imc

Rolling a pair of fair $6$-sided dice, the probability of getting a sum of $7$ is $\frac{1}{6}:$ Regardless what the first die shows, the second die has exactly one outcome to make the sum $7.$ We consider the complement: The probability of not getting a sum of $7$ is $1-\frac{1}{6}=\frac{5}{6}.$ Rolling the pair of dice $n$ times, the probability of getting a sum of $7$ at least once is $1-{\left(\frac{5}{6}\right)}^n.$ Therefore, we have $1-{\left(\frac{5}{6}\right)}^n>\frac{1}{2},$ or $${\left(\frac{5}{6}\right)}^n<\frac{1}{2}.$$ Since ${\left(\frac{5}{6}\right)}^4<\frac{1}{2}<{\left(\frac{5}{6}\right)}^3,$ the least integer $n$ satisfying the inequality is $\boxed{4}.$