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 ?
Thanks in advance;