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  #21  
Old 08-16-2005, 01:00 PM
pzhon pzhon is offline
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Default Re: A Similar Problem

[ QUOTE ]

What is the probability that you will get through the whole deck without ever dealing the same 2 cards simultaneously?

Then for an encore what is the probability for n Decks
(1 < n < 52)

[/ QUOTE ]
I'm not sure what your encore question means. Do you want the probability that on all 52 cards, every deck has a different card, or do you want the probability that not all decks coincide?

If the former, it is the number of 52xn Latin rectangles.
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  #22  
Old 08-16-2005, 01:07 PM
BruceZ BruceZ is offline
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Default Re: A Similar Problem

[ QUOTE ]
[ QUOTE ]

What is the probability that you will get through the whole deck without ever dealing the same 2 cards simultaneously?

Then for an encore what is the probability for n Decks
(1 < n < 52)

[/ QUOTE ]
I'm not sure what your encore question means. Do you want the probability that on all 52 cards, every deck has a different card, or do you want the probability that not all decks coincide?

If the former, it is the number of 52xn Latin rectangles.

[/ QUOTE ]

I assumed that we have 2 stacks with n decks in each stack, and that we want to go through all the cards, dealing a card from each stack, without ever dealing a pair of the same card.
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  #23  
Old 08-16-2005, 01:54 PM
LetYouDown LetYouDown is offline
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Default Re: A Similar Problem

For the initial problem...here's an approach to the entire rank-derangement problem. I haven't thoroughly looked at it yet, but it seemed sound. Font is a bit ugly and hard to read, however.

http://math.dartmouth.edu/~doyle/doc...rank/rank.html
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  #24  
Old 08-16-2005, 02:48 PM
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Default Re: A Similar Problem

Great link LetYouDown. I enjoyed reading the historical background. I guess I'm not the only one who likes this problem. Funny thing too is that I've considered pitching this game to a casino. I've come up with several ideas for it's presentation.

Anyway, Marilyn is a smart woman and I concur that the expected number of matches in 52 is 4. That was another thing I did with the program along with the earlier post I had with probability of no matches for each card. It goes like this

Matches %
0 = 1.6%
1 = 6.89%
2= 14.44%
3 = 19.8%
4 = 20.1%
5 = 16.1%
6 = 10.5%
7 = 5.8%
8 = 2.7%
9 = 1.1%
10 = 0.4%
11 = .14%
12 = .04%
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  #25  
Old 08-16-2005, 04:52 PM
BruceZ BruceZ is offline
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Default Re: A Similar Problem

[ QUOTE ]
Anyway, Marilyn is a smart woman and I concur that the expected number of matches in 52 is 4.

[/ QUOTE ]

That's obvious since the probability of a match on each card is 1/13, the average number of matches is 52*1/13 = 4.
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  #26  
Old 08-16-2005, 08:54 PM
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Default Re: A Similar Problem

[ QUOTE ]
I assumed that we have 2 stacks with n decks in each stack, and that we want to go through all the cards, dealing a card from each stack, without ever dealing a pair of the same card.

[/ QUOTE ]As the decks get large, doesn't this approach 1-1/e? (Assuming that "same card" means same rank and suit.)
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