Difference between revisions of "2019 AMC 10B Problems/Problem 22"
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==Problem== | ==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 | + | Raashan, Sylvia, and Ted play the following game. Each starts with <math> \$1</math>. A bell rings every <math>15</math> 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 <math>\$1</math> to that player. What is the probability that after the bell has rung <math>2019</math> times, each player will have <math>\$1</math>? (For example, Raashan and Ted may each decide to give <math>\$1</math> to Sylvia, and Sylvia may decide to give her her dollar to Ted, at which point Raashan will have <math>\$0</math>, Sylvia will have <math>\$2</math>, and Ted will have <math>\$1</math>, 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 <math> \$1</math> to, and the holdings will be the same at the end of the second round.) |
− | + | <math>\textbf{(A) } \frac{1}{7} \qquad\textbf{(B) } \frac{1}{4} \qquad\textbf{(C) } \frac{1}{3} \qquad\textbf{(D) } \frac{1}{2} \qquad\textbf{(E) } \frac{2}{3}</math> | |
− | + | ==Solution== | |
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− | ==Solution | ||
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On the first turn, each player starts off with <math>\text{\$1}</math> each. Each turn after that, there are only two situations possible: either everyone stays at <math>\text{\$1}</math> <math>\text{(1-1-1)}</math>, or the distribution of money becomes <math>\text{\$2-\$1-\$0}</math>, in any order <math>\text{(2-1-0)}</math>. | On the first turn, each player starts off with <math>\text{\$1}</math> each. Each turn after that, there are only two situations possible: either everyone stays at <math>\text{\$1}</math> <math>\text{(1-1-1)}</math>, or the distribution of money becomes <math>\text{\$2-\$1-\$0}</math>, in any order <math>\text{(2-1-0)}</math>. | ||
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We can see that there will always be a <math>\textstyle\frac{1}{4}</math> chance that the money is distributed <math>\text{1-1-1}</math> (as long as the bell rings once), so the answer is <math>\boxed{\textbf{(B) }\frac{1}{4}}</math>. | We can see that there will always be a <math>\textstyle\frac{1}{4}</math> chance that the money is distributed <math>\text{1-1-1}</math> (as long as the bell rings once), so the answer is <math>\boxed{\textbf{(B) }\frac{1}{4}}</math>. | ||
− | + | ~~Edited by IYN~~ | |
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− | + | Edited by [[User:DottedCaculator|DottedCaculator]] | |
==See Also== | ==See Also== |
Revision as of 00:25, 15 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 . 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
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 .
~~Edited by IYN~~
Edited by DottedCaculator
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.