Mock AIME 2 2006-2007 Problems/Problem 13


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 $m+n.$


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

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