Well, there should be half as many cases of multiples-of-9-bits ran out than for multiples-of-8-bits.

I don't think this is enough of a reason, though.

If you're deciding between using 8 bits or 16 bits, you might pick 16 because 8 is too small. But making the same decision between 9 and 18 bits could lead to picking 9 because it's good enough at the time. So no I don't think there would be half as many cases. They'd be different cases.