Source: USATST-2017-P4
Category: Operations and Strategies

Problem:
You are cheating at a trivia contest. For each question, you can peek at each of the $n > 1$ other contestant's guesses before writing your own. For each question, after all guesses are submitted, the emcee announces the correct answer. A correct guess is worth $0$ points. An incorrect guess is worth $-2$ points for other contestants, but only $-1$ point for you, because you hacked the scoring system. After announcing the correct answer, the emcee proceeds to read out the next question. Show that if you are leading by $2^{n-1}$ points at any time, then you can surely win first place.

Solution:
We will prove the result with $2^{n-1}$ replaced by the stronger bound $2^{n-2}+1$. We first make the following reductions. First, change the weights to be $+1$, $-1$, $0$ respectively for a correct answer, an incorrect answer by another contestant, and an incorrect answer by you (rather than $0$, $-2$, $-1$); this has no effect on the relative score differences. Without loss of generality, assume that all contestants except you initially have a score of zero, and that your score exceeds $2^{n-2}$. We may ignore rounds in which all answers are the same. Finally, we can ignore rounds in which you get the correct answer, since that leaves you at least as well off relative to the other contestants as before. In other words, we will assume your score is always fixed, but you can pick any group of people with the same answer and ensure they lose $1$ point, while the group of people with the correct answer gains $1$ point.

The key observation is the following. Consider two rounds $R_1$ and $R_2$ such that: in round $R_1$, some set $S$ of contestants gains a point; in round $R_2$, the set $S$ of contestants all have the same answer. Then, if we copy the answers of contestants in $S$ during $R_2$ (which is an incorrect answer, as we ignore rounds where we answer correctly), they lose a point in $R_2$, which cancels out the point they gained in $R_1$. For any contestant not in $S$, they lost a point in $R_1$ and gain at most one point in $R_2$, so their net score change across the two rounds is at most $0$. In other words, the combined effect of $R_1$ and $R_2$ does not increase any contestant's score, so we can ignore $R_1$ and $R_2$ forever.

We thus consider the following strategy. We keep a list $\mathcal{L}$ of subsets of $\{1, \dots, n\}$, initially empty. On each round, we apply the following rules:
Suppose there exists a set $S$ of people with the same answer such that $S \in \mathcal{L}$. (If multiple such sets exist, choose one arbitrarily.) Then, copy the answer of $S$, causing them to lose a point. Delete $S$ from $\mathcal{L}$. (Importantly, we do not add any new sets to $\mathcal{L}$ in this case.)

Otherwise, copy the answer of any set $T$ of contestants, selecting $T$ such that $|T| \ge n/2$ if possible. Let $S$ be the set of contestants who answer correctly (if any), and add $S$ to the list $\mathcal{L}$. Note that $|S| \le n/2$; if a set of size at least $n/2$ exists, we chose $|T| \ge n/2$, and since $S$ is disjoint from $T$ (as we assume we did not copy the correct answer), $|S| \le n - |T| \le n/2$. If no set of size at least $n/2$ exists, then $|S| < n/2$ holds automatically.

By construction, $\mathcal{L}$ has no duplicate sets. So the score of any contestant $c$ is bounded above by the number of times that $c$ appears among sets in $\mathcal{L}$. The number of sets in $\mathcal{L}$ containing $c$ is at most the number of subsets of $\{1, \dots, n\}$ containing $c$ of size at most $n/2$. This is equivalent to the number of subsets of an $(n-1)$-element set of size at most $(n-2)/2$, which is at most $\frac{1}{2} \cdot 2^{n-1} = 2^{n-2}$. So, if you lead by $2^{n-2}+1$, then you ensure victory. This completes the proof.


Grading guidelines:
(Partial) 1. Makes the standard reduction to an equivalent "+1/-1/0" scoring (or otherwise argues that only score differences matter) and assumes WLOG that only other contestants’ scores need to be controlled while your score stays fixed. 2. States and correctly justifies the key two-round cancellation idea: if a set S gained in some earlier round and later all contestants in S give the same answer, then by copying that answer in the later round one can ensure the earlier gain of S is neutralized while no other contestant’s total increases across the two rounds. 3. Introduces the bookkeeping strategy of maintaining a list/multiset of previously correct sets S and: (i) when a current equal-answer group matches an S on the list, copy them and delete S; (ii) otherwise copy some large equal-answer group and add the current correct set S to the list, with an explanation why the added sets have size at most n/2. (Almost) 1. Presents the full list-of-sets strategy and proves an upper bound on every opponent’s score via counting appearances in the list, but has a small gap in the combinatorial estimate (e.g., bounding the number of subsets containing a fixed contestant of size ≤ n/2) or mishandles the parity/boundary case when n is odd. 2. Completes the argument that no opponent can ever exceed your initial lead (hence you win) but does not fully justify one reduction step (e.g., “ignore rounds where all answers are the same” or “ignore rounds where you answer correctly”) or does not clearly argue that duplicates cannot appear in the maintained list. 3. Solution is almost complete, but made minor mistakes which are not negligible.