At Least One Low Dice Sum in Five Rolls
Two fair six-sided dice are rolled 5 times independently, and each time the sum of the two dice is recorded. What is the probability that at least one of the 5 sums is no greater than 3? Round to 2 decimal places.
Hints
- Use the complement: $P(\text{at least one sum} \leq 3) = 1 - P(\text{no sum} \leq 3)$. This is almost always simpler than summing over all 'at least one' cases directly.
- First find the probability that a single roll produces a sum $\leq 3$. List the outcomes: only $(1,1), (1,2), (2,1)$ qualify out of 36 total, giving $p = 3/36 = 1/12$.
- Since the 5 rolls are independent, $P(\text{no sum} \leq 3) = (1 - 1/12)^5 = (11/12)^5$. Subtract from 1.
Worked Solution
How to Think About It: 'At least one' is almost always easier to compute via the complement. Instead of summing over 'exactly 1 low sum, exactly 2 low sums, ...,' compute one minus the probability that none of the 5 rolls produces a sum of 3 or less. This reduces a complex union calculation to a simple product of independent probabilities.
Quick Estimate: The probability of getting a sum $\leq 3$ on a single roll is small -- only $(1,1), (1,2), (2,1)$ qualify, so $P(\text{sum} \leq 3) = 3/36 = 1/12 \approx 0.083$. Over 5 independent rolls, the chance of never seeing a low sum is $(1 - 1/12)^5 = (11/12)^5 \approx 0.91^5 \approx 0.64$. So the probability of at least one low sum is about
Approach: Compute the complement via independence of the 5 rolls.
Formal Solution:
Step 1: P(single roll sum $\leq$ 3).
Outcomes with sum $\leq 3$: $(1,1)$ gives 2, $(1,2)$ and $(2,1)$ give 3. That is 3 outcomes out of $6^2 = 36$: $p = P(\text{sum} \leq 3) = \frac{3}{36} = \frac{1}{12}$
Step 2: P(all 5 sums
Step 3: Complement. $P(\text{at least one sum} \leq 3) = 1 - \left(\frac{11}{12}\right)^5 = 1 - \frac{161051}{248832} = \frac{87781}{248832} \approx 0.3527$
Answer: $\approx \mathbf{0.35}$
Intuition
The complement trick -- $P(\text{at least one}) = 1 - P(\text{none})$ -- is one of the most powerful and reusable shortcuts in probability. It works whenever the 'none' event factors nicely over independent trials, turning a complicated union into a single product. In contrast, computing $P(\text{at least one})$ directly requires summing $P(\text{exactly } k)$ for $k = 1, 2, 3, 4, 5$, each of which involves combinatorial factors.
This structure appears constantly in finance: the probability that a portfolio suffers at least one large drawdown, that at least one counterparty defaults, or that at least one of several signals fires during a trading day. In all these cases, if the events are approximately independent, the complement approach gives a quick and accurate answer. When events are correlated, the inclusion-exclusion principle extends the idea, though it becomes rapidly more complex.