Difference between revisions of "2019 AMC 10B Problems/Problem 22"
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On the first turn, each player starts off with <math>\$1</math> each. There are now only two situations possible, after a single move: either everyone stays at <math>\$1</math>, or the layout becomes <math>\$2-\$1-\$0</math> (in any order). Only <math>2</math> combinations end up with this outcome: <math>S-T-R</math> and <math>T-R-S</math>. On the other hand, given the interchangeability of the three people, <math>S-R-R</math>, <math>T-R-R</math>, <math>S-R-S</math>, <math>S-T-S</math>, <math>T-T-R</math>, and <math>T-T-S</math> can all be reproduced. Since each one of the scenarios is equally likely, there is a <math>\frac{2}{8} = \frac{1}{4}</math> chance to get the <math>2-1-0</math> type of format. | On the first turn, each player starts off with <math>\$1</math> each. There are now only two situations possible, after a single move: either everyone stays at <math>\$1</math>, or the layout becomes <math>\$2-\$1-\$0</math> (in any order). Only <math>2</math> combinations end up with this outcome: <math>S-T-R</math> and <math>T-R-S</math>. On the other hand, given the interchangeability of the three people, <math>S-R-R</math>, <math>T-R-R</math>, <math>S-R-S</math>, <math>S-T-S</math>, <math>T-T-R</math>, and <math>T-T-S</math> can all be reproduced. Since each one of the scenarios is equally likely, there is a <math>\frac{2}{8} = \frac{1}{4}</math> chance to get the <math>2-1-0</math> type of format. | ||
− | If one arrives at <math>1-1-1</math> at any point in time, we have essentially "cycled" back to the beginning. Similarly, if the setup becomes <math>2-1-0</math> (again, with <math>\frac{3}{4}</math> probability), assume <math>\textrm{WOLOG}</math> that <math>R</math> has <math>\$2</math>, player <math>S</math> received a <math>\$1 | + | If one arrives at <math>1-1-1</math> at any point in time, we have essentially "cycled" back to the beginning. Similarly, if the setup becomes <math>2-1-0</math> (again, with <math>\frac{3}{4}</math> probability), assume <math>\textrm{WOLOG}</math> that <math>R</math> has <math>\$2</math>, player <math>S</math> received a <math>\$1</math> amount, and participant <math>T</math> gets <math>\$0</math>. We can say that the possibilities are <math>S-T</math>, <math>S-R</math>, <math>T-R</math>, and <math>T-T</math>, which lead to the following combinations of <math>1-1-1</math>, <math>2-1-0</math>, <math>2-0-1</math>, and <math>1-0-2</math>, respectively. |
If one of the latter three are true, we return to the normal result, where the map remains at <math>\$2-\$1-\$0</math> or some variation thereof. If the first case holds, then their game simply returns to its initial base of <math>1-1-1</math>/. Either way, the probability of getting the <math>1-1-1</math> mixture has a <math>\frac{1}{4}</math> probability beyond round <math>n \in \mathbb N \geq 1</math>. | If one of the latter three are true, we return to the normal result, where the map remains at <math>\$2-\$1-\$0</math> or some variation thereof. If the first case holds, then their game simply returns to its initial base of <math>1-1-1</math>/. Either way, the probability of getting the <math>1-1-1</math> mixture has a <math>\frac{1}{4}</math> probability beyond round <math>n \in \mathbb N \geq 1</math>. | ||
− | + | The bell must ring at least once for this to be true, thus the correct response is <math>\boxed{\textbf{(B) }\frac{1}{4}}</math>. <math>\square</math> | |
− | <math> | ||
--anna0kear. | --anna0kear. |
Revision as of 22:46, 14 February 2019
- 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 $1. A bell rings every 15 seconds, at which time each of the players who currently has mondey simultaneously chooses one of the other two players independently and at random and gives $1 to that player. What is the probability that after the bell has rung 2019 times, each player will have $1?
For example, Raashan and Ted may each decide to give $1 to Sylvia, and Sylvia may decide to give her dollar to Ted, at which point Raashan will have $1, Sylvia will have $2, and Ted will have $1, and that is the end of the first round of play. In the second round Raashan has no money to give, but Sylvia and Ted might choose each other to give their $1 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. There are now only two situations possible, after a single move: either everyone stays at , or the layout becomes (in any order). Only combinations end up with this outcome: and . On the other hand, given the interchangeability of the three people, , , , , , and can all be reproduced. Since each one of the scenarios is equally likely, there is a chance to get the type of format.
If one arrives at at any point in time, we have essentially "cycled" back to the beginning. Similarly, if the setup becomes (again, with probability), assume that has , player received a amount, and participant gets . We can say that the possibilities are , , , and , which lead to the following combinations of , , , and , respectively.
If one of the latter three are true, we return to the normal result, where the map remains at or some variation thereof. If the first case holds, then their game simply returns to its initial base of /. Either way, the probability of getting the mixture has a probability beyond round .
The bell must ring at least once for this to be true, thus the correct response is .
--anna0kear.
Solution 2
On the first turn, each player starts off with each. Each turn after that, there are only two situations possible: either everyone stays at , or the distribution of money becomes , in any order .
(Note: means that gives his money to , gives her money to , and gives his money to .)
From the state, there are two ways to distribute the money so that it stays in a state: and . There are 6 ways to change the state to : , , , , , and . This means that the probability that the state stays is , and the probability that the state changes to is .
From the state, there is one way to change the state back to : . (We can assume that has , has , and has since only the distribution of money matters, not the specific people.) There are three ways to keep the state: , , . This means that the probability that the state changes to is , and the probability that the state stays is .
We can see that there will always be a chance that the money is distributed (as long as the bell rings once), so the answer is .
Solution 3
After each bell's ring, there are two situations: either they each have each, or one of them has , another has , and the third has . In each of these cases, we need to calculate the probability of returning to the state.
Case 1: Each player has . WLOG, let Raashan give his dollar to Sylvia. Then Sylvia must give her dollar to Ted and Ted must give his dollar to Raashan, which happens with probability.
Case 2: One player has , another has , and the third has . WLOG, let Raashan have , Sylvia have , and Ted have . Then Raashan must give his dollar to Sylvia and Sylvia must give her dollar to Ted, which happens with probability.
Since the probability of returning to the state is no matter what the situation is, the probability that each player will have after the bell rings times is .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.