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 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?
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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 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.)
  
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.
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<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>
  
<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>
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==Solution 1==
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On the first turn, each player starts off with <math>\$1</math>. Each turn after that, there are only two possibilities: either everyone stays at <math>\$1</math>, which we will write as <math>(1-1-1)</math>, or the distribution of money becomes <math>\$2-\$1-\$0</math> in some order, which we write as <math>(2-1-0)</math>. (<math>(3-0-0)</math> 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 nothing to start with). We will consider these two states separately.
  
==Solution 1==
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In the <math>(1-1-1)</math> state, each person has two choices for whom to give their dollar to, meaning there are <math>2^3=8</math> possible ways that the money can be rearranged. Note that there are only two ways that we can reach <math>(1-1-1)</math> again:
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1. Raashan gives his money to Sylvia, who gives her money to Ted, who gives his money to Raashan.
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2. Raashan gives his money to Ted, who gives his money to Sylvia, who gives her money to Raashan.
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Thus, the probability of staying in the <math>(1-1-1)</math> state is <math>\frac{1}{4}</math>, while the probability of going to the <math>(2-1-0)</math> state is <math>\frac{3}{4}</math> (we can check that the 6 other possibilities lead to <math>(2-1-0)</math>)
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In the <math>(2-1-0)</math> state, we will label the person with <math>\$2</math> as person A, the person with <math>\$1</math> as person B, and the person with <math>\$0</math> as person C. Person A has two options for whom to give money to, and person B has 2 options for whom to give money to, meaning there are total <math>2\cdot 2 = 4</math> ways the money can be redistributed. The only way that the distribution can return to <math>(1-1-1)</math> is if A gives <math>\$1</math> to B, and B gives <math>\$1</math> to C. We check the other possibilities to find that they all lead back to <math>(2-1-0)</math>. Thus, the probability of going to the <math>(1-1-1)</math> state is <math>\frac{1}{4}</math>, while the probability of staying in the <math>(2-1-0)</math> state is <math>\frac{3}{4}</math>.
  
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 <math>\frac{2}{8} </math>\= <math>\frac{1}{4}</math>. to get the 2-1-0 type.
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No matter which state we are in, the probability of going to the <math>(1-1-1)</math> state is always <math>\frac{1}{4}</math>. This means that, after the bell rings 2018 times, regardless of what state the money distribution is in, there is a <math>\frac{1}{4}</math> probability of going to the <math>(1-1-1)</math> state after the 2019th bell ring. Thus, our answer is simply <math>\boxed{\textbf{(B) } \frac{1}{4}}</math>.
  
Similarly, if the setup becomes 2-1-0 (again, with <math>\frac{3}{4}</math> 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.
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==Solution 2 (Symmetry)==
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After the first ring, either nothing changes, or someone has <math>\$2</math>. No one can have <math>\$3</math>, since in that hypothetical round, that person would have to give away <math>\$1</math>. Thus, the outcome is either <math>1-1-1</math> or six symmetrical cases where one person gets <math>\$2</math> (e.g. a <math>1-2-0</math> or <math>2-1-0</math> split).  
  
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. <math>\boxed{\textbf{(B) }\frac{1}{4}}</math>,
 
  
==Solution 2==
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'''Case 1: Probability of returning to 1-1-1 from 1-1-1'''
  
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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There are two ways for the three people to exchange dollars to get to the same <math>1-1-1</math> result. To see this, seat R, S, and T in a circle. Each person gives their dollar to either the person at left, or at right, to result in again 1 dollar for each person. There are 8 overall possibilities (since each person has 2 choices when giving away his or her dollar, therefore <math>2^3</math> total possibilities). So, there is <math>1/4</math> chance of returning to <math>1-1-1</math>.  
  
(Note: <math>\text{S-T-R}</math> means that <math>\text{R}</math> gives his money to <math>\text{S}</math>, <math>\text{S}</math> gives her money to <math>\text{T}</math>, and <math>\text{T}</math> gives his money to <math>\text{R}</math>.)
 
