# 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!

• There's nothing special about the $10^{th},11^{th}$ trials...you could pick any two numbers. We have no information about the other trials. If it helps with intuition, imagine the cards are spread out in a circle. Now it's clear, I think, that referring to trials "$10,11$" is just an arbitrary convention. – lulu Apr 19 '17 at 18:03
• As another way to convince yourself that your text has it right: do the problem your way, only work it for the second and third cards. Now your calculation is perfectly manageable. Persuade yourself that you get the same result as for the first and second cards. – lulu Apr 19 '17 at 18:06
• Hi Lulu -- Thank you for your comments. I absolutely understand what you're saying -- I just need to reconcile this manner of reasoning with my understanding of the total theory of probability. – Scentless Apprentice Apr 19 '17 at 20:52
• That was the point of my second suggestion. You can work your method completely in that case. Then do it for the third and fourth slots. I expect that's as far as you'll want to go though. – lulu Apr 19 '17 at 20:54
• To do it by your method in general: I'd proceed inductively. Suppose you had $k$ Kings and $n$ non Kings and let $P(i,n,k)$ denote the probability that you get the pair $K,N$ on the $i^{th},(i+1)^{st}$ draws. Go inductively on the sum $i+k+n$. Assuming $i≤k+n-1$ I say the answer is $\frac k{n+k}\times \frac n{n+k-1}$. Easy to verify for small cases. Now follow your approach to get a recursive definition for $P(i,n,k)$ in terms if $P(i-1,n-1,k),P(i-1,n,k-1)$. – lulu Apr 19 '17 at 21:08

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 $1/13 \times 16/17 = 16/221$.