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

#1**+12 **

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

#3**+10 **

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

#7**+5 **

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"

Nauseated Apr 21, 2015