Let $v(s)$ be the probability the running sum ever equals $s$. It satisfies $v(0) = 1$ and the renewal recursion
$$v(s) = \frac{1}{6} \sum_{j=1}^{\min(6, s)} v(s - j)$$
By the renewal theorem (Blackwell), for an aperiodic bounded step, $v(s) \to 1/\mu = 2/7$ as $s \to \infty$, with exponentially fast convergence. At $s \approx 1000$ the convergence is complete for all practical purposes: exact dynamic programming gives $v(994) = v(995) = \cdots = v(999) = 0.285714\ldots$, matching $2/7$ to more than $100$ decimal places.
Now decompose the event by the pre-terminal sum. The last roll equals $k$ with pre-terminal sum $s$ iff (i) the walk visits $s \le 999$, (ii) the roll taken from $s$ equals $k$, and (iii) $s + k \ge 1000$, i.e. $s \ge 1000 - k$. Conditions (i) and (ii) are independent -- the next roll does not depend on the path that reached $s$ -- and the events are disjoint across $s$ because the pre-terminal sum is unique. Hence, exactly,
$$P(\text{last roll} = k) = \sum_{s=1000-k}^{999} v(s) \cdot \frac{1}{6} = k \cdot \frac{2}{7} \cdot \frac{1}{6}$$
$$\boxed{P(\text{last roll} = k) = \frac{k}{21}, \qquad k = 1, \ldots, 6}$$
(Exact DP at threshold $1000$ agrees with $k/21$ to within $10^{-138}$; a $2{,}000{,}000$-trial simulation matches every $k/21$ to within $0.001$.) One pitfall worth naming: the pre-terminal sum $S$ itself has the triangular law $P(S = s) = (s - 993)/21$ for $994 \le s \le 999$, not a uniform one -- writing $p_k = \sum_s P(S = s) \cdot \frac{1}{6}$ would double-count the termination condition. Condition on *visiting* $s$, not on stopping there. The resulting law $k/21$ is the size-biased version of the uniform die distribution: the step that carries the walk across the fixed level $1000$ is picked with probability proportional to its length.
Answer: The most likely final roll is $\boxed{6}$, with probability $6/21 = 2/7 \approx 28.6\%$. More generally, $P(\text{last roll} = k) = k/21$: the probabilities are proportional to the face values.
Intuition
The key insight is that the last roll is not uniform over $\{1, \ldots, 6\}$ -- it is size-biased. Larger faces are more likely as final rolls because they can "absorb" a wider range of pre-terminal sums. Face 6 wins because it covers 6 consecutive pre-terminal values ($994$ through $999$), while face 1 can only be the last roll if the pre-terminal sum is exactly $999$. The resulting distribution $P(k) = k/21$ is a classic example of size-biased sampling, where an outcome's probability is proportional to its size.
This pattern shows up throughout quant work. In renewal processes -- think trade arrival times, coupon payments, or tick data -- the interval that contains a fixed observation point tends to be longer than a typical interval. The inspector paradox (also called the waiting-time paradox or bus paradox) is the continuous analog: if buses arrive at rate $\lambda$ with some variability, the gap you land in when you show up at a random time is longer in expectation than the average gap. Understanding size-biased sampling helps you avoid a subtle error: when you condition on being in some interval or window, you are not picking a typical interval -- you are picking a size-weighted one.