What's the chance that exactly 2 2s will have appeared before the 2nd 8? A standard deck of 52 cards is randomly shuffled. Bob keeps drawing from the deck until he has drawn two 8s.
What is the probability that, when, he draws the second 8, he has already drawn exactly two 2s?
I think this could be a combinations problem because the order of the drawing two 2s does not matter. However, I am not sure how to account for the fact that he could draw an 8 or a second 8 at any point.
 A: Hint:  one approach is to recognize that all the other cards are just fluff.  There are only ${8 \choose 4}=70$ possibilities for the order of 2's and 8's, so list them and count.  You can make it less by listing the orders that get to 3 2's or 2 8's and assessing the probability.  The first are failures, the second are successes if you have two 2's already.  For example, one is that you start with three 2's.  This is $\frac 12 \cdot \frac 37 \cdot \frac 26=\frac 1{14}$.  Another is 2828.  This is $\frac 12 \cdot \frac 47 \cdot \frac 12  \cdot \frac 35=\frac 3{35}$
A: As was observed by Ross Millikan, there are only $8$ relevant cards. The first $4$ of these can appear in the dealing in $(8)(7)(6)(5)$ orders, all equally likely.
We count the number of "favourable" orders. These are given by any one of three patterns, which we call $2288$, $2828$, and $8228$. 
The pattern $2288$ can occur in $(4)(3)(4)(3)$ ways, as can the other two patterns. Thus the required probability is 
$$\frac{(3)(4)(3)(4)(3)}{(8)(7)(6)(5)}.$$
If we want to simplify a bit, we get $\dfrac{9}{35}$. 
Remark: The problem seems to have been solved correctly in the comments both by you and by Ross Millikan. Then there appears to have been a change of mind.  
