Since the selection from the hat is assumed to be random, it seems that the presence or absence of you and your friend at the drawing is irrelevant to the problem.
One approach to the problem is to consider how many ways 12 people can be selected from a hat. Since this is a draft order, then this must be a permutation, and all 12 are being selected. So this number of possibilities is 12! = 12·11·10·9·8·7·6·5·4·3·2·1 = 479,001,600. This would be the total number of possible selections, which would be the denominator of the probability fraction.
Next we would need to consider how many ways you and your friend would not be in the first 10 selected. This would mean that, for the first 10 drawings, you and your friend would not be part of the total "selectable" group. So there would be only 10 people to choose from for 10 drawings without replacement. This would be 10! = 10·9·8·7·6·5·4·3·2·1 = 3,628,800.
There is still one remaining factor. The last figure is how many ways 10 people can be arranged, but it doesn't consider the order of the 11th and 12th selection. There are 2 ways that can happen, and for each of these there are 3,628,800 ways for the 10 prior selections to happen. So the 3,628,800 figure must then be multiplied by 2, to get 7,257,600.
So then the probability is 7,257,600 / 479,100,600 ≈ 0.015.
There is probably another way to do this, but sometimes getting back to the essence of probability is useful and instructive.
