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Basic Probability Question

Posted on 2006-06-29
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My wife is going back to college for a Master's degree and is taking some probability courses.  I took a lot of Probability and statistics back in school, but I haven't used any of it in 5 years.  She has a question, and I want to help, but I think I'm missing something.

[Question]

Helen and Adiana are playing cards.  They each shuffle a standard deck of cards thoroughly, demonstrating their fanciest casino-style moves.  What is the probability that both decks are in the EXACT same order after all that shuffling?


My current answers are 1/(52!*52!) or 52/(52!*52!), but I feel I am doing something wrong.  This is a pretty basic course.  So far they have only used the basic permutation formula and some tree diagrams.  I don't they have gotten into conditional probabilities yet.



One final thought... I tried to work backwards and figure out what the probability of them *NOT* matching was and then subract it from one.   I don't think that worked so well for me.



If Helen has a probability of 1/52! of getting a specific order (any order), then Adiana has a probability of (52!-1)/52! of getting anything EXCEPT the order that helen got.

if (52!-1)/52! is anything EXCEPT the order that Helen got, then 1 - [(52!-1)/52!] is the exact order that Helen got.  
   Which reduces to -->  1/52!  and that can't be right.
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Question by:_TAD_
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d-glitch earned 500 total points
ID: 17012499
It is just 1/(52!)

It doesn't matter what the first deck is.  The second deck has to match it.
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by:d-glitch
ID: 17012530
The probability of the first card matching in each deck is 1/52.

Given that the first card did match, the probability of the second card matching is 1/51.

And so on...
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by:_TAD_
ID: 17012719

Gosh... and to think all this time I thought I forgot everything I learned in college.


I came up with the 1/52! in the first two minutes, and then spent the next 40 min thinking I was wrong (and trying to figure out the "right" answer).


Thanks for the clarification!
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