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#1
10-02-2007, 06:19 PM
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Card Deck Arrangement Problem
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? 
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#2
10-02-2007, 06:39 PM
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Re: Card Deck Arrangement Problem
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!
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#4
10-02-2007, 10:07 PM
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Re: Card Deck Arrangement Problem
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.
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#6
10-03-2007, 01:20 AM
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Re: Card Deck Arrangement Problem
[ 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.
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#7
10-03-2007, 11:39 AM
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Re: Card Deck Arrangement Problem
[ 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 ] Shouldn't that be a resulting 44 card deck, making the probability quite a bit smaller ? (44*45 = 1980 times smaller I guess). The method is perfectly correct, of course [img]/images/graemlins/smile.gif[/img] MarkW
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#8
10-03-2007, 12:16 PM
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Re: Card Deck Arrangement Problem
[ QUOTE ]
[ 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 ] Shouldn't that be a resulting 44 card deck, making the probability quite a bit smaller ? (44*45 = 1980 times smaller I guess). [/ QUOTE ] Taking out the As and Ks leaves you with 44 cards, but putting the "glued" cards back in gives you 46. You have to pick two places among the remaining 44 cards to stick the As and Ks back in. Siegmund is right.
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#9
10-03-2007, 12:18 PM
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Re: Card Deck Arrangement Problem
[ QUOTE ]
[ 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 ] Shouldn't that be a resulting 44 card deck, making the probability quite a bit smaller ? (44*45 = 1980 times smaller I guess). The method is perfectly correct, of course [img]/images/graemlins/smile.gif[/img] MarkW [/ QUOTE ] No. There are 46 cards. There are 44 non-ace/non-kings, 1 super-ace, and 1 super-king to shuffle.
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#10
10-04-2007, 02:43 PM
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Re: Card Deck Arrangement Problem
[ 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 ] I couldn't see why this worked. So I tried it the hard way. Could understand 4!4!44!. But not 46 X 45. 52 cards. 49 ways to place the 4 aces. AAAA 1234 2345 3456 4567 5678 etc. 1234 now 45 ways for the kings. The same 45 ways for 49,50,51,52 slots. 2345 now 44 ways for the kings. 3456 now 43 ways for the kings. 4567 now 42 ways. 5678 also 42 ways. There were 43 ways for aces to have 42 ways for kings. AAAA 1234; 45 X 2 = 90 2345; 44 X 2 = 88 3456; 43 X 2 = 86 4567; 42 X 43 = 1806 Sum the four numbers 2070. Amazingly that's the same as 46 X 45 = 2070
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