  
From the <math>\text{1-1-1}</math> state, there are two ways to distribute the money so that it stays in a <math>\text{1-1-1}</math> state: <math>\text{S-T-R}</math> and <math>\text{T-R-S}</math>. There are 6 ways to change the state to <math>\text{2-1-0}</math>: <math>\text{S-R-R}</math>, <math>\text{T-R-R}</math>, <math>\text{S-R-S}</math>, <math>\text{S-T-S}</math>, <math>\text{T-T-R}</math>, and <math>\text{T-T-S}</math>. This means that the probability that the state stays <math>\text{1-1-1}</math> is <math>\textstyle\frac{2}{8}=\frac{1}{4}</math>, and the probability that the state changes to <math>\text{2-1-0}</math> is <math>\textstyle\frac{6}{8}=\frac{3}{4}</math>.
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'''Case 2: Probability of returning to 1-1-1 from 2-1-0'''
  
From the <math>\text{2-1-0}</math> state, there is one way to change the state back to <math>\text{1-1-1}</math>: <math>\text{S-T-0}</math>. (We can assume that <math>\text{R}</math> has <math>\text{\$2}</math>, <math>\text{S}</math> has <math>\text{\$1}</math>, and <math>\text{T}</math> has <math>\text{\$0}</math> since only the distribution of money matters, not the specific people.) There are three ways to keep the <math>\text{2-1-0}</math> state: <math>\text{S-R-0}</math>, <math>\text{T-R-0}</math>, <math>\text{T-T-0}</math>. This means that the probability that the state changes to <math>\text{1-1-1}</math> is <math>\textstyle\frac{1}{4}</math>, and the probability that the state stays <math>\text{2-1-0}</math> is <math>\textstyle\frac{3}{4}</math>.
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Without loss of generality, take the <math>2-1-0</math> case. Only 2 people can give money, so there are now <math>2^2=4</math> possible outcomes after the bell rings. It either decomposes back into <math>1-1-1</math>, <math>2-1-0</math> (remained unchanged), <math>2-0-1</math>, <math>1-0-2</math>. Thus, there is a <math>1/4</math> chance of returning to <math>1-1-1</math>. Notice that this works for any of the 6 cases, as each is symmetrical to the others.
  
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>.
 
  
==Solution 3==
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'''Answer'''
  
After each bell's ring, there are two situations: either they each have <math>\text{\$1}</math> each, or one of them has <math>\text{\$2}</math>, another has <math>\text{\$1}</math>, and the third has <math>\text{\$0}</math>. In each of these cases, we need to calculate the probability of returning to the <math>\text{1-1-1}</math> state.
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Since the starting state has a <math>1/4</math> chance of remaining unchanged, and each of the different 6 symmetric states all also have a <math>1/4</math> chance of reverting back to <math>1-1-1</math>, the chance of it being <math>1-1-1</math> after any state is always <math>\boxed{\textbf{(B) } \frac{1}{4}}</math>.
  
Case 1: Each player has <math>\text{\$1}</math>. 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 <math>\frac12 \cdot \frac12 = \frac14</math> probability.  
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==Solution 3==
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The two possible scenarios are they all have <math>1</math> dollar, or one person has <math>2</math> dollars, another has <math>1</math>, and the last has none. We will consider the second scenario all to be the same no matter who has the <math>2</math> dollars, <math>1</math> dollar or <math>0</math> dollars from symmetry.
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Let's consider all possible scenarios when the bell rings if they currently all have 1 dollar.
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<cmath>SRR -> 2, 1, 0</cmath>
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<cmath>SRS -> 2, 1, 0</cmath>
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<cmath>STR -> 1, 1, 1</cmath>
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<cmath>STS -> 2, 1, 0</cmath>
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<cmath>TRR -> 2, 1, 0</cmath>
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<cmath>TRS -> 1, 1, 1</cmath>
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<cmath>TTR -> 2, 1, 0</cmath>
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<cmath>TTS -> 2, 1, 0</cmath>
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We see that <math>\frac{2}{8}</math> or <math>\frac{1}{4}</math> of the cases lead to them continuing to all of them having <math>1</math> dollar.
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Now, let's consider all the possible scenarios when the bell rings if they currently have <math>2, 1, 0</math> dollars. Without loss of generality, let's say that R and S have <math>2</math> and <math>1</math> dollar respectively. (We can switch the names, our answer won't change).
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<cmath>SR -> 2, 1, 0</cmath>
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<cmath>ST -> 1, 1, 1</cmath>
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<cmath>TR -> 2, 1, 0</cmath>
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<cmath>TT -> 2, 1, 0</cmath>
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We see that <math>\frac{1}{4}</math> of the cases lead to them changing to all have <math>1</math> dollar.
  
