The 2022 Jeopardy! National College Championship is working slightly differently from past Jeopardy! tournaments. 36 quarter-finalists have all played. We’re now down to 12 semifinalists playing over the next four games.

How were the contestants seeded for the semifinals?

I believe that it was quarter-finalist Jeric Brual who mentioned this on Reddit, but he said that the contestants were seeded based on their quarterfinal scores going into Final Jeopardy, and, judging by the pairings, this appears to be correct.

Pot A: Raymond Goslow (25,200), Emmey Harris (21,000), Lauren Rodriguez (20,300), Jaskaran Singh (19,900)

Pot B: Nam Vu (17,000), Joey Kornman (16,400), Kristin Donegan (15,200), Stephen Privat (14,300)

Pot C: Isaac Applebaum (13,800), Neha Seshadri (8,600), Liz Feltner (8,400), Megan Sullivan (5,400)

One contestant from each pot was placed in each semifinal.

Here are the four semifinals:

Semifinal #1: Nam Vu (Georgetown), Neha Seshadri (Harvard), Raymond Goslow (Kennesaw State)

Semifinal #2: Lauren Rodriguez (Pomona), Isaac Applebaum (Stanford), Stephen Privat (LSU)

Semifinal #3: Kristin Donegan (Carnegie Mellon), Liz Feltner (Northeastern), Emmey Harris (Minnesota)

Semifinal #4: Joey Kornman (Brandeis), Jaskaran Singh (Texas), Megan Sullivan (Virginia).

The non-winning semifinal contestants receive $20,000, and the semifinal winner with the lowest score (not making the finals and qualifying for the Second Chance Tournament) will receive $35,000.

How will this affect strategy?

The major strategic consideration: Just winning your game won’t be enough to qualify for the final. Your score needs to be in the top 3 in order to qualify.

So, what score will advance?

Much like the wild cards for past tournaments, we can look at past data to predict what scores have a good chance of advancing. I used the following methodology to construct my JNCC wild card model:

1. Look back at the semifinal games for the 30 previous College Championships and get both the scores going into Final Jeopardy and the outcome distribution in Final Jeopardy (RRR/RRW/etc).
2. Find the average ($12,694) and standard deviation ($6,494) in the scores, as well as enumerate the specific distribution of the possible Final Jeopardy! outcomes.
3. From this, randomly generate thousands of game outcomes using the previous data, assuming that every contestant feels the need to maximize their score by betting everything in Final Jeopardy, and record the winning score for each game.
   • Is this the best way of going about it? It’s certainly possible that the contestants aren’t going to bet everything. But I also think it’s best to prepare for the “worst case scenario”, and to me, that “worst case scenario” is “everybody else doubles their score”.
4. The average winning final score turned out to be approximately $27,200, and the standard deviation turned out to be approximately $15,200.
5. From this, we can use the same calculations that are used in my normal tournament wild card prediction models to determine the chance of each score advancing.

Here’s the graphical version, for each score between $0 and $50,000:

Is it viable to stand pat and hope for one of the other games to be a Triple Stumper?

The simple answer: Do you like coin flips that are slightly weighted against you?

The complicated answer: 12 of the 86 past College Championship semifinal games (with 3 players) have had Triple Stumper Final Jeopardy clues. Taking that 13.95% chance into account gives you a 54.82% chance of there being 0 Triple Stumpers (thus, a 45.18% chance of there being at least 1). There’s also a 9.62% chance of there being at least 2 Triple Stumpers. If you like those odds, then go for it!

So, there you have it. It’s certainly going to be an interesting couple of days—and at least with the Second Chance Tournament, there isn’t going to be a contestant seeing their Jeopardy! journey end with 2 wins and 0 losses. Enjoy the semi-finals!

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