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 2007-01-06, 15:39 #1 TimSorbet Account Deleted     "Tim Sorbera" Aug 2006 San Antonio, TX USA 11·389 Posts Card Sorting Probability Given $x$ 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 $x^{x}$? 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?
 2007-01-06, 16:28 #2 Wacky     Jun 2003 The Texas Hill Country 44116 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 2007-01-06 at 16:29
 2007-01-08, 19:34 #3 ATH Einyen     Dec 2003 Denmark 19·181 Posts A deck with 52 cards can, as Wacky said, be sorted in 52! = 8*1067 ways. So if you want the chance for it to be sorted in 1 specific way, the chance is 1 in 8*1067 or 1 in 80 million trillion trillion trillion trillion trillion.
 2007-01-09, 00:40 #4 jasong     "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.
2007-01-09, 07:15   #5
davieddy

"Lucan"
Dec 2006
England

2·3·13·83 Posts

Quote:
 Originally Posted by jasong 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.
I don't think he was asking this , but...

There are 39! ways of sorting the non-hearts, and 40 positions to
insert the heart suit.
Probability = 40!/52! or 1 in 1020

David

Last fiddled with by davieddy on 2007-01-09 at 07:22

2007-01-09, 13:50   #6
TimSorbet
Account Deleted

"Tim Sorbera"
Aug 2006
San Antonio, TX USA

11×389 Posts

Actually, I was talking about the 13-card hand you are dealt each round in the game of Hearts being sorted in some specific way.
Quote:
 Originally Posted by ATH A deck with 52 cards can, as Wacky said, be sorted in 52! = 8*1067 ways. So if you want the chance for it to be sorted in 1 specific way, the chance is 1 in 8*1067 or 1 in 80 million trillion trillion trillion trillion trillion.
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.

 2007-01-11, 10:53 #7 ATH Einyen     Dec 2003 Denmark 19×181 Posts You can "pick" 13 cards out of 52 in: $\large\left(\begin{array}\\52\\13\end{array}\right)=\frac{52*51*50*49*48*47*46*45*44*43*42*41*40}{1*2*3*4*5*6*7*8*9*10*11*12*13}=635,013,559,600 ways$ So if you want 1 specific 13-card hand its 1 in 635 billion.
2007-01-11, 17:08   #8
S485122

"Jacob"
Sep 2006
Brussels, Belgium

1,907 Posts

Since the order is important:
Quote:
 the 13-card hand you are dealt each round in the game of Hearts being sorted in some specific way
you should not have divided by 13!

The result is then 52! / (52-13)! = 3954242643911239680000 or roughly 4*1021

But I am stil not completely sure about the exact question.

2007-01-11, 18:26   #9
ATH
Einyen

Dec 2003
Denmark

19·181 Posts

Quote:
 Actually, I was talking about the 13-card 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.
Mini-Geek is talking about Hearts game in windows, and there you get all your 13 cards at once, so I don't think he means the order in which you recieve the 13 cards. I think he means the chance you get all 13 hearts for example, and that is 1 in 635 billion.

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 2007-01-11 at 18:43

 2007-01-11, 23:03 #10 TimSorbet 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 1-52 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 13-card 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.
 2007-01-12, 07:02 #11 ATH 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 4-suit 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!/(52-13)! = 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 $\left(\begin{array}\\52\\13\end{array}\right)$ = 635013559600, which is the number of unsorted 13-card 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 13-card 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 2007-01-12 at 07:20

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