Bloomberg interview question

Say I have a deck of 52 cards, regular deck of cards. I put a joker in the deck somewhere and shuffle it up. Now I start dealing you cards until the joker shows up. Once it shows up, I stop dealing you cards. What is the probability that you have, in your set of cards, all 4 aces?

Interview Answers

Anonymous

27 June 2013

You are complicating it too much... There are 5 cards... 4 ACes and 1 Joker. and there are 5 places.. What is the probability that joker occupies the 5th place? It is 1/5 = 20%

9

Anonymous

12 Mar 2013

Select 5 positions from the 53 cards; assign the 4 aces to the first 4 positions, and the joker to the 5th position. The 4 aces and the rest 48 cards can have full permutations. The probability is C(53,5)*4!*48!/53!=0.2

4

Anonymous

19 Feb 2013

It's a hyper-geometric distribution. There is no single answer since the probability of having all 4 aces depends on how many cards were chosen. If all cards were chosen, the probability of you having all four aces is 1. So just make an expression with n equal to the amount of draws from the deck until reaching joker. This is what I got. (49 C (n-4)) / (53 C n)

2

Anonymous

17 Nov 2014

I dont think the answer is 20%. You could get joker on the 5th card or on the 49th card.So the chance of joker occupying the 5th place is 1/5 , occupying sixth place is 1/6 and so on till the joker occupies 49th place is 1/49 .. this sum 1/5+1/6+...+1/49 will give u the answer. correct me if i'm wrong...

1

Anonymous

10 Oct 2012

4/53

1

Anonymous

24 Oct 2012

sum (i+1)(i+2)(i+3)(i+4)/ (53*52*51*50*49) from i =0 to 48 we got 1/240 ~ 0.00416666...

Anonymous

19 Dec 2012

Guys!! the 0.2 is the correct answer. Actually I have come across an even tougher question based on this one. it asked what is the expected number of pokers that you have delivered when you dealing all 4 Aces out.

1

Anonymous

7 Dec 2012

it should be a function of the position of joker

Anonymous

29 June 2012

20%

5

Anonymous

12 Sept 2012

The joker shows up at the nth position. Then the answer is 1 - Probability of getting one ace in (n-1) cards.