Anyway, I have been thinking a little bit about the game theory of Final Jeopardy betting. If you don't know, contestants (1) learn the category of the question, (2) place bets anywhere from $0 up to all of their current money, (3) hear the question and submit their answers, after which they either gain (if correct) or lose (if incorrect) the amount of their bid. So in FJ, you can accomplish anything from losing all your money to doubling it...and so can the other players.

As you know, I normally like Rose and Colin for my rows and columns, but let's use Joon and Franny today. Joon has $14,200 going into Final Jeopardy, while Franny has $17,600. To simplify this, let's talk about betting "big" or "small," by which I loosely mean betting quantities that do and don't bridge the gap between their scores. Yes, this is casual...umm let's say betting small means betting $0 and big means betting everything. Just go with it!

Here's a breakdown of the possible outcomes:

- If they both bet big, Joon wins only when he's right and she's wrong.
- If they both bet small, Joon will never win.
- If Joon bets big and Franny bets small, he will win whenever he's right.
- If Joon bets small and Franny bets big, he wins whenever she's wrong.

*p*for both of them, and let's say those are independent (see below for more on this tho). And let's say they only care about winning (i.e. not winning

**s**). Oh man. Do you feel the powahhhhh? You know the drill, game theory types! We just got ourselves one of these:

Unsurprisingly, the unique equilibrium is a mixed strategy. If my algebra is correct (and that is a big if), Joon plays Small with probability r=(p^2)/(1-p+p^2), and Franny plays Small with probability q=(1-2p+p^2)/(1-p+p^2). (Damn you Blogger for not enabling TeX! Or maybe it is good to disincentivize math in blog posts)

Anyway, if p=.5, this works out to r=q=1/3 (they are each more likely to bet Big) and the probability of Joon winning is 1/3, not bad for someone who is behind going into the finals. If p=2/3, we have r=4/7, q=1/7, and the probability of Joon winning is 2/7, slightly worse. As we move toward p=0 or 1, Franny becomes heavily favored; indeed, if p=0 or 1, she can

*guarantee*a win by betting 0 or everything, respectively.

The gameshow determines the difficulty of FJ questions and thus has control over

*p*. Where do they set it?

*

One interesting detail ignored above: whether the contestants' answers are right or wrong is likely to be

*correlated*. A question can be easy or hard, and if Joon gets the answer right, it's more likely that Franny will get it right as well. Interestingly, when Joon answered tonight but before Franny's answer had been revealed, the subjective (as perceived by me) probability of him winning momentarily

*fell*even though he got it right!

Why? If they both bet small, Joon has no chance. So if Joon bets small, he needs Franny to bet big. And if she bets big, Joon wins precisely when she answers incorrectly. But the fact that he got the question right made it much more likely (in my mind) that Franny would get it right too!

**Added:**Here's an article on the events of the week.

I also watch the Final Jeopardy betting to see if it fits a certain pattern. And, 9 times out of 10, it does... though the rationale I have assigned to it does not involve a mixed strategy, or it does. Maybe some equilibrium in a game with heterogeneous types and incomplete information. My story is loose and relies on heuristics at times and I haven't worked it out, but I figured I would share it here to see what you think.

ReplyDeleteHere's the pattern I have seen. If the leader does not have a "run away" victory, there is always some amount that the leader could bet to *guarantee* winning as long as s/he gets the question right.

For example, if Franny has 20,000 and Joon has 14,000 before Final Jeopardy, Franny can wager 8,001 to ensure that getting the question right == a win. Why bid just enough to ensure a win? There's a crazy Joon-type who will always wager everything in Final Jeopardy, and the only thing more embarassing than losing to a crazy person is losing to a crazy person when *you* got the final question right! Put enough weight on that embarassment and you can peg Franny to her bet.

What does the follower usually wager in this 14K to 20K situation? Taking as given an 8001 wager by Franny, Joon wagering less than 2000 will ensure winning if both get the question wrong. But, 2000 isn't enough to overtake Franny's initial ammount. Hence, it doesn't provide protection in the right-right situation against the crazy Franny-type who bets nothing. So, Joon will bet "big" to avoid this embarassment... and if you're betting big, you might as well go all in (or everything except $2).

It's not always the case that the follower bets it all. Sometimes the follower places a small bet, but this usually happens in a very peculiar circumstance (based on my anecdotal observation; I wonder if systematic data support this view).

Suppose that Joon has 16000 while Franny has 20000. Take as given that Franny will guarantee a win with her bet, so she'll wager 12001. Now, Joon is in a nice situation, regardless of which Franny he thinks he faces. He can beat the crazy Franny type who bets nothing by wagering more than 4000 (and both getting it right), but he can beat the scared Franny type (if both get the question wrong) if he only bets 8000 or less.

There's a window of opportunity for Joon to bet small (between 4K and 8K), and based on my Jeopardy watching, that's usually what happens. So, why doesn't Franny take that small bet as given? There's still the possibility that she is actually playing the game against crazy Joon. Crazy Joon will wager everything and -- in the event that everyone gets the question right -- that is embarassing enough to peg Franny at her 12001 wager.