A die is loaded in such a way that an even number is twice as likely to occur as an odd number.

$P[E]=2P[O]\\ P[E]+P[O]=1\\ P[E]=\dfrac 2 3,~P[O]=\dfrac 1 3\\ P[1]=P[3]=P[5]=\dfrac 1 3 \cdot \dfrac 1 3 =\dfrac 1 9\\ P[2]=P[4]=P[6]=\dfrac 1 3 \cdot \dfrac 2 3 = \dfrac 2 9\\ P[roll < 4]=P[1]+P[2]+P[3] = \dfrac 1 9 + \dfrac 2 9 + \dfrac 1 9 = \dfrac 4 9$