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 Replies: 678 / Views: 16,159 |
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Valued Member
Canada
287 Posts |
Quote:
It's as if you cut the deck (or even can), you would find the cards within 2 of each other. It's like you laid out the cards in a circle with #1 next to #52. Anywhere in the circle you are looking for the two cards side by each, or separated by one card at the most. Correct? Edited by rikcando
10/24/2010 01:31 am
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Pillar of the Community
 United States
4000 Posts |
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Valued Member
Canada
287 Posts |
I have been working on the math and it just does not come anywhere close to what I am experiencing in person.
Looking for 4 possible arrangements in each of the 52! possibilities (using 2 cards leaves 50 cards in different arrangements) gives me 4*52*50!/52!= 7.8% chance of it happening.
Trying it 30 times it failed on only 4 shuffles.
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Pillar of the Community
United States
4000 Posts |
I think using 52! could be wrong. Wouldn't that be taking into consideration the suits?
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Valued Member
Canada
287 Posts |
This is the 4 possibilities I see
A=card #1
B=card #2
x=any other card
AxBx
BxAx
xABx
xBAx
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Valued Member
Canada
287 Posts |
Quote:
Wouldn't that be taking into consideration the suits? It would be 13! if we accounted for the suits. 52! includes all cards regardless of suit. |
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Valued Member
Canada
287 Posts |
I am thinking the it should be based on 54! since you allow a two card overlap at the beginning/end of the deck.
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Valued Member
Canada
287 Posts |
The 52! is the number of different ways to arrange the 52 cards.
The overlap would not give me more arrangements as I still have 52 cards, not 54. Cancel my previous comment.
Edited by rikcando
10/24/2010 01:48 am
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Pillar of the Community
United States
4000 Posts |
Quote:
AxBx
BxAx
xABx
xBAx
But that is limiting it to being within the first 4 cards of the deck. xxxxxxxxxxxxxAxBxxxxxxxxxxxx is also a winner. So, does this give you a rolling ppApp window 48 times through the deck? (p=poss 2nd card) |
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Valued Member
Canada
287 Posts |
Not 48. There are 52 different places at which you can find the 'A' due to the circle overlap.
The numerator in my formula has 4 possible arrangements of the two cards time the 52 places it may occur times the 50! arrangements of the other cards in the deck. This should give me the number of winning (favorable) arrangements of the cards.
The denominator give the total number of possible arrangements (52!).
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Pillar of the Community
United States
4000 Posts |
But, you've accounted for 4 positions already. xAxB where x = any card = 1, so 1*4*1*4 leaves 48.
And since xAxB and xBxA mean the same thing, times 2. Right?
Edited by Scooby Due
10/24/2010 02:18 am
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Valued Member
Canada
287 Posts |
Would it be correct to say that since there are 4 six's in the deck the possibility needs to be multiplied by 4. It could happen 4 time per arrangement. This would raise the chances of it happening up to 31.2% if so.
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Valued Member
Canada
287 Posts |
Quote:
But, you've accounted for 4 positions already. xAxB where x = any card = 1, so 1*4*1*4 leaves 48. I could have account for 5 positions if I typed ABxxx instead. Fact is there are only 2 card positions defined and the x is included in the random combination of cards. It may have been more correct to use AxB and only use 3 three spaces. |
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Valued Member
Canada
287 Posts |
I think I'll check back in in the morning. I need to get to bed. I'm hoping someone can come up with a (legitimate) formula that gives a value higher than 75%.
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Pillar of the Community
United States
4000 Posts |
I'm looking at it from this angle:
There are only 13*12 possible combinations that they can choose to be the winning combo.
13*12 = 156 possible selections. Of which, can come in 2 ways. AB or BA. so, 13*12*2 = 312 possible combos to look for.
Since AxB (and BxA) are also winners, are there are 4*1*4 more winning combinations?
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Replies: 678 / Views: 16,159 |
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Posted 10/24/2010 01:32 am
