Pythagoras’ Comma

Here's how Julian from Wilson's School reasoned :

The first note in the scale is 1
The second note is 1 $\times$ 2/3 = 2/3
The third is 2/3 $\times$2/3 = 4/9, but that is less than 1/2, so we multiply it by 2, giving the answer 8/9
The fourth is 8/9 $\times$2/3 = 16/27

This process needs to be repeated 12 times in total.
That means that 1 will be multiplied by 2/3 12 times, so we can do the following calculation:

$(\frac{2}{3})^{12} = 0.0077073466$ (10dp)

We are also know that 7 doublings are required, so 0.0077073466... $\times$ 128 = 0.9865403685 (10dp)

Therefore, they were off by 1 - 0.9865403685... = 0.0134596315 (10dp)

Thanks Julian