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# Carson flips over the cards of a standard 52-card deck one at a time. What is the probability that he flips over the ace of spades before an

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Carson flips over the cards of a standard 52-card deck one at a time. What is the probability that he flips over the ace of spades before any face card (jack, queen or king)?

Mellie  Apr 18, 2015

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I just used MathCad to work out your expression Melody.

Alan  Apr 18, 2015
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This is what I have come up with.  :/   [no guarantees that it is correct though]

Where on Earth are you digging these questions up from Mellie. :))

Now I have to work out how to calculate this without doing every bit separately.

Got any ideas Alan?  Do you know how to get the web2 calc to do the numerator?

$$\displaystyle\sum_{x=1}^{40}\left(\frac{39!}{(40-x)!}\times(52-x)!\right)\div 52!$$

Melody  Apr 18, 2015
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I got it, it is 1/13. i lost the solution.

Mellie  Apr 18, 2015
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There is 1 ace of spades and there are 12 face cards.  The other 39 cards are irrelevant!  So this reduces to the probability of choosing the ace of spades first out of a pack of 13.  This probability is just 1/13.

As it happens, Melody, your complicated expression evaluates to 1/13.

Alan  Apr 18, 2015
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Thanks Alan that makes sense.

How did you work out my expression?

Melody  Apr 18, 2015
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I just used MathCad to work out your expression Melody.

Alan  Apr 18, 2015
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ok thanks Alan :)

Melody  Apr 18, 2015
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Melody, if you like doing the Rube Goldberg of math equations, this script will solve it for you.

sum ((39!*(52-i)!)/(40-i)!) /(52!) from i=1 to 40

Click here to execute Melody’s Rube Goldberg math script

Here are some cool Rube Goldberg contraption videos

2014 winner Rube Goldberg contest

London's Science Museum most elaborate Rube Goldberg machine, "On The Move"

Nauseated  Apr 21, 2015
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Thanks Nauseated :))

"An intentionally delightful waste time and effort"  I love it.

That is what I do with my time Folks.   Isn't that a wonderful description.  :))

These Rude Goldburg displays are definitely worth the watch.   :))

Thanks for the Wolfram|Alpha solution too.

Melody  Apr 21, 2015