# Perverse Incentives

I do watch a couple of Let’s Plays on YouTube, and one game that a couple of channels have been playing is Trivia Murder Party from the Jackbox Party Pack. I’ve found the format to be interesting and relatively well-designed for building tension, though there are a few weird edge cases because of how the rules are laid out.

The final round has rules as follows:

• One player is in the lead (the player with the fewest number of spaces left to the finish line).
• Players are presented with a category, and need to decide if objects presented fit that category or not. The player in the lead has two objects; other players have three (this is a rubberbanding measure used to encourage them to catch up).
• For example, a category could be “Asian Countries” with objects Russia, Shanghai and the USA. In this case players should only select Russia (Shanghai is a city, and the USA is in North America).
• For each object correctly classified, players advance one space on the board. No penalties are issued for objects that were incorrectly classified.
• The player with the lead moves first. If he is across the finish line, he wins (independent of what non-leading players do).
• In the event of a tie, the highest ranked player (based on the previous rounds) who is not in the lead takes the lead. Furthermore, all tying players apart from the new leader (including the old leader, if relevant) move back 1 space.

One strategy of playing the final round, and an arguably sensible one is to always try and answer the questions to the best of one’s ability. Yet, the aforementioned rubberbanding mechanism means this isn’t always optimal – suppose you’re playing against an incredibly knowledgeable opponent who implements the maximising strategy. Furthermore, suppose you’re in the lead, with 5 spaces to go before finishing, while your opponent has 7. Then, consider:

• If you advance 2 spaces, you’ll have 3 spaces left to go whilst your opponent has 4. Then, no matter what you do, your opponent will take the lead on your next turn, and you’ll lose.
• If you advance 1 space and your opponent advances 3, you’ll both be at 4 spaces to go. Because of the tie rule, he leads and you get pushed back one space. This brings you to (5, 4). Now, as a non-leader advance 3 spaces; no matter what happens you retake the lead. Finally, advance 2 spaces and win.

We can generalise this to solve the two-player case, by considering what is a winning position – in the case where both players have very strong trivia knowledge, but the opponent always advances the maximum number of spaces.

• $(1, n), n \geq 2$ is clearly a winning position. Simply advance and exit.
• $(2, n), n \neq 1$ is also winning for the same reasons (and $(2, 1)$ is losing).
• $(3, n)$ is a bit more interesting. $n \leq 2$ is clearly lost. But $(3, 4)$ is actually lost too, because that becomes $(2, 1)$ or $(3, 1)$ after one turn, both of which are losing positions.  $(3, n), n \geq 5$ and greater are won; advance to $(1, n-3)$ and win.

Having considered a few cases, we can use what’s called induction to determine the rest of the winning positions. What this means is that we are defining winning positions in terms of whether other positions are winning. This can be dangerous if we produce circular definitions, but the relationships we use here are well-founded in that we maintain that the total number of spaces left to go for both players always decreases, for each of the sub-problems we look at.

We thus say that a position $(p, q)$ is winning, if and only if any of the following are true:

• $p = 1$, or $p = 2, q \neq 1$.
• $p > q + 1$ and at least one of $(p-3, q-2)$, $(p-2, q-2)$ and $(p-1, q-2)$ is winning.
• $p = q + 1$ and at least one of $(p-3, q-1)$, $(p-2, q-2)$ and $(p-1, q-2)$ is winning.
• $p = q - 1$ and at least one of $(p-1, q-3)$ and $(p, q-3)$ is winning.
• $p = q - 2$ and at least one of $(p-2, q-3)$ and $(p, q - 3)$ is winning.
• $p = q - 3$ and at least one of $(p-2, q-3)$, $(p-1, q-3)$ and $(p + 1, q-3)$ is winning.

We can derive a similar inductive rule for cases where one always maximises the number of answers: for each of the “at least one of” choices, always pick the first option. We can then consider when the (probably) “intended” strategy of trying to answer to the best of one’s knowledge fails to be optimal. In terms of implementation, the inductive definition above swiftly lends itself to a dynamic-programming implementation, whether “top-down” or “bottom-up”.

I’ve only considered the 2 player case because it’s relatively simple to analyse… things get substantially more complex once we go beyond that, especially factoring in the tiebreaking rules based on first-round performance. In practice, given that the questions can be difficult and unpredictable, at least for me it would probably be too risky to try to take a dive like this – but I can see this becoming relevant if expert players played this.