I personally prefer your method, but the one they suggest gives an equal distribution also... It just looks awkward
Roll any die, odd use a 1d20 and even 1d10+20... so say we roll a d4 initially, on a 1 or 3 you get an equal chance of a 1-20... if the d4 was a 2 or 4, a 1d10 and add 20 still...