Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 13"
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Revision as of 14:31, 3 April 2012
Contents
Problem
In his spare time, Richard Rusczyk shuffles a standard deck of 52 playing cards. He then turns the cards up one by one from the top of the deck until the third ace appears. If the expected (average) number of cards Richard will turn up is 
where 
and 
are relatively prime positive integers, find 
Solution
Let the aces divide the 48 other cards into 5 "urns", with a, b, c, d, and e non-aces in them, respectively.
The position of the third ace is equal to 
, and thus the expected value of its position is ![$E[a+b+c+3]$](http://latex.artofproblemsolving.com/d/9/d/d9d98e65f916f52c0de462117f090a33593dc9e0.png)
.
By linearity of expectation, this is ![$E[a]+E[b]+E[c]+3$](http://latex.artofproblemsolving.com/2/0/3/20315b5f13e725c498168e8d62a27f8112f109f5.png)
.
Because the setup is symmetric between the five "urns", ![$E[a]=\ldots = E[e]$](http://latex.artofproblemsolving.com/7/8/d/78d39d939ecd054d20972c97f2af2f5f1a137646.png)
.
Since these must add to ![$E[a+b+c+d+e]=48$](http://latex.artofproblemsolving.com/e/a/4/ea4c2e606d51d721da8e4261179e607e6da59484.png)
, ![$E[a]=\ldots=E[e]= \frac{48}{5}$](http://latex.artofproblemsolving.com/3/b/4/3b4fde1608a2c87e7cb9275d664b6456e6582748.png)
.
The result is 
See also
Problem Source
4everwise thought of this problem when watching Round 4 of the Professional Poker Tour. (What else can one do during the commercial breaks?) 
