Question

probability question:
this question is based on a famous card magic trick in which the probability of any 2 cards(king and ace ; seven and three of any suit) in a well shuffled deck being adjacent is quite high.
Here was my method.
I placed 4 kings first

_ K _ K _ K _ K _
we need atleast 1 ace adjacent to one of the kings.
There are 5 places where an ace can go. So what I did was I selected an ace , and then a position , say we get this :

_ AK _ K _ K _ K _
Now we have remaining 47 cards which can be filled in these 5 spaces . Total number of ways of doing that = C(51,4) . Also these 47 cards can be permuted in 47! ways ; and the kings in 4! ways.
So according to me the probability should have been (C(51,4) * C(4,1) *C(5,1) * 47! *4!) /52!
which yields 5/13 , which is surely wrong.
I can say this because when I extended this to total of 13n cards where n is number of kings, aces etc, I get final probability as (n+1)/13 , which is clearly absurd as according to it probability>1 for n=13 and above.
My query is what am I missing ?

Answer 1

This discussion of the problem may offer insights:

http://math.stackexchange.com/questions/456347/probability-of-2-cards-being-adjacent

Also,
http://www.osaka-ue.ac.jp/zemi/nishiyama/math2010/cardshuffle.pdf

Answer 2

what is wrong in my logic?

Answer 3

I don't understand your logic to begin with, so I have no idea where you are wrong. How can you assume you have 4 of a kind?

Answer 4

I ordered 4 kings first and then arranged the other cards accordingly :|

Answer 5

Can you describe the question better? Give an example of what wouldn't count and what would?

Answer 6

I'll rephrase the question.
It says, 'what is the probability that a king and an ace will be adjacent to each other in a well shuffled deck of 52 cards?'

Answer 7

Oh ok, I see.

Answer 8

See, w hat if all the kinds are clumped together? What if the kings are on an edge?

Answer 9

This is actually a very tough problem, I think.

Answer 10

that C(51,4) comes from the arrangement of 47 cards which we have to fill in 5 places
a+b+c+d+e = 47

I think this does account for kings being together or on the edges when either of the variables is 0

Answer 11

But it treats the clumped up case as identical to the spread out case when it is not. In the spread out case there are 8 places for an ace to go to get adjacency. For the clumped up at an edge case, there is only one place.

Answer 12

Is it just ace and king or any two adjacent ranks?

Answer 13

a spectator names any 2 cards, and we have to find the probability of them being adjacent.
in this case, i have taken ace and king

Answer 14

Oh ok.

Answer 15

I'm not sure of the elegant solution for this, to be honest.

Answer 16

I can brute force it, but that is a pain.

Answer 17

Well the answer is close to 1/2.
It is discussed over here, but I don't understand it.

Answer 18

http://recursed.blogspot.in/2010/01/neat-problem-on-card-arrangements.html

Answer 19

It's very complicated.

Answer 20

Good luck.