Expected Number of Drinks in 'Ride the Bus' drinking game So there is a drinking game called 'Ride the Bus', played with one deck of cards, and the 'loser' has to ride the bus, as follows:

Place ten cards face down, and reveal them one by one. If a Jack is
  shown, have one drink and add one more face down card. If a Queen, two
  drinks and two cards. King three, Ace four. You stop once all the
  cards have been flipped.

So I wondered, how many drinks are you expected to drink? I couldn't figure it out on paper, but I coded a simulation and determined that mean is about 22.5 drinks, and the probability distribution function seems to be constant except for a large spike at 40 drinks. So in that case:


*

*Can we theoretically determine the expected value and probability distribution function of the number of drinks when riding the bus?

*Can we generalise this to different card values? For example, a 10 draws/drinks 1, a Jack 3, the rest 0.

 A: This is not an answer. Think about states and transitions between them. A state is $(n,j,q,k,a)$, where $n$ is the number of face-down cards you have, $j$ is the number of jacks flipped, $q$ is the number of queens flipped etc.
Your game starts in state $(10,0,0,0,0)$. With probability $\frac{4}{52}$ it transitions into state $(10,1,0,0,0)$ which involves a drink. With probability $\frac{4}{52}$ it transitions into state $(11,0,1,0,0)$ which involves two drinks. The same probability applies to states $(12,0,0,1,0)$ and $(13,0,0,0,1)$ and to two and three drinks respectively. With the remaining probability you transition to state $(9,0,0,0,0)$ with no drink.
Hence value of $(10,0,0,0,0)$is $$v_{(10,0,0,0,0)}=\tfrac{4}{52}(1+v_{(10,1,0,0,0)})+\tfrac{4}{52}(2+v_{(11,0,1,0,0)})+\tfrac{4}{52}(3+v_{(12,0,0,1,0)})+\tfrac{4}{52}(4+v_{(13,0,0,0,1)})+\tfrac{52-16}{52}v_{(9,0,0,0,0)}$$ If you know the value of the states in the expression, you know the expected number of drinks. You can proceed working backwards from the states that are final. For example, $v_{(n,4,4,4,4)}=0$ for any $n$. Practical problem with calculating the values is that, say, the probability of drawing jack changes depending on how many you have drawn already.
If you are willing to change the game such that 'cards are drawn with replacement' your calculation would be easier. That is, assume that you have deck of 52 cards and you start knowing that you have to flip $10$. You draw a card. You apply your rules adding to the number of card you have to flip if you draw jack/queen/etc. and put the card back among the $52$.
With this, state is described simply by the number of card you have to draw.
