Linearity of Expectation
Your approach looks good. Let's just write it clearly. Let's look at $E[X_1]$. 

With probability $1/10$ the first coin was used to buy the coffee, thus the value is $X_1=0$. 

With probability $9/10$ we are back in the situation of part II (we can add a new "Gold" coin - the party that gets this coin will not get his dollar back), so the expectation of this case is just like in Part II. 

Adding both cases (with the appropriate weight) should solve your question, for $X_1$, and by the linearity of expectations, and since all parties are symmetric, the answer is $10 E[X_1]$.