10 cards drawn, probability of drawing non-king on 11th draw given king on 10th Here's the full text of the problem:
"Draw from a deck without replacement.  Find the probability that the $10^{th}$ card is a king and the $11^{th}$ is a non-king."
The text gives the answer as "$P$(10th is king, 11th is non-king) = $P$(1st is king, 2nd is non-king) $=\frac{4}{52}\cdot\frac{48}{51}$".  But doesn't this assume that no kings had been drawn by the 10th draw?  I don't see any justification for that in the problem statement.  
My attempt at a solution approached this as two-stage problem.  The first stage has four possible starting states: 1) 10 card hand with 1 king; 2) 10 card hand with 2 kings; 3) 10 card hand with 3 kings; 4) 10 card hand with 4 kings. The second stage, or 11th card draw, has two possible outcomes: king or non-king.  So I applied the total theorem of probability to get:
\begin{align}
P(\text{non-king}) &= P(\text{10 cards drawn with 1 king})P(\text{11th draw is non-king|10 cards drawn with 1 king})
 + P(\text{10 cards drawn with 2 kings})P(\text{11th draw is non-king|10 cards drawn with 2 kings}) + P(\text{10 card hand with 3 kings})P(\text{11th draw is non-king|10 cards drawn with 3 kings}) + P(\text{10 card hand with 4 kings})P(\text{11th draw is non-king|10 cards drawn with 4 kings})
\end{align}
Am I on the right track, or have I made this problem into something more complex than it is?  Any help resolving this would be greatly appreciated!  
 A: Perhaps surprisingly, it does not assume that no kings had been drawn by the tenth draw, and so no unjustified assumption is required from the problem statement.
The reason why this might be surprising is that you are imagining situations where you see any number of kings in the first nine cards.  The probabilities of a king and non-king on the tenth and eleventh draws would obviously be higher when there are no kings drawn in the first nine cards than if all four kings were drawn in the first nine cards (in which case the probability would be zero).
But imagine you don't look at any of the first nine cards.  You only look at the tenth card and eleventh card.  Clearly, this doesn't change the probability being asked for, and the tenth card might be any of the $52$ cards in the deck.  The probability that it is a king is therefore $4/52 = 1/13$.  Suppose that happens.  Then the eleventh card might be any of the $51$ remaining cards, of which $48$ are non-kings, so the probability that it is a non-king is $48/51 = 16/17$.  Thus, the desired probability is $1/13 \times 16/17 = 16/221$.
