2019 AMC 10B Problems/Problem 22
- The following problem is from both the 2019 AMC 10B #22 and 2019 AMC 12B #19, so both problems redirect to this page.
Problem
Raashan, Sylvia, and Ted play the following game. Each starts with 
. A bell rings every 
seconds, at which time each of the players who currently have money simultaneously chooses one of the other two players independently and at random and gives to that player. What is the probability that after the bell has rung 
times, each player will have ? (For example, Raashan and Ted may each decide to give to Sylvia, and Sylvia may decide to give her her dollar to Ted, at which point Raashan will have 
, Sylvia will have 
, and Ted will have , and that is the end of the first round of play. In the second round Rashaan has no money to give, but Sylvia and Ted might choose each other to give their to, and the holdings will be the same at the end of the second round.)
Solution 1
On the first turn, each player starts off with . Each turn after that, there are only two possibilities: either everyone stays at , which we will write as 
, or the distribution of money becomes 
in some order, which we write as 
. (
cannot be achieved since either(1)the person cannot give money to himself or (2)there are a maximum of 2 dollars being distributed and the person has $0 to start with)We will consider these two states separately.
In the$ (Error compiling LaTeX. Unknown error_msg)(1-1-1)
2^3=8
(1-1-1)$again:
1. Raashan gives his money to Sylvia, who gives her money to Ted, who gives his money to Raashan.
2. Raashan gives his money to Ted, who gives his money to Sylvia, who gives her money to Raashan.
Thus, the probability of staying in the$ (Error compiling LaTeX. Unknown error_msg)(1-1-1)
\frac{1}{4}
(2-1-0)\frac{3}{4}
(2-1-0)$)
In the$ (Error compiling LaTeX. Unknown error_msg)(2-1-0)
$2
$1
$0
2\cdot 2 = 4
(1-1-1)
$1
$1
(2-1-0)
(1-1-1)\frac{1}{4}
(2-1-0)\frac{3}{4}$.
No matter which state we are in, the probability of going to the$ (Error compiling LaTeX. Unknown error_msg)(1-1-1)
\frac{1}{4}
\frac{1}{4}
(1-1-1)
\boxed{\textbf{(B) } \frac{1}{4}}$.
Solution 2 (Symmetry)
After the first ring, either nothing changes, or someone has . No one can have 
, since in that hypothetical round, that person would have to give away .
Thus, the outcome is either 
or six symmetrical cases where one person gets (e.g. a 
or 
split). There are two ways for the three people to exchange dollars to get to the same result. As such, there are 8 overall possibilities (which make sense, since each person has 2 choices when giving away his or her dollar, resulting in 
total possibilities). As such, from the case, there is a 
chance of returning to .
Without loss of generality, take the case. Only 2 people can give money, so there are now 
possible outcomes after the bell rings. It either decomposes back into , remains unchanged, turns into 
, or turns into 
. As such, from the case, there is a chance of returning to . Notice that this works for any of the 6 cases.
Since the starting state has a chance of remaining unchanged, and each of the different 6 symmetric states all also have a chance of reverting back to , the chance of it being 1-1-1 after any state is always 
Video Solution
See Also
| 2019 AMC 10B (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 10 Problems and Solutions | ||
| 2019 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 18 |
Followed by Problem 20 |
| 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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