2017 AMC 12B Problems/Problem 22

Problem 22

Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In each round, four balls are placed in an urn---one green, one red, and two white. The players each draw a ball at random without replacement. Whoever gets the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins?

$\textbf{(A)}\quad \dfrac{7}{576} \qquad \qquad \textbf{(B)}\quad \dfrac{5}{192} \qquad\qquad \textbf{(C)}\quad \dfrac{1}{36} \qquad\qquad \textbf{(D)}\quad \dfrac{5}{144} \qquad\qquad\textbf{(E)}\quad \dfrac{7}{48}$

Solution

It amounts to filling in a $4 \times 4$ matrix. Columns $C_1 - C_4$ are the random draws each round; rows $R_A - R_D$ are the coin changes of each player. Also, let $\%R_A$ be the number of nonzero elements in $R_A$ .

WLOG, let $C_1 = \begin{pmatrix} 1\\-1\\0\\0\end{pmatrix}$ . Parity demands that $\%R_A$ and $\%R_B$ must equal $2$ or $4$ .

Case 1: $\%R_A = 4$ and $\%R_B = 4$ . There are $\binom{3}{2}=3$ ways to place $2$ $-1$ 's in $R_A$ , so there are $3$ ways.

Case 2: $\%R_A = 2$ and $\%R_B=4$ . There are $3$ ways to place the $-1$ in $R_A$ , $2$ ways to place the remaining $-1$ in $R_B$ (just don't put it under the $-1$ on top of it!), and $2$ ways for one of the other two players to draw the green ball. (We know it's green because Bernardo drew the red one.) We can just double to cover the case of $\%R_A = 4$ , $\%R_B = 2$ for a total of $24$ ways.

Case 3: $\%R_A=\%R_B=2$ . There are three ways to place the $-1$ in $R_A$ . Now, there are two cases as to what happens next.

Sub-case 3.1: The $1$ in $R_B$ goes directly under the $-1$ in $R_A$ . There's obviously $1$ way for that to happen. Then, there are $2$ ways to permute the two pairs of $1, -1$ in $R_C$ and $R_D$ . (Either the $1$ comes first in $R_C$ or the $1$ comes first in $R_D$ .)

Sub-case 3.2: The $1$ in $R_B$ doesn't go directly under the $-1$ in $R_A$ . There are $2$ ways to place the $1$ , and $2$ ways to do the same permutation as in Sub-case 3.1. Hence, there are $3(2+2 \cdot 2)=18$ ways for this case.

There's a grand total of $45$ ways for this to happen, along with $12^3$ total cases. The probability we're asking for is thus $\frac{45}{(12^3)}=\frac{5}{192}.$

Solution 2 (Less Casework)

We will proceed by taking cases based on how many people are taking part in this "transaction." We can have 2, 3, or 4 people all giving/receiving coins during the 4 turns. Basically, (like the previous solution), we are thinking this as filling out a 4x2 matrix of letters, where a letter on the left column represents this person gave, and a letter on the right column means this person recieved. We need to make sure that for each person that gave a certain amount, they received in total from other people that same amount, or in other words there are an equal number of A's, B's, C's, and D's on both columns of the matrix.

Case 1: $2$ people. In this case, we can 4C2 ways to choose the two people, and 6 ways to get order them to get a count of 6 * 6 = 36 ways.

Case 2: $3$ people. In this case, we have 4*(3C2)*4! = 288 ways to order 3 people.

Case 3: $4$ people. In this case, we have 3*3*4! = 216 ways to order 4 people.

So we have a total of 36+288+126=540 ways to order the four pairs of people. Now we divide this by the total number of ways - (4*3)^4 ( 4 times, 4 ways to choose giver, 3 to choose receiver). So the answer is 5/192.

~ccx09 (NOTE: Due to the poor quality of this solution, please PM me and I will explain the numbers, I have some diagrams but I can't show it here)

2017 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

Page

Toolbox

Search

2017 AMC 12B Problems/Problem 22

Contents

Problem 22

Solution

Solution 2 (Less Casework)

See Also