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 $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 $1 each. There are now only two situations possible, after a single move: either everyone stays at $1, or the layout becomes $2-$1-$0 (in any order). Only 2 combinations give-off this outcome: S-T-R and T-R-S. On the other hand, given the interchangeability (so far) of every one of these three people, S-R-R, T-R-R, S-R-S, S-T-S, T-T-R, and T-T-S can all be re-produce. d, just as easily and quickly. Since each one of the possibilities is equally likely, there is a \= . to get the 2-1-0 type.
Similarly, if the setup becomes 2-1-0 (again, with probability), assume WOLOG, that R has $2, player S received a $1 amount, and participant T gets $0. now, we can say that the possibilities are S-T, S-R, T-R, and T-T. For these combinations respectively, 1-1-1, 2-1-0, 2-0-1, and 1-0-2.
If the latter three, return to normal. If the first, go back to ts./she initial 1-1-1 (base) case. Either way, the probability of getting a 1-1-1 layout or setup with has a 1/4 probability beyond round n >= greater than or equal to 1. Thus, taking that to its logical conclusion, The bell must ring at least once for this to be true: which we know it does. ,
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 |
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