Probability the Last Die Roll Is Two When Summing Past 100
You throw a fair six-sided die repeatedly and keep a running total. The moment the total exceeds $100$, you stop.
What is the probability that the final throw -- the one that pushed the total past $100$ -- was a $2$?
Hints
- Think about which running totals, just before the final throw, would allow a roll of $2$ to push the sum past $100$.
- The Key Renewal Theorem tells you how often a random walk with mean step size $\mu$ visits any particular integer -- for large integers, that probability converges to $1/\mu$.
- You need the total to be at $99$ or $100$ just before the final throw. Each of those positions is visited with probability approximately $2/7$, and you roll a $2$ with probability $1/6$. Combine these.
Worked Solution
How to Think About It: The total is climbing by $1$-$6$ each throw, so just before the crossing throw the total sits at one of the values $95, 96, \ldots, 100$. For the last throw to be a $2$, you need the total to be at $99$ or $100$ right before that throw (because only then does adding $2$ push you past $100$). So the question reduces to: how likely is the random walk to visit each of those positions? For large thresholds the renewal theorem tells us that each position is visited with roughly the same probability $1/\mu$, which makes the answer proportional to $k$.
Quick Estimate: The mean of a fair die is $\mu = 3.5$, so the renewal theorem says each integer is visited with probability approximately $1/3.5 = 2/7 \approx 0.286$. For the last throw to be $2$, we need the running total to be $99$ or $100$ just before throwing. Each of those two states is visited with probability about $2/7$, and from each the probability of rolling a $2$ is $1/6$. So $P \approx 2 \times \frac{2}{7} \times \frac{1}{6} = \frac{4}{42} = \frac{2}{21} \approx 0.095$.
Approach: Use the Key Renewal Theorem for a discrete random walk with step distribution $\{1,2,3,4,5,6\}$, each with probability $1/6$.
Formal Solution:
Let $S_n = X_1 + X_2 + \cdots + X_n$ where each $X_i$ is a fair die roll. We stop at the first $n$ with $S_n > 100$. The last throw is $X_n$.
Define $p_j = P(S_m = j \text{ for some } m \geq 0)$ as the probability the random walk ever visits position $j$. By the Key Renewal Theorem, since the step distribution has $\gcd$ of its support equal to $1$ (aperiodic) and finite mean $\mu = 7/2$:
$$p_j \to \frac{1}{\mu} = \frac{2}{7} \quad \text{as } j \to \infty$$
For the last throw to equal $k$, we need $S_{n-1} \leq 100$ and $S_{n-1} + k > 100$, which means $S_{n-1} \in \{101 - k, \, 102 - k, \, \ldots, \, 100\}$. That is $k$ possible positions.
The probability is:
$$P(\text{last throw} = k) = \sum_{j=101-k}^{100} p_j \cdot \frac{1}{6}$$
For $N = 100$ (large enough for the renewal approximation to be excellent), each $p_j \approx 2/7$, so:
$$P(\text{last throw} = k) \approx k \cdot \frac{2}{7} \cdot \frac{1}{6} = \frac{k}{21}$$
Sanity check: $\sum_{k=1}^{6} \frac{k}{21} = \frac{1+2+3+4+5+6}{21} = \frac{21}{21} = 1$. The probabilities sum to $1$.
For $k = 2$:
$$P(\text{last throw} = 2) = \frac{2}{21}$$
Note that larger die values are more likely to be the crossing throw, which makes sense: a $6$ can push you over from any of six positions ($95$-$100$), while a $1$ can only do so from position $100$ itself.
Answer: $P(\text{last throw} = 2) = \dfrac{2}{21} \approx 0.0952$.
Intuition
This problem is a clean application of the Key Renewal Theorem from probability. When a random walk with positive integer steps crosses a large threshold, the probability that the crossing step has value $k$ is proportional to $k$ -- larger steps are more likely to be the one that pushes you over because they can reach the boundary from farther away. Specifically, a step of size $k$ works from any of $k$ different positions near the boundary, and by the renewal theorem each of those positions is visited equally often (with probability $1/\mu$). So the probability is just $k/(\text{number of faces} \times \mu) = k/21$.
This shows up in real quantitative work whenever you model threshold crossings -- for instance, in limit order books where a sequence of trades pushes the price past a level, or in risk management where cumulative losses breach a VaR threshold. The general lesson is that crossing events are biased toward large increments, not uniform across step sizes. A common mistake is assuming each die face is equally likely to be the last throw (giving $1/6$), but that ignores the geometry of the boundary.