We take turns rolling a fair die until the sum $S$ of the numbers obtained exceeds 100. What is the most likely value of $S$?
Solution. Let $S_{1}$ denote the sum of the numbers before the last roll. Since $S_{1} \leqq 100$ and $S>100$, it follows that $S_{1} \geqq 95$. Thus, $S_{1}$ can have six possible values.

1. $S_{1}=100$. In this case, the value of $S$ could have been 101, 102, ..., 106, depending on whether the last roll was 1, 2, 3, 4, 5, or 6, with equal probability.
2. $S_{1}=99$. In this case, the last roll could only have been 2, 3, 4, 5, or 6, with equal probability, so the value of $S$ (with equal probability) could have been 101, 102, 103, 104, or 105.
3. $S_{1}=98$. This time, the last roll could have been 3, 4, 5, or 6, so the value of $S$ could have been 101, 102, 103, or 104, with equal probability.
4. $S_{1}=97$. This time, only 4, 5, or 6 could have been rolled last, so the possible values of $S$ are 101, 102, or 103.
5. $S_{1}=96$. This time, the last roll could have been 5 or 6, so $S$ could have been 101 or 102.
6. $S_{1}=95$. This time, only a 6 could have been rolled last, so the value of $S$ could only have been 101.

From the list, it is already clear that the most likely value of $S$ is 101. More precisely, the probability of 101 occurring is exactly as much greater than the probability of 102 as the probability of $S_{1}=95$. Similarly, the probability of 102 is greater than the probability of 103 by the probability of $S_{1}=96$, and so on. The possible values of $S$ in decreasing order of probability are: $101, 102, 103, 104, 105, 106$.

Remark. Using the same line of reasoning, for any natural number $n(\geqq 6)$, we find that when rolling a die until the sum exceeds $n$, the most likely value of this sum is $n+1$. This result does not change if, instead of rolling a die, we randomly select from the numbers $1, 2, \ldots, k$ with equal probability. (Of course, this is only valid for $n \geqq k$.)