The lowest of five dice equals exactly two

Question

You roll five fair six-sided dice. What is the probability that the smallest number showing is exactly 2?

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"Exactly 2" means the minimum reaches 2 but not 3. Turn each into an "all dice clear the bar" event: P(	ext{min} \ge 2) = \left(\frac{5}{6}ight)^5 = \frac{3125}{7776}, \qquad P(	ext{min} \ge 3) = \left(\frac{4}{6}ight)^5 = \frac{1024}{7776}. The minimum equals exactly 2 on the difference: P(	ext{min} = 2) = \frac{3125}{7776} - \frac{1024}{7776} = \frac{2101}{7776} \approx 0.270. Takeaway. For an exact minimum value, subtract two tail probabilities: P(	ext{min} = v) = P(	ext{min} \ge v) - P(	ext{min} \ge v+1).

The lowest of five dice equals exactly two

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