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# Card Counting

Best Answer bonanova, 29 April 2013 - 11:39 AM

Spoiler for Comparing the two approaches:

Go to the full post

19 replies to this topic

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Posted 25 April 2013 - 10:09 PM

A standard pack of cards is thrown into the air in such a way that each card, independently, is equally likely to land face up or face down. The total value of the cards which landed face up is then calculated. (Card values are assigned as follows: Ace=1, 2=2, ... , 10=10, Jack=11, Queen=12, King=13. There are no jokers.)

What is the probability that the total value is divisible by 13?

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### #2 k-man

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Posted 25 April 2013 - 10:37 PM

Spoiler for first thought without doing any calculations

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Posted 25 April 2013 - 11:00 PM

I am extremely tempted to mark your answer solved. It is very close. I never looked at the approach like that before for this problem. It gives a very very good approximation one that you have to go out about 8 decimal places to see a difference. I will wait and see if someone can find the more precise answer.

Spoiler for first thought without doing any calculations

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### #4 bonanova

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Posted 26 April 2013 - 09:10 AM

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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

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Posted 26 April 2013 - 08:00 PM

The answer is even closer so if no one can find an even more precise answer then this will be the answer.  And before I get some guessess with a bunch of nines, the answer that i found is not consisting solely of running nines after the decimal.

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### #6 bonanova

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Posted 26 April 2013 - 08:50 PM

The answer is even closer so if no one can find an even more precise answer then this will be the answer.  And before I get some guessess with a bunch of nines, the answer that i found is not consisting solely of running nines after the decimal.

It would repeat, being rational.
I just tried to investigate its closeness to 1/13.
I'm still thinking about the small excess count for multiples.
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

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Posted 26 April 2013 - 09:34 PM

Spoiler for hint

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### #8 superprismatic

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Posted 26 April 2013 - 11:09 PM

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Posted 27 April 2013 - 03:07 AM

Spoiler for hint 2

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### #10 bonanova

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Posted 27 April 2013 - 10:47 AM

Spoiler for Better approximation

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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
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