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Difference between revisions of "2019 AMC 10B Problems/Problem 22"

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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>.
 
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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==Solution 2==
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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). Two different combinations produces that outcomes: only S-T-R, OR T-R-S turci kto do the turn. By working for a magical sorve? LS.
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==Solution 3==
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On the first turn, each player starts off with <math>1 each. There are now only two situations possible, after a single move: either everyone stays at </math>1, or the layout becomes <math>2-</math>1-<math>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}{3}<math>. to get the 2-1-0 type.
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Similarly, if the setup becomes 2-1-0 (again, with </math>\frac{3}{4}<math> probability), assume WOLOG, that R has </math>2, player S received a <math>1 amount, and participant T gets </math>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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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. QED <math>\square</math>
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==Solution 4==
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On the first turn, each player starts off with <math>1 each. There are now only two situations possible, after a single move: either everyone stays at </math>1, or the layout becomes <math>2-</math>1-<math>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 \frac{2}{8} \= \frac{1}{3} to get the 2-1-0 type.
 +
Similarly, if the setup becomes 2-1-0 (again, with #\frac{3}{4}# probability), assume WOLOG, that R has </math>2, player S received a <math>1 amount, and participant T gets </math>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 siestup with has a 1/4 probability beyond round n >= greater than or equal to 1. Thus, taking it to the's logical conclusion. that bell must ring at least once for this to be true, which we know it does. QED <math>\square</math>.
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==solution #5.==
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-==Solution 2==
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Either each person can have \$1 (case 1), or one person has \$2, one person has \$1, and one person has \$0 (case 2). No person can have \$3 because then they would have had to receive their own dollar which is not possible.<br>
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{|
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|
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! Case 1
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! Case 2
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|-
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! Case 1
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| <math>\frac{2}{27}</math>
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| <math>\frac{25}{27}</math>
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|-
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! Case 2
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| <math>\frac{1}{4}</math>
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| <math>\frac{3}{4}</math>
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|}
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Given that you start on the case in the left column, the number in the cell is the chance that you go to the case in the top row after 1 bell. Since both cases have a <math>\frac{1}{4}</math> chance or less to make case 1 the next case, the first two answer choices can be used. Since case 1 has less than a <math>\frac{1}{4}</math> chance to make case 1 the next case, the answer is <math>\textbf{(A) }\frac{1}{7}</math>.<br>
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<math>Q.E.D\blacksquare</math> <br>
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Solution by [[User:a1b2|a1b2]]
  
 
==See Also==
 
==See Also==

Revision as of 19:38, 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 $$1$. A bell rings every $15$ seconds, at which time each of the players who currently have money simultaneously chooses one of the other two players independently and at random and gives $$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 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 $\text{$1}$ each. Each turn after that, there are only two situations possible: either everyone stays at $\text{$1}$ $\text{(1-1-1)}$, or the distribution of money becomes $\text{$2-$1-$0}$, in any order $\text{(2-1-0)}$.

(Note: $\text{S-T-R}$ means that $\text{R}$ gives his money to $\text{S}$, $\text{S}$ gives her money to $\text{T}$, and $\text{T}$ gives his money to $\text{R}$.)

From the $\text{1-1-1}$ state, there are two ways to distribute the money so that it stays in a $\text{1-1-1}$ state: $\text{S-T-R}$ and $\text{T-R-S}$. There are 6 ways to change the state to $\text{2-1-0}$: $\text{S-R-R}$, $\text{T-R-R}$, $\text{S-R-S}$, $\text{S-T-S}$, $\text{T-T-R}$, and $\text{T-T-S}$. This means that the probability that the state stays $\text{1-1-1}$ is $\textstyle\frac{2}{8}=\frac{1}{4}$, and the probability that the state changes to $\text{2-1-0}$ is $\textstyle\frac{6}{8}=\frac{3}{4}$.

