The probability of dealing a second king on the sixth card

Question

I have a question that states cards are dealt at random without replacement from a standard deck. What is the probability that you deal a second king on the sixth card?

Things I know:

• The first king must occur on one of the first five draws.
• 4 non-kings are to be drawn in total.
• The second king occurs on the sixth draw.

How I approach this issue. Since the first king can be dealt on any of the first five draws, there are ${5 \choose 1}$ ways this can occur. We aren't too concerned about what the other cards are, apart from them not being a king.

$p($second king on sixth draw$)$ = ${5 \choose 1}{4\over 52}{48\over51}{47\over50}{46\over49}{45\over48}{3\over47}$

Apart from the ${5\choose 1}$ term, the ${4\over 52}$ represents the first king (there are four out of the 52 cards), then we choose four separate cards out of the whole deck, and finally the ${3\over47}$ term represents the second king being chosen from the rest of the deck.

I think this is a correct way to attack the problem, but I'm currently dealing with hypergeometric distributions, which is why I have some doubt (there may be an alternate way using these to solve too that I have not thought of).

Answer 1

but I'm currently dealing with hypergeometric distributions,

Yes, that's a clue that the solution probably involves such.

You require the probability of selecting 1 from 4 kings and 4 from 48 non-kings in the first five draws (from 52 cards), then 1 from 3 kings in the next draw (from the 47 remaining cards).   Can you see the hypergeometric expression in that?

$$\dfrac{\binom 4 1\binom {48}{4}}{\binom {52}5}\cdot\dfrac{3}{47}$$

But, yes, ${5 \choose 1}{4\over 52}{48\over51}{47\over50}{46\over49}{45\over48}{3\over47}$ is also the answer.   Different approach to the same destination.