# Wagering Strategy – Day 2 Of A Final

Hey everyone! Andy here again. Being that we’re in the middle of a tournament, I thought that it’d be a good idea to go over wagering strategy for the second day of a tournament final. The math is more difficult, and mistakes have been made, even by some of the best players in the show’s history. But, between the glory and the extra money, the stakes are so much higher!

Because the math is more difficult, it’s easier to work from “first principles” (as opposed to trying to memorize scenarios), meaning that more math is involved, and practice can be very, very helpful!

I’m going to walk through a couple of situations, starting off using the 2000 Tournament of Champions final as an example.

The scores after Day 1:

Robin Carroll \$8,000
Jeremy Bate \$7,000
Steve Fried \$4,500

The scores going into Final Jeopardy on Day 2:
Robin Carroll \$4,600
Jeremy Bate \$1,500
Steve Fried \$7,000

In order to figure out who can win with a correct response and a sufficient wager, double everybody’s Day 2 score and then add that to their Day 1 score.

In our example:

Robin: 4,600 * 2 = 9,200 + 8,000 = 17,200
Jeremy: 1,500 * 2 = 3,000 + 7,000 = 10,000
Steve: \$7,000 * 2 = 14,000 + 4,500 = 18,500

In this example, Steve controls his own destiny.

Your next priority should be to determine the leader’s optimum wager. This should be the wager that wins the tournament by \$1 if everybody answers correctly and the two trailing players bet everything.

Therefore, Steve should wager to ensure that his two-day total is \$17,201.

To arrive at this wager, take that target total, subtract your first day total, and then subtract your current score going into Final Jeopardy.

Therefore, Steve’s optimum wager is:

17,201 – 4,500 = 12,701 – 7,000 = 5,701

From here, the other two players need to realize that the only way the leader can lose is if the leader misses Final Jeopardy. Therefore, they should wager to best take advantage of this situation, after ensuring that they have covered any player behind them.

As Robin in second place, she needs to assume that Steve will miss and Jeremy will get the question correct and wager everything.

In our situation, if Steve misses Final Jeopardy, he will fall to:

7,000 – 5,701 = 1,299 + 4,500 = 5,799

This bodes well for Robin, as her Day 1 total is 8,000. However, she still needs to worry about Jeremy behind her.

If Robin wagers nothing, her score sits at:

4,600 + 0 = 4,600 + 8,000 = 12,600

This is higher than Jeremy’s maximum possible total of \$10,000. Therefore, Robin needs to make sure that her score does not fall below \$10,001.

In a situation where you can’t afford to fall below a certain number, arrive at it as follows: Take the score that you would finish with if you wagered nothing and subtract the target score. Robin’s optimum wager is:

12,600 – 10,001 = 2,599

In Jeremy’s situation, the only way he can win is if Robin makes a wagering blunder. As most wagering blunders from second-placed contestants consist of wagering either everything or nearly everything, Jeremy should wager to make certain he does not fall behind Robin’s score from the first game. A good target score for Jeremy would be \$8,001 (though wagering \$0 in this situation is quite defensible as well).

Jeremy’s optimum wager is calculated the same as Robin’s was:

1,500 + 0 = 1,500 + 7,000 = 8,500 – 8,001 = 499

In summary, the thought process should be:

1. Determine who is leading by doubling everybody’s second-day scores and adding that total to the first day.
2. Determine what the leader’s optimum wager is.
3. Determine what the 2nd-place player’s optimum wager is based on the score of 3rd place and the leader’s wager.
4. Determine what the trailer’s optimum wager is based on the wagers of the players ahead.

One important note: Bet the maximum amount you can without jeopardizing your chances of victory in this situation. This may work in your favor if the player in front of you makes a mathematical error.

Returning to our example, here’s what the players actually wagered:

Robin: \$4,600
Jeremy: \$1,500
Steve: \$5,701.

And here’s how the tournament turned out:

Robin: \$4,600 – \$4,600 = \$0 + \$8,000 = \$8,000
Jeremy: \$1,500 – \$1,500 = \$0 + \$7,000 = \$7,000
Steve: \$7,000 – \$5,701 = \$1,299 + \$4,500 = \$5,799

Here’s how the tournament would have turned out with optimum wagers:

Robin: \$4,600 – \$2,599 = \$2,001 + \$8,000 = \$10,001
Jeremy: \$1,500 – \$499 = \$1,001 + \$7,000 = \$8,001
Steve: \$7,000 – \$5,701 = \$1,299 + \$4,500 = \$5,799

Steve was the only player of the three to wager properly. Holding everything else constant, Jeremy would have won the tournament had he wagered less. Robin was very lucky that her own wagering blunder did not come back to haunt her.

Here is one more wagering scenario for you to ponder:

2003 Tournament of Champions Final:

 Player Day 1 Score Pre-Final Day 2 Score Brian Weikle 15,000 22,000 Mark Dawson 22,400 17,200 Eric Floyd 7,300 6,200

Brian: 22,000 * 2 = 44,000 + 15,000 = 59,000
Mark: 17,200 * 2 = 34,400 + 22,400 = 56,800
Eric: 6,200 * 2 = 12,400 + 7,300 = 19,700

We then determine Brian’s optimum wager first.
Target score: \$56,801

56,801 – 15,000 = 41,801 – 22,000 = 19,801

Now we move onto Mark.

If Brian is wrong, he falls to:

22,000 – 19,801 = 2,199 + 15,000 = 17,199

Since this total and Eric’s maximum possible total are below Mark’s Day 1 score of 22,000, the optimal wager is to bet everything, get the question right, and hope that Brian either misses Final or shanks the wager.

As for Eric, he can not catch Mark, but he can catch Brian if Brian misses. Eric’s minimum target score becomes \$17,200. His optimum wager is therefore at least:

17,200 – 7,300 = 9,900 – 6,200 = 3,700

However, as with Mark, Eric might as well bet everything as he needs to be correct in order to catch Brian, and he can’t catch Mark no matter what happens.

Here’s what actually happened, in case you don’t know:

 Player Day 1 Score Pre-Final Day 2 Score Wager Day 2 Total Final Score Brian Weikle 15,000 22,000 19,601 41,601 56,601 Mark Dawson 22,400 17,200 17,200 34,400 56,800 Eric Floyd 7,300 6,200 6,199 12,399 19,699

When Brian was doing his math, he accidentally wrote the 8 in 56,800 as a second 6. The mistake ended up being a \$193,399 one, and it brings this suggestion:

Brian was in a situation where a wrong answer would have not garnered him victory, and his destiny for 2nd or 3rd would have been controlled solely by whether Eric was correct or not (We’ll assume that Eric knows how to wager — this is the Tournament of Champions final, after all.) If you’re in this situation where you’re guaranteed of losing if you miss Final Jeopardy, it’s not a bad idea to add \$1,000 or so to your wagered total, just in case.

In terms of the tournament:

Semifinalists:

Kevin Yang
Eliza Mancuse
Ben Greenho
Eliza Scruton

Wild card standings:
Catherine Briley 21000
Anshika Niraj 17401
Rose Schaefer 15400
Krishna Bharathala 15001
Morgan Flood 8348
Caleb Olson 6800
Sam Leanza 2200
Jeff Haylon 200