2019 AMC 12B Problems/Problem 16

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 $\square$

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2019 AMC 12B Problems/Problem 16