Cards between the first two and the first ace

A standard $52$-card deck is shuffled and dealt face up one card at a time. Suppose the first $2$ comes out before the first ace. Under this condition, what is the expected number of cards dealt strictly between the first $2$ and the first ace?

$\tfrac{379}{45}$ (about 8.42). The 8 aces and 2s split the other 44 cards into 9 gaps of average size $\tfrac{44}9$. Given a 2 leads, the remaining three 2s fall before the first ace on average $\tfrac35$ times, so $\tfrac85$ gaps are spanned. The answer is $\tfrac85\cdot\tfrac{44}9+\tfrac35=\tfrac{379}{45}$.