# Probability on a Die

The normal probability of a number in a regular die (6 faces) is $\dfrac{1}{6}$. Let in an addicted [that is, "loaded"] die, the probability of a even number (2, 4 and 6) be twice the normal probability;

I've got such outcome: $regular\space probability \space on \space evens \space is \space \dfrac{3}{6}$, doubling it, it would become $\dfrac{6}{6}$, in other words, a certain event, it sounds strange to me, is that right ?

You might check the wording of the question. If the probability of each even number is twice that of a normal die, you are correct. If the probability of each even number is twice that of each odd number, the result is different-then the evens come up $2/9$ each for a total probability of $6/9=2/3$