Difference between revisions of "2019 AMC 12B Problems/Problem 19"
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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> | ||
| − | ==Solution== | + | ==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 <math>\frac{2}{8} </math>\= <math>\frac{1}{3}</math>. to get the 2-1-0 type. | 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}{3}</math>. to get the 2-1-0 type. | ||
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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> | 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> | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | 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> | ||
| + | {| | ||
| + | | | ||
| + | ! Case 1 | ||
| + | ! Case 2 | ||
| + | |- | ||
| + | ! Case 1 | ||
| + | | <math>\frac{2}{27}</math> | ||
| + | | <math>\frac{25}{27}</math> | ||
| + | |- | ||
| + | ! Case 2 | ||
| + | | <math>\frac{1}{4}</math> | ||
| + | | <math>\frac{3}{4}</math> | ||
| + | |} | ||
| + | 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> | ||
| + | <math>Q.E.D\blacksquare</math> <br> | ||
| + | Solution by [[User:a1b2|a1b2]] | ||
==See Also== | ==See Also== | ||
| + | {{AMC10 box|year=2019|ab=B|before=unknown|after=unknown}} | ||
{{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}} | ||
Revision as of 17:02, 14 February 2019
Contents
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. QED 
Solution 2
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.
| Case 1 | Case 2 | |
|---|---|---|
| Case 1 | 
|
|
| Case 2 | 
|
|
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 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 chance to make case 1 the next case, the answer is 
.
Solution by a1b2
See Also
| 2019 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by unknown |
Followed by unknown | |
| 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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