If I understand it correctly your question is: What is the probability that I roll a 5 or a 6 in two rolls of one dice, tossing again if the first time was not a 5 or a 6.
Indeed, the chance of rolling a 5 or a 6 the first time is 2/6 = 1/3
In the other case (2/3) you will roll again, where you once more have a 1/3 chance. Because you will need to both NOT roll a 5 or 6 on the first roll AND roll a 5 or 6 on the second roll, you'll need to multiply these chances: 2/3 * 1/3 = 2/9 (which indeed does convert to 22%)
Add these two up to get the total chance being: 1/3 + 2/9 = 5/9 = 55%