### 7.21.2004

## How to stop Ken Jennings

I bet what the Jeopardy! crew did was bring in Turd Ferguson and Sean Connery to verbally pummel him and possibly ass-rape him.

Since the June 2 show, he's won 35 straight games. Only five times has the outcome of the game been in doubt heading into Final Jeopardy. The rest of the time, Ken's post-double Jeopardy total has been more than twice the nearest challenger's total.

Actually, the way Ken lost is probably that some guy was within striking distance of Ken after Double Jeopardy and simply didn't wager very much on Final Jeopardy. Ken then got the answer wrong. (Ken is, as of this writing, only 24/35 on Final Jeopardy questions). According to this Jeopardy archive, Ken always assumes that the nearest challenger will answer correctly. This can be inferred by observing that always bets (2y - x + 1) if the outcome is in doubt. I think this might not be optimal strategy if Ken is only a 69% Final Jeopardy shooter and Ken's opponents are presumably worse; if y > (2x - 1)/3, then the challenger's best strategy, given this knowledge of Ken's strategy, is to just wager nothing and hope that Ken loses. Of course, this information might now have been available to Ken's opponents when the shows were being filmed.

Hmm. I'm really tempted to work out the optimal game-theoretic strategies for Final Jeopardy now.

UPDATE: Working out the game theory is proving to be quite painful. Not surprisingly, there are no Nash equilibria. Mixed strategies (strategies where each player chooses a certain wager with a certain probability) are necessary; it looks like it's been worked out here.

As an example of the results contained in the paper, suppose the challenger (hereafter known as "Chump") has 2/3 x dollars and Ken has x dollars. Chump needs to adopt a strategy where he threatens to wager either 0 or 2/3 x, and Ken needs to adopt a strategy where he threatens to wager either 0 or (1/3 x + epsilon).

The kicker here is that the probability that Chump and Ken assign to their strategies depends on their chances of answering correctly; Chump should assign the weight (p(both right) / p(Ken wrong), 1) to the wager (0, 2/3 x) and Ken should assign the weight (p(both wrong) / p(Ken right), 1) to the wager (0, 1/3 x + epsilon). So if the Final Jeopardy question sounds like it'll be hard, Ken should give more thought to wagering 0. If it sounds like it'll be an easy question, Chump doesn't have much of a chance, but he should consider wagering 0 and hope that Ken loses.

As Chump's score y varies in the range (2/3 x, x), the optimal strategies become more complex; Chump has to threaten a range of wagers in evenly spaced increments from 0 to y, and Ken has to threaten a range of wagers from 0 to (2y-x + epsilon).

Since the June 2 show, he's won 35 straight games. Only five times has the outcome of the game been in doubt heading into Final Jeopardy. The rest of the time, Ken's post-double Jeopardy total has been more than twice the nearest challenger's total.

Actually, the way Ken lost is probably that some guy was within striking distance of Ken after Double Jeopardy and simply didn't wager very much on Final Jeopardy. Ken then got the answer wrong. (Ken is, as of this writing, only 24/35 on Final Jeopardy questions). According to this Jeopardy archive, Ken always assumes that the nearest challenger will answer correctly. This can be inferred by observing that always bets (2y - x + 1) if the outcome is in doubt. I think this might not be optimal strategy if Ken is only a 69% Final Jeopardy shooter and Ken's opponents are presumably worse; if y > (2x - 1)/3, then the challenger's best strategy, given this knowledge of Ken's strategy, is to just wager nothing and hope that Ken loses. Of course, this information might now have been available to Ken's opponents when the shows were being filmed.

Hmm. I'm really tempted to work out the optimal game-theoretic strategies for Final Jeopardy now.

UPDATE: Working out the game theory is proving to be quite painful. Not surprisingly, there are no Nash equilibria. Mixed strategies (strategies where each player chooses a certain wager with a certain probability) are necessary; it looks like it's been worked out here.

As an example of the results contained in the paper, suppose the challenger (hereafter known as "Chump") has 2/3 x dollars and Ken has x dollars. Chump needs to adopt a strategy where he threatens to wager either 0 or 2/3 x, and Ken needs to adopt a strategy where he threatens to wager either 0 or (1/3 x + epsilon).

The kicker here is that the probability that Chump and Ken assign to their strategies depends on their chances of answering correctly; Chump should assign the weight (p(both right) / p(Ken wrong), 1) to the wager (0, 2/3 x) and Ken should assign the weight (p(both wrong) / p(Ken right), 1) to the wager (0, 1/3 x + epsilon). So if the Final Jeopardy question sounds like it'll be hard, Ken should give more thought to wagering 0. If it sounds like it'll be an easy question, Chump doesn't have much of a chance, but he should consider wagering 0 and hope that Ken loses.

As Chump's score y varies in the range (2/3 x, x), the optimal strategies become more complex; Chump has to threaten a range of wagers in evenly spaced increments from 0 to y, and Ken has to threaten a range of wagers from 0 to (2y-x + epsilon).