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

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==Solution== | ==Solution== | ||

− | Let the aces divide the 48 other cards into 5 "urns", with a, b, c, d, and e aces in them, respectively. | + | 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 <math>E[a+b+c+3]</math>. | + | 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>. |

By linearity of expectation, this is <math>E[a]+E[b]+E[c]+3</math>. | By linearity of expectation, this is <math>E[a]+E[b]+E[c]+3</math>. | ||

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The result is <math>3 \cdot \frac{48}{5} + 3 = \frac{159}{5} \implies \boxed{164}</math> | The result is <math>3 \cdot \frac{48}{5} + 3 = \frac{159}{5} \implies \boxed{164}</math> | ||

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==See also== | ==See also== |

## Revision as of 20:33, 14 December 2009

## 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 .

By linearity of expectation, this is .

Because the setup is symmetric between the five "urns", .

Since these must add to , .

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?)