What is the probability of picking 3 aces and 2 kings out of a 54-card deck (52-card standard with jokers)?

Question

A fundamental exercise is to calculate the probability of picking 3 aces and 2 kings while randomly picking a 5-card hand out of a 52-card deck. Our sample space would be $52\cdot51\cdot50\cdot49\cdot48\over5!$, and the event would be $4! \over3!$ $\cdot$ $4!\over2!$. So far so good.

But what if we add the jokers in the deck? The probability for the hand to contain a joker would be the same as any other card (and of course the sample space should increase to 54, etc.,) but since the joker can be anything, I can't find a way to handle the rest.

Can I assume a $6!\over3!$ event probability for the aces, let's say, since there are now 6 "possible aces" (the 4 "real" aces and the 2 jokers) in the deck? But if that holds true, I can't concurrently have 6 possible kings.

Answer 1

There are 2 ways

Here NcR represent N!/{R!(N-R)! Or ways to select R things from N things

1) let's let the joker can be anything with equal probability

a) 3 aces and 2kings have 4c3.4c2 cases

b) 2aces and 2king and 1 joker (joker need to be ace having 1/13 probability of that). 4c2 . 4c2 . 2c1 .1/13 cases (It is wrong to say cases u know cases but actually it is the probability of event with respect to 54c5)

c) 1 king and 3 aces and 1 joker(need to king) 4c1. 4c3 . 2c1. 1/13

d) 0 king , 3 aces and 2 jokers (both jokers should be king and each joker is independent of each other so of this probability is 1/169) 4c3.2c2.1/169 cases

E) 1 ace , 2king and 2 joker (both ace) 4c1.4c2.2c2.1/169

F) 1king, 2 ace and 2 joker(here joker need to be king and one ace which have probability of 2/169) 4c1.4c2.2c2.2/169

For the answer add all cases and divide it by 54c5. Actual sample space is very large but we need probability so both give correct answer

2) we can convert joker according to our need

Which means u don't have to account the probability of joker because just take it 1

A) 4c3.4c2, B)4c2.4c2.2c1, C)4c1.4c3.2c1, d)4c3.2c0.2c2 , E)4c1.4c2.2c2 , F) 4c1.4c2.2c2

For the answer just add these and divide by 54c5

Answer 2

Here is how I would do this problem: One way of getting "3 aces and 2 kings" is "AAAKK".

Since there are 54 cards, four of them aces, the probability the first card drawn is an ace is 4/54. Then there are 53 cards left, 3 of them aces. The probability tshe second card drawn is also an ace is 3/53. Now there are 52 cards left, 2 of them aces. The probability the third card drawn is an ace is 2/52. Now there are 51 cards left, 4 of them kings. The probability the fourth card drawn is a king is 4/51. Finally there are 50 cards left, three of them kings. The probability the fifth card drawn is a king is 3/50.

The probability of "AAAKK" in that order is (4/54)(3/53)(2/52)(4/51)(3/50).

It is not difficult to show that "three aces and two kings" in any order is the same (you get different fractions but with the same numerators and denominators so the same product). There are $\frac{5!}{3!2!}= 10$ ways to order "AAAKK" so the probability of "3 aces and 2 kings" drawn from a 54 card deck is 10(4/54)(3/53)(2/52)(4/51)(3/50). That can be written as $\frac{10(4!)^2(50!)}{2(54!)}$.