# How much should you wager in Final Jeopardy?

I think you all know how this one works, but let me sketch it out again anyway. You’ve been answering trivia questions, and now you have some dollar score going into the last round. Your two opponents have been doing the same. In the last round, you’re asked one question, for which you write down your answer, and how much you’re willing to bet on it, up to the amount of money you have; your opponents do the same. You’re trying to get as much money as you can from this game, so winning is not all that matters: you want to both win and have a higher score. The question is: how much should you wager? Your mother, Trebek.

We can model this by positing that each of players 1,2,3 have scores of $D_{1}, D_{2}, D_{3}$, respectively, going into the final round. They expect to answer correctly with probabilities $p_{1}, p_{2}, p_{3}$, respectively; we assume that these values are common knowledge. The wagers of each player are $w_{1}, w_{2}, w_{3}$, respectively, where for each i, $w_{i}\leq D_{i}$.

We describe $F_{i}(N_{i}|D_{-i},s_{-i})$ to be the probability that player i wins with a certain final score $N_{i}$ (after the question has been answered correctly or incorrectly), given the strategies $s_{-i}$ that the other players use, and the scores they have $D_{-i}$ going into Final Jeopardy. Obviously, since the person with the highest score wins, you have a better chance of winning if you have a higher final score. Thus, $F_{i}(\cdot|D_{-i},s_{-i})$ is nondecreasing in $N_{i}$, no matter what $s_{-i}$ and $D_{-i}$ are.

Claim: If $p_{i}\geq\frac{1}{2}$, then it is a dominant strategy to bid all your money; namely, $w_{i}=D_{i}$.

Proof of the claim:Fix strategies $s_{-i}$ for the other players. Suppose you wager $w_{i}$. Then your expected amount of money that you win is

$\Pi_{i}(w_{i},D_{i}|D_{-i},s_{-i})=(D_{i}-w_{i})F_{i}(D_{i}-w_{i}|D_{-i},s_{-i})(1-p_{i})+(D_{i}+w)F_{i}(D_{i}+w|D_{-i},s_{-i})p_{i}$.

Meanwhile, if you wager $D_{i}$, you can expect to win $2D_{i}F_{i}(2D_{i}|D_{-i},s_{-i})p_{i}$. Notice, though, that since $F_{i}(\cdot|D_{-i},s_{-i})$ is nondecreasing, and $p_{i}\geq\frac{1}{2}$,

$\Pi_{i}(w_{i},D_{i}|D_{-i},s_{-i})\leq(D_{i}+w)F_{i}(D_{i}+w|D_{-i},s_{-i})p_{i}+(D_{i}-w_{i})F_{i}(D_{i}+w_{i}|D_{-i},s_{-i})p_{i}$

$=2D_{i}F_{i}(D_{i}+w|D_{-i},s_{-i})p_{i}\leq2D_{i}F_{i}(2D_{i}|D_{-i},s_{-i})p_{i}$.

Hence, regardless of what $D_{-i}, s_{-i}$ are, you can expect to get the most amount of money by bidding all you’ve got. Indeed, this argument is so general that it applies no matter how many people are playing Final Jeopardy against you – as long as you think you’re more likely to get it right than wrong, you should bid everything.

Now what should you do if you’re not so certain you’re going to get it right? If both other players are still more likely than not to answer correctly, then what you should do is pretty straightforward – you know what they’re going to wager (which is everything they’ve got), and so you can precisely calculate $F_{i}(\cdot|D_{-i},s_{-i})$. All you then have to do is choose the value of $w_{i}$ which maximizes $\Pi_{i}(w_{i},D_{i})$, as defined above.

Unfortunately, the equilibrium analysis is in general not very tractable for a 3-player round of Final Jeopardy. That is, it doesn’t give a very nice closed-form solution (or at least, I can’t find one). So, while you should go by the advice in the previous paragraph (assuming these probabilities are commonly known), I can’t say much about what’s going to happen in general. Also, be aware that the advice doesn’t hold if the probability is not commonly known, in which case you will need to consider the entire hierarchy of beliefs.

In the spirit of academic honesty, I am heavily indebted to the online Google Books edition of “Strategy and Games: Theory and Practice,” by Prajit K. Dutta for his analysis for when $p_{i}\geq\frac{1}{2}$, though the proof I use of the strategic dominance of wagering everything is more formal than his. The analysis for when $p_{i}<\frac{1}{2}$ is my own, and so should probably be taken with a grain of salt.

Edit: A previous edition of this post said that Final Jeopardy violated the conditions of the Kakutani fixed point theorem, and so there was likely no equilibrium. While it is true that these conditions are violated, it is not for the reason claimed, and so there actually will be an equilibrium, as demonstrated by Dasgupta & Maskin (1986). It will likely involve ties in some cases, for reasons that I’m too lazy to show.