AMC 10 · 2019 · #19
Grade 7 probabilityPick an answer.
Tool #9 (Easier Problem): 2019 is intimidating; replace with 1 ring, then 2, then ask whether the answer depends on the ring count. Tool #2 (Systematic List): list all 2 × 2 × 2 = 8 outcomes from state (1,1,1) to see which return to (1,1,1). Tool #15 (Reorganize): group the 6 ordered (2,1,0)-type states into one bucket by symmetry. Tool #5 (Pattern): once we compute the transition probabilities, the answer is the same regardless of starting state — so it is the same for all n ≥ 1.
Start from the even state
There are eight outcomes.
Just enumerate the 8 simultaneous choices and see which return to (1,1,1).
7.SP.C.8Make A Systematic ListEven to even
The chance of returning is one quarter.
Each player must pass in the same circular direction — only 2 of the 8 patterns do.
7.SP.C.8Make A Systematic ListUneven to even
That is one quarter as well.
From (2,1,0) only one of 4 simultaneous choices (A → B, B → C) restores equal shares.
7.SP.C.8Organize Information In More WaysThe same from any state
From anywhere the value is the same.
Whatever state we are in just before a ring, the chance of (1,1,1) immediately after is 1/4.
Whatever state the game is in just before a ring, the chance of an even split right after is the same.
▸ Why?
Every simultaneous choice is just as likely as any other, so the chance is a count over a count.
▸ Why?
Since the same chance appears at every ring, the number of rings cannot change the answer.
The count does not matter
Any number of rings gives the same answer.
Number of rings is a red herring — the per-step return probability is always 1/4.
7.SP.C.7Solve An Easier Related ProblemRead the answer
The probability is one quarter.
Match to the answer choice.
7.SP.C.7Eliminate PossibilitiesThis AMC 12 problem only needs Grade 7 probability lists — list the 8 possible exchanges from (1,1,1) and the 4 from (2,1,0), see that each lands back in (1,1,1) with probability 1/4, so 2019 doesn't matter — the answer is 1/4.