How to wager in Final Jeopardy!: Part Four
Part Four looks at special three-player scenarios – particularly, in which the leader MUST wager for the tie. We analyzed special two-player scenarios in Part Two; a third player adds a new dimension – and additional possibilities – to the wagering.
Note: if you’re unfamiliar with basic wagering strategy, you’ll want to start at my tutorial page.
Part Four is split in halves. The first half is intended for players who still need some practice with wagering techniques. I walk you through the steps required to calculate the proper wagers for three players – in this case, resulting in a situation where the leader should wager for the tie.
The second introduces the two “bet-to-tie” scenarios I think every player should know going in to the game, so that if you find a Daily Double on the last clue of Double Jeopardy!, you can make an optimal wager. (I call this a “Penultimate Wager” situation.) If you’re comfortable with calculating three-player scenarios, you can jump right in here.
Jeopardy! fans have long known many algebraic formulas under which the leader should go for the tie. (There’s a good list here, relating to Watson’s strategies.)
You can figure out these scenarios when you’re taking as much time as you need to calculate your own wager. But I suggest that every player memorize two formulas on the off-chance he receives a Penultimate Wager situation.
In the second video, I walk you through these two, and set up a Penultimate Wager scenario for you to try.
Note: you can click on an image to jump directly to the corresponding part of the video.
Here’s the example of evenly-spaced scores:
And here’s the example of first equals second plus third:
For the Penultimate Wager situation we saw in the very first game of this season, James, who had the Daily Double, had three “good” wagers to end up in one of those two situations. (If he wagers 8,200 and gets it right, he’ll be in the lead, which is the ultimate goal, if possible.)