Card Deck Arrangement Problem

tshort · #1 10-02-2007, 06:19 PM

Random 52 Card Deck. Probability all 4 aces are adjacent and all 4 kings are adjacent.

I am not seeing a simple way of counting the arrangements without dealing with a number of various cases. Does anyone have a simple approach to this problem?

tshort · #2 10-02-2007, 06:39 PM

Cases approach:

Case 1: C(2,1) * C(2,1) * C(45,1) * 4! * 4! * 44!

1) C(2,1) Choose whether you are going to place aces or kings.
2) C(2,1) Place that set of four cards at the end of the deck or beginning of the deck.
3) C(45,1) Choose where to place the other set of four cards.
4) 4! Number of arrangements of one set of four (aces or kings).
5) 4! Number of arrangements of other set of four (aces or kings).
6) 44! Number of arrangements of rest of cards.

There will be a total of four special cases to get:

(2*45 + 2*44 + 2*43 + 46*42) * C(2,1) * 4! * 4! * 44!

ALawPoker · #3 10-02-2007, 10:00 PM

(3/51)*(2/50)*(1/49)*(3/47)*(2/46)*(1/45) = answer

Siegmund · #4 10-02-2007, 10:07 PM

Glue the 4 aces together. (Any of 24 suit-sequences.)
Glue the 4 kings together. (Any of 24 suit-sequences.)
Shuffle the resulting 46-card deck.

so, 4!4!46! possible sequences with the necessary cards adjacent, out of 52! possibilities, which simplifies to 1 / 20358520.

sirio11 · #5 10-03-2007, 12:23 AM

Pretty nice Siegmund !

tshort · #6 10-03-2007, 12:38 AM

[ QUOTE ]
(3/51)*(2/50)*(1/49)*(3/47)*(2/46)*(1/45) = answer

[/ QUOTE ]

3/51*2/50*1/49*3/47*2/46*1/45 = 2.961 x 10^-9

Whereas I believe the answer is:

(44!*4!*4!*(45*2 + 44*2 + 43*2 + 46*42))/52! = 4.16875 x 10^-8

How did you approach the problem?

madnak · #7 10-03-2007, 12:55 AM

I'm with Sieg. He takes the special cases into account and as far as I can tell he covers every specific permutation possible, so those / 52! should be correct. I'm not quite sure what you're doing, it seems like you're trying to separate out the kings and aces from the other 44 cards and deal with special cases individually, but doesn't the "glue solution" cover it?

tshort · #8 10-03-2007, 01:20 AM

[ QUOTE ]
Glue the 4 aces together. (Any of 24 suit-sequences.)
Glue the 4 kings together. (Any of 24 suit-sequences.)
Shuffle the resulting 46-card deck.

so, 4!4!46! possible sequences with the necessary cards adjacent, out of 52! possibilities, which simplifies to 1 / 20358520.

[/ QUOTE ]

Thanks siegmund. Great solution. Makes me feel dumb not thinking of it that way!

Our answers differ by a very small amount which surely is due to me miscalculating something in my solution.

ALawPoker · #9 10-03-2007, 02:10 AM

A better question would be WHY did I approach the problem, since I haven't been in a math class since high school, and since I wasn't really even sure what you were asking. So I'm not sure what I thought I could contribute. I'd blame it on the wine, but the wine and I had a good evening together, and I wouldn't ruin it like that.

I thought the question was that you have a deck of cards in a random order, and what are the chances that you have 4 aces and 4 kings in a row. So in my mind, I just imagined flipping through the deck. OK, here's an ace. 3 in 51 the next card is an ace. 2 in 50, 1 in 49. Then from the first king it's 3/47 that the next will be a king, etc. That's all I did. It was too simple though, and I guess you were either asking something else or I'm overlooking something dumb.

tshort · #10 10-03-2007, 03:05 AM

[ QUOTE ]
A better question would be WHY did I approach the problem, since I haven't been in a math class since high school, and since I wasn't really even sure what you were asking. So I'm not sure what I thought I could contribute. I'd blame it on the wine, but the wine and I had a good evening together, and I wouldn't ruin it like that.

[/ QUOTE ]

Where would we be if people didn't attempt problems they weren't sure they would get right?

[ QUOTE ]
I thought the question was that you have a deck of cards in a random order, and what are the chances that you have 4 aces and 4 kings in a row. So in my mind, I just imagined flipping through the deck. OK, here's an ace. 3 in 51 the next card is an ace. 2 in 50, 1 in 49. Then from the first king it's 3/47 that the next will be a king, etc. That's all I did. It was too simple though, and I guess you were either asking something else or I'm overlooking something dumb.

[/ QUOTE ]

This appraoch leaves quite a few cases unaccounted for. Think about the number of places you could find that first ace or king. Your method is calculating one possible placement of the first ace and king.