Probability Deck Question (4 kings)

Question

A standard deck of 52 cards contains 4 kings. Suppose you chose a random ordering, all 52! permutations being equally likely. Compute the following:

(i) the probability that all of the top 4 cards are kings

(ii) the probability that none of the 4 cards are kings.

(iii) the expected number of kings among the top 4 cards in the deck.

Just to double check: for:

(i) I have 49!4!/52!

for (ii) I have 1-49!4!/52!

for (iii) I got that since each slot on top has a 4/52 chance of being a king then E(number of kings) =4/52+4/52+4/52+4/52=16/52?

I am confused if (iii) is correct and I do not have an intuitive explanation for it?

Answer 1

For $(i)$ and $(ii)$ you may find it simpler to directly multiply probabilities.

$(i): \dfrac4{52}\cdot\dfrac3{51}\cdot\dfrac2{50}\cdot\dfrac1{49}$

$(ii): \dfrac{48}{52}\cdot\dfrac{47}{51}\cdot\dfrac{46}{50}\cdot\dfrac{45}{49}$

$(iii):$ Ok, by linearity of expectation, $4\times\dfrac4{52} = \dfrac4{13}$

Answer 2

(i) When you consider $4$ kings as a single card, you are not shuffling the kings but only shuffling the remaining $48$ cards. You always place the "single king" on top so you cannot shuffle them.

(ii) The possibility of events "all of the top 4 cards are kings" and "none of the top 4 cards are kings" does not add up to $1$. For example "three kings are in the top four cards" does not belong to any of these two events.

(iii) If you want a more intuitive explanation, you can think as $1\cdot P(1)+2\cdot P(2)+3\cdot P(3)+4\cdot P(4)=1\cdot {{4\choose1}{48\choose3}\over{52\choose4}}+2\cdot {{4\choose2}{48\choose2}\over{52\choose4}}+3\cdot {{4\choose3}{48\choose1}\over{52\choose4}}+4\cdot {{4\choose4}\over{52\choose4}}$