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Subject: "playing with permutations..."     Previous Topic | Next Topic
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zander
Charter Member
1 posts
Jul-07-01, 08:58 AM (EST)
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"playing with permutations..."
 
   First let me say that this is a neat site, I wish we would have had access like this when I was in school. We still did pretty good though...

My question has to do with probability and statistics...and a deck of playing cards.

Procedure is simple.

1) Count and Organize Cards to ensure 52 cards, 4 suits, 13 cards per suit

2) Shuffle Cards at least 10 times to try and generate as 'random' a deck as possible

3) Lay each card down into a grid whose axii are yaxis=suit , xaxis=card i.e.

Suit1
Suit2
Suit3
SUit4
a 2 3 4 5 6 7 8 9 10 j q k


Now starting with the first card on the top of the deck...

Turn the card and state it's position in the layout above as marked on the x axis and put card1 into (suit1,a), proceed with card2 in (suit1,2) ... (suit1,k).

Continue with (Suit2,a)...(suit(4,k)

When the entire deck has been laid out like this, count the number of times in the layout that he ACTUAL x-axis value of the card matches it's position. Relating anything to the y-axis is not involved in this question.

Here is my question: What are the odds, etc. of performing the above noted actions and find 0, (zero) matches at all?

I have done so exactly one time in move than I would estimate over 5 thousand casual attempts across the past few years.

p.s. Imagine the incredible jump the odds would take if the requirement were increased by including the data on the y axis as well!

Thanks for the help, I love playing with mathematics and concepts which can be 'brain-warping' at times... but who says that is bad


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Omar
Charter Member
4 posts
Jul-31-01, 11:02 PM (EST)
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1. "RE: playing with permutations..."
In response to message #0
 
   In Chapter Three of The Unexpected Hanging and Other Mathematical Diversions, Martin Gardner briefly discussed the case in which the card and the location must match on both rank and suit. He computed the odds of no matches to be approximately 1/e, and, of course, the odds of at least one hit to be approximately 1-1/e, that is about .6321. Unfortunately, his calculation was done by means of an (apparently well known) result from combinatorics that does all the hard work and hides the guts of the problem.


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alexb
Charter Member
672 posts
Jul-31-01, 11:22 PM (EST)
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2. "RE: playing with permutations..."
In response to message #1
 
   e-1 is an approximation to the relative number of derangements, i.e. permutations that move all elements with no exceptions.


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