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#11
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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#12
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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#13
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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#14
10-03-2007, 12:57 PM
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Re: Card Deck Arrangement Problem
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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. [/ QUOTE ] Does it matter though? As you move down the deck, when you do find the first A or K, it becomes more likely (since there are less remaining cards) that the next ones will follow. Think about it as if all the cards were arranged in a circle. The first one has to be somewhere, right? Isn't it always 3 in 51 that one of the next ones will be right behind it? (I don't doubt that my elementary approach to this problem is probably wrong for some reason, but even sober, I'm still not really seeing why. Also I don't understand the fancy math jargon, with the exclamation points and stuff, lol. So if you could explain your method in words, that would be appreciated.)
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#15
10-03-2007, 01:49 PM
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Re: Card Deck Arrangement Problem
[ QUOTE ]
[ 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. [/ QUOTE ] My mistake - even more embarassing as I can lay claim to a Math degree (a long time ago). Good stuff.
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#16
10-03-2007, 01:59 PM
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Re: Card Deck Arrangement Problem
[ QUOTE ]
Think about it as if all the cards were arranged in a circle. The first one has to be somewhere, right? Isn't it always 3 in 51 that one of the next ones will be right behind it? [/ QUOTE ] The problem is that they aren't arranged in a circle. If the "first" king is in position 52, then there is no chance that the next one will be right behind it. There is no card after 52.
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#17
10-03-2007, 09:46 PM
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Re: Card Deck Arrangement Problem
[ QUOTE ]
[ 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. [/ QUOTE ] Does it matter though? As you move down the deck, when you do find the first A or K, it becomes more likely (since there are less remaining cards) that the next ones will follow. Think about it as if all the cards were arranged in a circle. The first one has to be somewhere, right? Isn't it always 3 in 51 that one of the next ones will be right behind it? (I don't doubt that my elementary approach to this problem is probably wrong for some reason, but even sober, I'm still not really seeing why. Also I don't understand the fancy math jargon, with the exclamation points and stuff, lol. So if you could explain your method in words, that would be appreciated.) [/ QUOTE ] You are somewhat on the right track and I will make the problem simpler for explanation. Lets say we have 8 cards say A1, A2, K1, K2, Q1, Q2, J1, J2. Lets say we want the probability the two Aces are next to each other in the deck. Here is an answer: 2! * 7! / 8! = 1/4 In the denominator, we have 8! or the total number of ways to arrange a deck of 8 cards (or any 8 distinguishable objects put into 8 slots). 8! is "eight factorial" ( Wikipedia). 7! represents the possible arrangements of the new deck after we "glue together the 2 aces". 2! represents the number of ways to glue together the 2 aces. 2! = 2. Along your reasoning, you would say the chance the second ace is after the first ace is 1/7. The correct answer is 1/4 from above. If you are going to "look through the deck" the probability the second A follows the first A is conditional on where the first A is in the deck. Demonstration: 8 slots represent slots for cards in the deck: _ _ _ _ _ _ _ _ You have 7 cases to "condition" on where the first A is: 1) A ? _ _ _ _ _ _ The prob the first A is in slot 1 is 2/8. Prob the second A is in slot 2 is 1/7. The prob of this arrangement then is 2/8 * 1/7 = 2/56. + 2) _ A ? _ _ _ _ _ The prob there is not an A in slot 1 is (6/8) * prob there is an A in slot (2/7) * prob the remaining A is in slot 3 (1/6) = 2/56 + 3) _ _ A ? _ _ _ _ The prob there is not an A in slot 1 is (6/8) * prob there is not an A in slot 2 (given no ace in slot 1) is (5/7) * prob the first A is in slot 3 (2/6) * prob the second A is in slot 4 (1/5) = 2/56 + 4) _ _ _ A ? _ _ _ = 6/8 * 5/7 * 4/6 * 2/5 * 1/4 = 2/56 ... ... so you enumerate through the 7 possible cases and get 7 * (2/56) = 1/4. The probability the first Ace is in the say 30th slot is not 1/52 because to say so is making an assumption about what cards can be in the first 29 slots. Hope this helps, but if you want to learn more here are free probability books: Brief Wikibook on Probability Free Probability Book Chapters 1-3 of this book will serve as a good intro to probability (consider starting with 3 then going back to chatper 1).
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#18
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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#19
10-04-2007, 06:09 PM
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Re: Card Deck Arrangement Problem
Another way to get from 4!4!44! to 4!4!46!, which may make more sense to you...
Put the four aces in a pile on your left, the four kings on in a pile in front of you, and the 44 other cards in a pile on your right. Shuffle the cards within each pile. This gives you 4!4!44! ways. Now, take the four aces and stick them any of 45 places: on top of the 44-card pack, under the first card of the 44-card pack, ..., under the last card of the 44-card pack. (45x4!4!44!=4!4!45! ways, before the we choose where to put the four kings.) Now, take the four kings and stick them into in the 48-card pack. There would normally be 49 such places -- on top of the pack, under the first card, ..., under the last card, but three of these -- between the first and second aces, between the second and third aces, and between the third and fourth aces -- are off limits. so.. 4!4!44!x45x46 ways to assemble the 52-card deck from the three piles. Re your discovery that 45+44+43+42+...+42+43+44+45=2070. You might find this easier to understand if you count two separate ways. The kings can be UNDER the aces in 45 ways if the aces are 1234; in 44 ways if the aces are 2345; in 43 ways if the aces are 3456; ... in 3 ways if the aces are 43-46; in 2 ways if the aces are 44-47; in 1 way if the aces are 45-48; impossible if the aces are 46-49, 47-50, 48-51, 49-52. Total number of ways to have kings under aces: 45+44+43+...+3+2+1=(46x45/2) Then count the number of ways to kings over aces: 0+0+0+0+1+2+3+4+...+43+44+45=(46x45/2).
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