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

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==Solution==
 
==Solution==
{{solution}}
 
  
==See also==
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Let the aces divide the 48 other cards into 5 "urns", with a, b, c, d, and e non-aces in them, respectively.
*[[Mock AIME 2 2006-2007/Problem 12 | Previous Problem]]
 
  
*[[Mock AIME 2 2006-2007/Problem 14 | Next Problem]]
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The position of the third ace is equal to <math>a+b+c+3</math>, and thus the expected value of its position is <math>E[a+b+c+3]</math>.
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By linearity of expectation, this is <math>E[a]+E[b]+E[c]+3</math>.
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Because the setup is symmetric between the five "urns", <math>E[a]=\ldots = E[e]</math>.
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Since these must add to <math>E[a+b+c+d+e]=48</math>, <math>E[a]=\ldots=E[e]= \frac{48}{5}</math>.
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The result is <math>3 \cdot \frac{48}{5} + 3 = \frac{159}{5} \implies \boxed{164}</math>
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==See Also==
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{{Mock AIME box|year=2006-2007|n=2|num-b=12|num-a=14}}
  
*[[Mock AIME 2 2006-2007]]
 
  
  
 
== Problem Source ==
 
== 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?) [[Image:Razz.gif]]
 
4everwise thought of this problem when watching Round 4 of the Professional Poker Tour. (What else can one do during the commercial breaks?) [[Image:Razz.gif]]

Latest revision as of 10:53, 4 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

Mock AIME 2 2006-2007 (Problems, Source)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15


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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