Optimal Final Jeopardy wagering

On the 6/3/19 episode of Jeopardy, which ended up being James Holzhauer’s final episode, the following scenario arose for Final Jeopardy. The scores were $11,000 for Jay, $23,400 for James, and $26,600 for Emma, with final jeopardy category “Shakespeare’s Time.”

I created a game-theoretic model of this situation to determine the Nash equilibrium wagering strategies for all players. The model is based on the following assumptions and simplifications:

1. Assume that Emma and James will get the answer correct with probability 0.9, and Jay will with probability 0.7 (someone gave me these values as estimates).
2. Assume that the players only care about winning and not their final score. Obviously the players do have some incentive to also be maximizing their winnings, as the champion receives the amount of the current winning plus returns for possible future winnings; however, it is difficult to model this without knowing how much each player values the future. Clearly for James for example he expects the winnings in the future to radically dwarf winning an extra $10,000 in the current episode, though this may not necessarily be the case for the other contestants who are not as confident about their ability to succeed in the future. I believe that for all players the value of continuing on as champion is almost definitely significantly greater than the value of a slightly larger margin of victory, so I think in addition to simplifying the model this assumption is also pretty accurate.
3. Assume that the payoff for finishing in first place is 1 and second/third place is 0. If there is a tie for first then both players would get 1/2, and if there is a 3-way tie then 1/3. (This agrees with the rules according to Wikipedia, which states that only one contestant can continue on to the next episode): “Since November 2014,[29] ties for first place following Final Jeopardy! are broken with a tie-breaker clue, resulting in only a single champion being named, keeping their winnings, and returning to compete in the next show. The tied contestants are given the single clue, and the first contestant to buzz-in must give the correct question. A contestant cannot win by default if the opponent gives an incorrect question; that contestant must give a correct question to win the game. If neither player gives the correct question, another clue is given.[30]” https://en.wikipedia.org/wiki/Jeopardy!
4. For computational tractability I will divide all the values by $100 (so that the scores are $266, $234, and $110), and assume that players can wager any multiple of $1. While this does not fully model the real game, it should be pretty obvious how to translate the solution to this simplified model back to the original situation.

I have computed the following Nash equilibrium for this game:

Emma’s strategy: wager $203 with probability 0.992, wager $18 with probability 0.008.

James’ strategy: wager $13 with probability 0.843, wager $77 with probability 0.152, wager $234 with probability 0.005.

Jay’s strategy: wager $110 with probability 0.993, wager $46 with probability 0.007.

Note a few caveats about interpreting these results. If any of the modeling assumptions (e.g., the 0.9 and 0.7 correctness probabilities) are changed, the solution may change radically. Also, Nash equilibrium essentially assumes that all players are perfectly rational, and are assuming the others are also perfectly rational, etc. If the players have reason to think the other contestants are playing a particular “irrational” strategy (or have a different model of the game) then they should pick a strategy that they expect would perform best against those specific strategies of the opponents rather than playing Nash equilibrium. In reality it is likely better to play such an exploitative strategy, but I think that it is still interesting and informative to examine the Nash equilibrium regardless.

Also note that there is a small degree of approximation error in the solutions and it is not exact (no player can gain more than 0.0003 by deviating from these strategies).

Finally note that Nash equilibrium is a less compelling concept in three-player games than for two-player (zero-sum) games (such as two-player poker) in which a Nash equilibrium essentially guarantees “unbeatability.” In three-player games it is possible for collusion, for a player’s deviation to hurt other players in addition to him or herself, etc., and there is no theoretical guarantee of following a Nash equilibrium. (For elaboration on this distinction between 3-player and two-player zero-sum games, see this previous post https://medium.com/ganzfried-gleans/3-player-kuhn-poker-project-ce1b818b4d5e and journal paper https://www.mdpi.com/2073-4336/9/2/33).

In reality, Jay wagered $6000, James wagered $1399, and Emma $20,201. James and Emma’s strategies agreed with their maximal probability Nash equilibrium wager, while Jay chose a smaller size. (Note that James can’t always wager such a small amount, or it would allow Emma to theoretically exploit him by wagering $0 and ensure a victory, which explains why James must still wager the larger amounts of $77 and $234 with some probability.)

The full episode is available here (final jeopardy starts at 16:54) https://www.youtube.com/watch?v=CwgdVoDLHpY