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?
Solution
It amounts to filling in a 
matrix. Columns 
are the random draws each round; rows 
are the coin changes of each player. Also, let 
be the number of nonzero elements in 
.
WLOG, let 
. Parity demands that and 
must equal 
or 
.
Case 1: 
and 
. There are 
ways to place 
's in , so there are 
ways.
Case 2: 
and 
. There are ways to place the in , ways to place the remaining in 
(just don't put it under the on top of it!), and 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 , 
for a total of 
ways.
Case 3: 
. 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