From the $\text{2-1-0}$ state, there is one way to change the state back to $\text{1-1-1}$: $\text{S-T-0}$. (We can assume that $\text{R}$ has $\text{$2}$, $\text{S}$ has $\text{$1}$, and $\text{T}$ has $\text{$0}$ since only the distribution of money matters, not the specific people.) There are three ways to keep the $\text{2-1-0}$ state: $\text{S-R-0}$, $\text{T-R-0}$, $\text{T-T-0}$. This means that the probability that the state changes to $\text{1-1-1}$ is $\textstyle\frac{1}{4}$, and the probability that the state stays $\text{2-1-0}$ is $\textstyle\frac{3}{4}$.

We can see that there will always be a $\textstyle\frac{1}{4}$ chance that the money is distributed $\text{1-1-1}$ (as long as the bell rings once), so the answer is $\boxed{\textbf{(B) }\frac{1}{4}}$.

~~Edited by IYN~~

Edited by DottedCaculator

Solution 2

After each bell ring, there are two situations: either they each have $\text{$1}$ each, or one of them has $\text{$2}$, another has $\text{$1}$, and the third has $\text{$0}$. In each of these cases, we need to calculate the probability of returning to the $\text{1-1-1}$ state.

Case 1: Each player has $\text{$1}$. 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 $\frac12 \cdot \frac12 = \frac14$ probability.

Case 2: One player has $\text{$2}$, another has $\text{$1}$, and the third has $\text{$0}$. WLOG, let Raashan have $\text{$2}$, Sylvia have $\text{$1}$, and Ted have $\text{$0}$. Then Raashan must give his dollar to Sylvia and Sylvia must give her dollar to Ted, which happens with $\frac12 \cdot \frac12 = \frac14$ probability.

Since the probability of returning to the $\text{1-1-1}$ state is $\frac14$ no matter what the situation is, the probability that each player will have $\text{$1}$ after the bell rings $2019$ times is $\boxed{\textbf{(B) }\frac{1}{4}}$.



Solution 2

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). Two different combinations produces that outcomes: only S-T-R, OR T-R-S turci kto do the turn. By working for a magical sorve? LS.

Solution 3

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-$0 (in any order). Only 2 combinations give-off this outcome: S-T-R and T-R-S. On the other hand, given the interchangeability (so far) of every one of these three people, S-R-R, T-R-R, S-R-S, S-T-S, T-T-R, and T-T-S can all be re-produce. d, just as easily and quickly. Since each one of the possibilities is equally likely, there is a$\frac{2}{8}$\=$ (Error compiling LaTeX. Unknown error_msg)\frac{1}{3}$. to get the 2-1-0 type.

Similarly, if the setup becomes 2-1-0 (again, with$ (Error compiling LaTeX. Unknown error_msg)\frac{3}{4}$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. QED $\square$

Solution 4

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-$0 (in any order). Only 2 combinations give-off this outcome: S-T-R and T-R-S. On the other hand, given the interchangeability (so far) of every one of these three people, S-R-R, T-R-R, S-R-S, S-T-S, T-T-R, and T-T-S can all be re-produce. d, just as easily and quickly. Since each one of the possibilities is equally likely, there is a \frac{2}{8} \= \frac{1}{3} to get the 2-1-0 type. Similarly, if the setup becomes 2-1-0 (again, with #\frac{3}{4}# probability), assume WOLOG, that R has$ (Error compiling LaTeX. Unknown error_msg)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 siestup with has a 1/4 probability beyond round n >= greater than or equal to 1. Thus, taking it to the's logical conclusion. that bell must ring at least once for this to be true, which we know it does. QED $\square$.


solution #5.

-==Solution 2==

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− Either each person can have $1 (case 1), or one person has $2, one person has $1, and one person has $0 (case 2). No person can have $3 because then they would have had to receive their own dollar which is not possible.

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− − −

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Case 1

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Case 2

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Case 1

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$\frac{2}{27}$

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$\frac{25}{27}$

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Case 2

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$\frac{1}{4}$

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$\frac{3}{4}$

−

− Given that you start on the case in the left column, the number in the cell is the chance that you go to the case in the top row after 1 bell. Since both cases have a $\frac{1}{4}$ chance or less to make case 1 the next case, the first two answer choices can be used. Since case 1 has less than a $\frac{1}{4}$ chance to make case 1 the next case, the answer is $\textbf{(A) }\frac{1}{7}$.

− $Q.E.D\blacksquare$

− Solution by a1b2

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. AMC logo.png

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