Case 2: One player has <math>\text{\$2}</math>, another has <math>\text{\$1}</math>, and the third has <math>\text{\$0}</math>. WLOG, let Raashan have <math>\text{\$2}</math>, Sylvia have <math>\text{\$1}</math>, and Ted have <math>\text{\$0}</math>. Then Raashan must give his dollar to Sylvia and Sylvia must give her dollar to Ted, which happens with <math>\frac12 \cdot \frac12 = \frac14</math> probability.
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So, no matter what was the scenario when the bell had been rung <math>2018</math> times, when the bell is rung the <math>2019</math>th time, there is always a <math>\frac{1}{4}</math> chance that it will turn into (or stay as) <math>(1, 1, 1)</math>
  
Since the probability of returning to the <math>\text{1-1-1}</math> state is <math>\frac14</math> no matter what the situation is, the probability that each player will have <math>\text{\$1}</math> after the bell rings <math>2019</math> times is <math>\boxed{\textbf{(B) }\frac{1}{4}}</math>.
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==Video Solution==
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https://youtu.be/XT440PjAFmQ
  
 
==See Also==
 
==See Also==
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{{AMC12 box|year=2019|ab=B|num-b=18|num-a=20}}
 
{{AMC12 box|year=2019|ab=B|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}
remiving. sl; re-ordering. sl
 

Latest revision as of 12:07, 25 April 2021

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 $$1$. A bell rings every $15$ seconds, at which time each of the players who currently have money simultaneously chooses one of the other two players independently and at random and gives $$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 $$0$, Sylvia will have $$2$, and Ted will have $$1$, and that is the end of the first round of play. In the second round Rashaan has no money to give, but Sylvia and Ted might choose each other to give their $$1$ to, and the holdings will be the same at the end of the second round.)

$\textbf{(A) } \frac{1}{7} \qquad\textbf{(B) } \frac{1}{4} \qquad\textbf{(C) } \frac{1}{3} \qquad\textbf{(D) } \frac{1}{2} \qquad\textbf{(E) } \frac{2}{3}$

Solution 1

On the first turn, each player starts off with $$1$. Each turn after that, there are only two possibilities: either everyone stays at $$1$, which we will write as $(1-1-1)$, or the distribution of money becomes $$2-$1-$0$ in some order, which we write as $(2-1-0)$. ($(3-0-0)$ 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 nothing to start with). We will consider these two states separately.

In the $(1-1-1)$ state, each person has two choices for whom to give their dollar to, meaning there are $2^3=8$ possible ways that the money can be rearranged. Note that there are only two ways that we can reach $(1-1-1)$ 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 $(1-1-1)$ state is $\frac{1}{4}$, while the probability of going to the $(2-1-0)$ state is $\frac{3}{4}$ (we can check that the 6 other possibilities lead to $(2-1-0)$)


In the $(2-1-0)$ state, we will label the person with $$2$ as person A, the person with $$1$ as person B, and the person with $$0$ as person C. Person A has two options for whom to give money to, and person B has 2 options for whom to give money to, meaning there are total $2\cdot 2 = 4$ ways the money can be redistributed. The only way that the distribution can return to $(1-1-1)$ is if A gives $$1$ to B, and B gives $$1$ to C. We check the other possibilities to find that they all lead back to $(2-1-0)$. Thus, the probability of going to the $(1-1-1)$ state is $\frac{1}{4}$, while the probability of staying in the $(2-1-0)$ state is $\frac{3}{4}$.

