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Spread of $3$ dice

ProbabilityHard

Roll $3$ fair $6$ -sided dice independently. What is the probability that the largest result minus the smallest result is at least $4$ ? Give a fraction in lowest terms.

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Hint

Count outcomes with a given minimum and maximum, then sum over gaps $d = max - min$ .

Solution

For a gap $d \geq 1$ and fixed min and max, the number of ordered triples is $(d + 1)^{3} - 2 d^{3} + (d - 1)^{3} = 6 d$ .
The number of (min, max) pairs with gap $d$ is $6 - d$ .
For $d = 4$ there are $2 \times 24 = 48$ outcomes; for $d = 5$ there are $1 \times 30 = 30$ .
The total is $78$ out of $216$ outcomes.
The probability is $\frac{78}{216} = \frac{13}{36}$ .

Answer: $\frac{13}{36}$

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