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: . There are three ways to place the
in
. Now, there are two cases as to what happens next.
Sub-case 3.1: The in
goes directly under the
in
. There's obviously
way for that to happen. Then, there are
ways to permute the two pairs of
in
and
. (Either the
comes first in
or the
comes first in
.)
Sub-case 3.2: The in
doesn't go directly under the
in
. There are
ways to place the
, and
ways to do the same permutation as in Sub-case 3.1.
Hence, there are
ways for this case.
There's a grand total of ways for this to happen, along with
total cases. The probability we're asking for is thus
See Also
2017 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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