No matter which state we are in, the probability of going to the $(1-1-1)$ state is always $\frac{1}{4}$. This means that, after the bell rings 2018 times, regardless of what state the money distribution is in, there is a $\frac{1}{4}$ probability of going to the $(1-1-1)$ state after the 2019th bell ring. Thus, our answer is simply $\boxed{\textbf{(B) } \frac{1}{4}}$.

Solution 2 (Symmetry)

After the first ring, either nothing changes, or someone has $$2$. No one can have $$3$, since in that hypothetical round, that person would have to give away $$1$. Thus, the outcome is either $1-1-1$ or six symmetrical cases where one person gets $$2$ (e.g. a $1-2-0$ or $2-1-0$ split).


Case 1: Probability of returning to 1-1-1 from 1-1-1

There are two ways for the three people to exchange dollars to get to the same $1-1-1$ result. To see this, seat R, S, and T in a circle. Each person gives their dollar to either the person at left, or at right, to result in again 1 dollar for each person. There are 8 overall possibilities (since each person has 2 choices when giving away his or her dollar, therefore $2^3$ total possibilities). So, there is $1/4$ chance of returning to $1-1-1$.


Case 2: Probability of returning to 1-1-1 from 2-1-0

Without loss of generality, take the $2-1-0$ case. Only 2 people can give money, so there are now $2^2=4$ possible outcomes after the bell rings. It either decomposes back into $1-1-1$, $2-1-0$ (remained unchanged), $2-0-1$, $1-0-2$. Thus, there is a $1/4$ chance of returning to $1-1-1$. Notice that this works for any of the 6 cases, as each is symmetrical to the others.


Answer

Since the starting state has a $1/4$ chance of remaining unchanged, and each of the different 6 symmetric states all also have a $1/4$ chance of reverting back to $1-1-1$, the chance of it being $1-1-1$ after any state is always $\boxed{\textbf{(B) } \frac{1}{4}}$.

Solution 3

The two possible scenarios are they all have $1$ dollar, or one person has $2$ dollars, another has $1$, and the last has none. We will consider the second scenario all to be the same no matter who has the $2$ dollars, $1$ dollar or $0$ dollars from symmetry.

Let's consider all possible scenarios when the bell rings if they currently all have 1 dollar. \[SRR -> 2, 1, 0\] \[SRS -> 2, 1, 0\] \[STR -> 1, 1, 1\] \[STS -> 2, 1, 0\] \[TRR -> 2, 1, 0\] \[TRS -> 1, 1, 1\] \[TTR -> 2, 1, 0\] \[TTS -> 2, 1, 0\] We see that $\frac{2}{8}$ or $\frac{1}{4}$ of the cases lead to them continuing to all of them having $1$ dollar. Now, let's consider all the possible scenarios when the bell rings if they currently have $2, 1, 0$ dollars. Without loss of generality, let's say that R and S have $2$ and $1$ dollar respectively. (We can switch the names, our answer won't change). \[SR -> 2, 1, 0\] \[ST -> 1, 1, 1\] \[TR -> 2, 1, 0\] \[TT -> 2, 1, 0\] We see that $\frac{1}{4}$ of the cases lead to them changing to all have $1$ dollar.

So, no matter what was the scenario when the bell had been rung $2018$ times, when the bell is rung the $2019$th time, there is always a $\frac{1}{4}$ chance that it will turn into (or stay as) $(1, 1, 1)$

Video Solution

https://youtu.be/XT440PjAFmQ

See Also

2019 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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