Difference between revisions of "2019 AMC 12B Problems/Problem 16"
(Added solution; corrected redirect to wrong page number (Problem 19 on the test initially)) |
(Added solution; corrected redirect to wrong page number (Problem 19 on the test initially)) |
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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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− | <math> | + | 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. |
− | + | 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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Revision as of 20:43, 14 February 2019
On the first turn, each player starts off with 1, or the layout becomes
1-
\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}2, player S received a
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