imc

Since 3 out of 6 of the numbers are even, there is a $\frac{3}{6}$ chance that the first number we choose is even. Since the number rolled first is irrelevant, we don't have to consider it. Therefore there are 2 even numbers out of the 5 choices left. There is a $\frac{2}{5}$ chance that the next number that is distinct from the first is even. There is a $\frac{1}{4}$ chance that the next number distinct from the first two is even. (There is only one even integer left. ) With all the even integers taken, the next integer rolled must be odd. $\frac{3}{6}\cdot \frac{2}{5}\cdot \frac{1}{4}=\frac{1}{20}$, so the answer is $\boxed{\frac{1}{20}}.$