20070106, 15:39  #1 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
11·389 Posts 
Card Sorting Probability
Given cards, what is the probability that all the cards will be sorted in some specific way? (e.g. 13 in hearts, sorted the way that Hearts included with Windows sorts them)
Would it be ? With 13, as in hearts, this would make the chance just over one in 302 trillion. Is this correct? If not, what is the correct formula? 
20070106, 16:28  #2 
Jun 2003
The Texas Hill Country
441_{16} Posts 
A few minor points. You ask for the probability and then quoted the chance.
Probability is always a number in the closed interval 0 to 1. It is related to the chance A::B by p= A/B As for the card sorting, your universe is the number of permutations of the objects, in this case x things taken x at a time, and the number of successes is 1. The number of permutations of x things taken x at a time is x!. Thus the probability of a particular sort sequence, assuming that everything is chosen at random is 1/x! Last fiddled with by Wacky on 20070106 at 16:29 
20070108, 19:34  #3 
Einyen
Dec 2003
Denmark
19·181 Posts 
A deck with 52 cards can, as Wacky said, be sorted in 52! = 8*10^{67} ways. So if you want the chance for it to be sorted in 1 specific way, the chance is 1 in 8*10^{67} or 1 in 80 million trillion trillion trillion trillion trillion.

20070109, 00:40  #4 
"Jason Goatcher"
Mar 2005
3·7·167 Posts 
I think he asked a slightly different question. What are the odds that you can look through the pack and find all the hearts(in order?) right next to each other.

20070109, 07:15  #5  
"Lucan"
Dec 2006
England
2·3·13·83 Posts 
Quote:
There are 39! ways of sorting the nonhearts, and 40 positions to insert the heart suit. Probability = 40!/52! or 1 in 10^{20} David Last fiddled with by davieddy on 20070109 at 07:22 

20070109, 13:50  #6 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
11×389 Posts 
Actually, I was talking about the 13card hand you are dealt each round in the game of Hearts being sorted in some specific way.1 in 80 million trillion trillion trillion trillion trillion would be the chance that all four players in the game have a perfectly sorted hand dealt to them. I was thinking about one player.

20070111, 10:53  #7 
Einyen
Dec 2003
Denmark
19×181 Posts 
You can "pick" 13 cards out of 52 in:
So if you want 1 specific 13card hand its 1 in 635 billion. 
20070111, 17:08  #8  
"Jacob"
Sep 2006
Brussels, Belgium
1,907 Posts 
Since the order is important:
Quote:
The result is then 52! / (5213)! = 3954242643911239680000 or roughly 4*10^{21} But I am stil not completely sure about the exact question. 

20070111, 18:26  #9  
Einyen
Dec 2003
Denmark
19·181 Posts 
Quote:
Look at it this way: You have 52 cards and have to pick out 13. First card you can choose from 52 cards, 2nd time 51, then 50, and so on. The 13th card you can choose from 40 cards. Now you don't care if you get 2 of spades and then 3 of hearts or 3 of hearts and then 2 of spades, so you divide by 13! which is the number of diffrent sequences the same 13 cards can be in. Last fiddled with by ATH on 20070111 at 18:43 

20070111, 23:03  #10 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
11·389 Posts 
I'm not talking about which cards you get, I'm talking about the order in which you receive them.
Let me clarify my question in the following way: A shuffled deck has 52 cards numbered 152 You take 13 of those cards and put them in a pile. What are the odds/chances/probability (not sure which term to use) that when you pick them up, your (the 13card pile is "yours") lowest will be on the left, your next lowest to the right of that, and so on, all the way to the highest number in your group of cards being on the far right. 
20070112, 07:02  #11 
Einyen
Dec 2003
Denmark
19×181 Posts 
Ok, heh, so easy to get confused in this.
So a deck with 52 cards numbered 1 to 52, not a normal 4suit deck. You get 13 cards, don't care which 13 cards, just that you get them in order lowest to highest. Well the chance is always "number of favoured combinations" out of "number of total combinations". Now that you care about the order in which you get them, the total number of combinations is 52*51*.....*40 = 52!/(5213)! = 3954242643911239680000 as S485122 said. That is the number of ways you can get 13 cards out of 52, when you care about the order too. The number of "favoured combination" is = 635013559600, which is the number of unsorted 13card hands. So the chance is: 635013559600 in 3954242643911239680000 which is: 1 in 6,227,020,800 (= 13! ) Another way to see this is: Each different 13card hand can be sorted in 13! ways, and you want the only one sorted from lowest to highest, so 1 in 13!. Last fiddled with by ATH on 20070112 at 07:20 
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