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Difference between revisions of "2017 AMC 12B Problems/Problem 22"

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==Solution==
 
==Solution==
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It amounts to filling in a 4x4 matrix M. Columns C1 - C4 are the random draws each round; rows RA -RD are the coin changes of each player. Also, let <RA>=number of non-0 elements in RA.
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It amounts to filling in a <math>4 \times 4</math> matrix. Columns <math>C_1 - C_4</math> are the random draws each round; rows <math>R_A - R_D</math> are the coin changes of each player. Also, let <math>\%R_A</math> be the number of nonzero elements in <math>R_A</math>.
  
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WLOG, let C1=[1,-1,0,0]. Parity demands that <RA> and<RB>= 1 or 3.  
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WLOG, let <math>C_1 = \begin{pmatrix} 1\\-1\\0\\0\end{pmatrix}</math>. Parity demands that <math>\%R_A</math> and <math>\%R_B</math> must equal <math>2</math> or <math>4</math>.  
  
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CASE 1. <RA>=3, and <RB>=3. 3 choices for -1in RA, then everything is determined. So 3 subcases here.
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Case 1: <math>\%R_A = 4</math> and <math>\%R_B = 4</math>. There are <math>\binom{3}{2}=3</math> ways to place <math>2</math> <math>-1</math>'s in <math>R_A</math>, so there are <math>3</math> ways.
  
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Case 2. <RA>=1, and <RB>=3. 3 Choices for the -1in RA, 2 choices for the rest of 1 -1 in RB, 2 choices between C and D to fill in the remaining pair of 1 and -1. Then double for symmetrical case of <RA>=3 and <RB>=1. Total 24 subcases.
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Case 2: <math>\%R_A = 2</math> and <math>\%R_B=4</math>. There are <math>3</math> ways to place the <math>-1</math> in <math>R_A</math>, <math>2</math> ways to place the remaining <math>-1</math> in <math>R_B</math> (just don't put it under the <math>-1</math> on top of it!), and <math>2</math> 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 <math>\%R_A = 4</math>, <math>\%R_B = 2</math> for a total of <math>24</math> ways.
  
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Case 3. <RA>=1, and <RB>=1. 3 Choices for -1in RA. Then (1)1 in RB goes directly under 1 in RA. Then 2 choices to assign remaining two pairs of 1, -1in RC and RD. Or (2) 1 in RB goes not directly under 1 in RA: 2 choices here. Then 2 choices to finish. Hence 3(2+2×2)=18.
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Case 3: <math>\%R_A=\%R_B=2</math>. 3 Choices for -1in RA. Then (1)1 in RB goes directly under 1 in RA. Then 2 choices to assign remaining two pairs of 1, -1in RC and RD. Or (2) 1 in RB goes not directly under 1 in RA: 2 choices here. Then 2 choices to finish. Hence 3(2+2×2)=18.
  
 
In sum, 3+24+18=45 cases. Probability= 45÷(12^3)=5/192.
 
In sum, 3+24+18=45 cases. Probability= 45÷(12^3)=5/192.

Revision as of 14:00, 10 September 2017

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: $\%R_A=\%R_B=2$. 3 Choices for -1in RA. Then (1)1 in RB goes directly under 1 in RA. Then 2 choices to assign remaining two pairs of 1, -1in RC and RD. Or (2) 1 in RB goes not directly under 1 in RA: 2 choices here. Then 2 choices to finish. Hence 3(2+2×2)=18.

In sum, 3+24+18=45 cases. Probability= 45÷(12^3)=5/192.

See Also

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

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