Note that the blank cards don't matter at all, so we can just remove them as they won't affect the probability. Now there are 8 cards: 3 skull, 5 coin. Say we win if we stop because of seeing 3 coins, and lose otherwise. So we want to find the probability we lose. Claim: We lose if and only if the last three cards are all coins. Proof: Note that if the last three cards are all coins, the third coin is in the first of these three cards, i.e. the sixth position. However there are three skulls in the first five cards, so we will see 3 skulls before we see the third coin. Thus we lose in this case. However, if at least one of the last three cards are skulls, then there are at most two coins in the last two positions, meaning that there are at least three coins in the first five positions. So we will see three coins in at most five cards, and we won't see the third skull before this happens. Thus we win in this case. As both cases are considered, the claim is proven. $\square$ Now we just have to find the probability that the last three cards are all coins, which by considering the possible positions of the skulls is equal to $\dfrac{\binom{5}{3}}{\binom{8}{3}} = \dfrac{10}{56} = \boxed{\dfrac{5}{28}}$. Yay! :DD