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

I just used MathCad to work out your expression Melody.  

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?

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

I got it, it is 1/13. i lost the solution.

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.

. 

Thanks Alan that makes sense.   

How did you work out my expression?

ok thanks Alan :)

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

https://www.youtube.com/watch?v=uF3nV0r87v8

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

https://www.youtube.com/watch?v=JmnRJ2M4eik

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.