Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 13"

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==See also==
 
==See also==
*[[Mock AIME 2 2006-2007/Problem 12 | Previous Problem]]
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*[[Mock AIME 2 2006-2007 Problems/Problem 12 | Previous Problem]]
  
*[[Mock AIME 2 2006-2007/Problem 14 | Next Problem]]
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*[[Mock AIME 2 2006-2007 Problems/Problem 14 | Next Problem]]
  
 
*[[Mock AIME 2 2006-2007]]
 
*[[Mock AIME 2 2006-2007]]

Revision as of 15:32, 3 April 2012

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 $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$

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 $a+b+c+3$, and thus the expected value of its position is $E[a+b+c+3]$.

By linearity of expectation, this is $E[a]+E[b]+E[c]+3$.

Because the setup is symmetric between the five "urns", $E[a]=\ldots = E[e]$.

Since these must add to $E[a+b+c+d+e]=48$, $E[a]=\ldots=E[e]= \frac{48}{5}$.

The result is $3 \cdot \frac{48}{5} + 3 = \frac{159}{5} \implies \boxed{164}$

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?) Razz.gif

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