The problem with your idea of just using 4/6 (not getting any 5s or 6s) would not be correct because that excludes the possibility of getting (a 5 and no 6) or (a 6 and no 5).
When calculating odds you don't need to calculate probability (although you can use it to get odds).
Odds = (# ways to succeed)/(# ways to fail)
For your question # ways to succeed = all possibilities - (no 5) - (no 6) + (no 5 and no 6)
= 65 - 55 - 55 + 45 =2550
# ways to fail = 65 - 2550 = 5226
odds of at least one 5 and at least one 6 = 2550/5226
If you did calculated probability of getting at least one 5 and at least one 6 say = a/b and you wanted odds the answer would be a/(b-a).
Any questions please comment back.