How to (probably) Win Jeopardy! Using Game Theory
This past monday marked the start of 35th season of Jeopardy! – one of the most popular and storied game shows of all time. The first few episodes of this season have ended in 1 runaway victory and 2 where the winner was decided in the last round, known as “Final Jeopardy!” In Final Jeopardy!, contestants wager some portion of their earnings on one final question. Before the last commercial break, they are given the category of the question. At this point they make their wagers. When the break ends, contestants are given 30 seconds to answer as the iconic Jeopardy Theme plays in the background. Once players have written down their answers, the correct answer is revealed and their wager is added to their total if they get it right, or subtracted if it’s wrong. Whoever has the most money at the end of Final Jeopardy! wins and is welcomed back the following day. As a trivia fan and daily watcher of Jeopardy, I’ve had numerous debates with friends and family about the best Final Jeopardy strategy. If you’re in the lead but it’s not a runaway, should you bet it all or play it safe? (and hope the other contestant got it wrong or didn’t bet enough) After learning a bit about Game Theory in Networks, I did some research to see if there was a Game Theory approach to final Jeopardy that could settle the debate about the ultimate strategy once and for all.
I found an interesting article from 1994 that analyses 170 games of Jeopardy and uses game theory to determine the best strategy. In the section of the paper that applies game theory, the researchers only consider a Final Jeopardy! contest with two players. There are normally three players in this round, but if a contestant has a negative score (players are deducted points for getting a question wrong) then you are unable to participate in Final Jeopardy! (Often times the player in last has next to no chance of winning anyway.) This was done for simplicity’s sake to avoid a three-dimensional payoff matrix. For players A (the leader going into Final Jeopardy) and B (2nd place, lower score than A), they created an AxB payoff matrix where A and B are the possible wagers of the respective players (ranging from 0 to A or 0 to B). Each entry in this payoff matrix represents a possible combination of wagers, denoted as i and j respectively. There are three possible scenarios for a Final Jeopardy between two players:
1) Player A has more than double the score of player B, giving player B no chance of winning, since B’s maximum wager + current winnings is less than A’s current score. Player A can bet nothing and still win. There is a clear dominant strategy for A: wager nothing. (Player A could also bet any value less than A-2B if they want to take a risk to increase their final score, but this paper only considers victories, not the amount by which a player wins)
2) Player A’s score is less than double B’s, but still greater than 1.5 times B’s score. In this scenario, player A can bet $(2B – A + 1) to win in 2 of 3 possible outcomes. If they get the question right, there is no way for B to win. For example, if player A has $12,000 and B has $7,000, player A can bet 2($7,000) – $12,000 + $1 = $2,001. If player A and B both get the question right, no matter what B’s wager is, their maximum possible score is $14,000. If player A gets it right with this wager, they will end up with $14,001 and win the game. If both players get it wrong, then Player A will win as well since $9,999 > $7000. The payoff matrix for this scenario shows that for player A, betting $(2B – A + 1) is the dominant strategy. It results in a win for A in 2 of 3 possible outcomes. A only loses if they get the question wrong and B bets enough. Since $(2B – A + 1) is the dominant strategy for Player A, player B should ensure that they can still win if B loses so player B’s dominant strategy is to wager everything.
3) Scenario three is where Player A’s score is less than 1.5 times B’s score. This is interesting because Player A is not guaranteed victory in 2 of 3 possible outcomes given a dominant strategy like in scenario 2, regardless of wager. In fact, the paper shows that there is no dominant strategy for this scenario. The paper instead turns to finding mixed strategy equilibria, assuming that the players each knew the probabilities of the other player wagering a certain amount. (This is quite the assumption since all players but the previous day’s winner are new to the show, but perhaps this could be the Tournament of Champions where previous winners have been called back to participate and all players are returning players, so the other players have an idea of how they bet.) I apologize if this is a bit anticlimactic, but the results showed 3 possible equilibria that are a bit too complicated to summarize in this blog post. In short, there really isn’t a clear strategy for this scenario if you don’t know the betting behaviors of your opponent. You sort of have to go with your gut and bet based on how you well you think you and your opponents know the category.
If you’d like to read more about the results for the final scenario, a payoff matrix with three contestants, or are generally interested in learning more, I highly recommend checking out this paper. I’ve cited and linked it below. Happy playing!
Gilbert, George T., and Rhonda L. Hatcher. “Wagering in Final Jeopardy!” Mathematics Magazine, vol. 67, no. 4, 1994, pp. 268–277. JSTOR, JSTOR, www.jstor.org/stable/2690846.