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| − | ==Problem==
| + | #REDIRECT[[2019_AMC_10B_Problems/Problem_22]] |
| − | 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 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 \$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}\textbf{(B) }\frac{1}{4}\textbf{(C) }\frac{1}{3}\textbf{(D) }\frac{1}{2}\textbf{(E) }\frac{2}{3}</math>
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| − | ==Solution==
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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). Only 2 combinations produce 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 giveof:: s2-1-0.s. Since each one of the possibilities is equally likely, there is a \frac{2}{8} \= \frac{1}{3} of completing an exercise-inside,
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| − | Similarly, if the setup becomes 2-1-0 (again, with \frac{3}{4} probability), assume WOLOG, that R has \$2, player S a \$1 amount, and participant T :broke> lks So now, we can say that the possibilities are only, again-STR, TRS, which would (always) definitely, keep the balance.
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| − | ==See Also==
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| − | {{AMC12 box|year=2019|ab=As|num-b=18|num-a=20}}
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