# Mock AIME 2 2006-2007 Problems/Problem 13

## 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 1

Let the aces divide the 48 other cards into 5 "urns", with a, b, c, d, and e non-aces between each of these 5 dividers, 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 $3 \cdot \frac{48}{5} + 3 = \frac{159}{5} \implies \boxed{164}$

## Solution 2 (sums)

Let the third ace be located in a certain deck position $m$. Then there are ${m-1\choose2}$ ways to place the first and second aces behind the third ace and $52-m$ ways to place the fourth ace after the third ace. We can use a summation to express this concisely: $\sum_{k=3}^{51}{k-1\choose2}(k)(52-k)$. This sum directly gives us the sum of all placements of all third aces between placements $3$ and $51$, inclusive. Note that we cannot place a third ace in places $1,2,52$, thus giving us the beginning and ending numbers for the summation. (The summation disregards suits and considers all aces to be the same. If we had regarded suits, we would have divided $4!$ by $4!$ at the end of the problem, so it would not have affected the end result.)

$\sum_{k=3}^{51}{k-1\choose2}(k)(52-k)=52\sum_{k=3}^{51}k{k-1\choose2}-\sum_{k=3}^{51}k^2{k-1\choose2}$

$=156\sum_{k=3}^{51}{k\choose3}-\sum_{k=3}^{51}\left((k+1)(k){k-1\choose2}-(k){k-1\choose2}\right)$

$=156\sum_{k=3}^{51}{k\choose3}-\sum_{k=3}^{51}(k+1)(k){k-1\choose2}+\sum_{k=3}^{51}(k){k-1\choose2}$

$=156\sum_{k=3}^{51}{k\choose3}-12\sum_{k=3}^{51}{k+1\choose4}+3\sum_{k=3}^{51}{k\choose3}$

$=159\sum_{k=3}^{51}{k\choose3}-12\sum_{k=3}^{51}{k+1\choose4}$

By the Hockey-Stick Identity, this simplifies nicely to $159{52\choose4}-12{53\choose5}$.

The total number of ways to place the four aces (again, disregarding suits) is ${52\choose4}$, so dividing the total sum by the total number of placements gives the answer.

$\frac{159{52\choose4}-12{53\choose5}}{{52\choose4}}=159-12\left(\frac{{53\choose5}}{{52\choose4}}\right)$

$=159-12\left(\frac{53!48!4!}{52!48!5!}\right)$

$=159-12\left(\frac{53}{5}\right)$

$=159-12\left(10\frac{3}{5}\right)$

$=159-12\cdot10-12\cdot\frac{3}{5}$

$=39-\frac{36}{5}$

$=31\frac{4}{5}$

$=\frac{159}{5}$.

Thus the answer is $159+5=\boxed{164}$.

~eevee9406