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Strategy Talk: College Tournament Wild Card Predictions

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Matthew Riggle, career statistics:
47 correct, 6 incorrect
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(Quarterfinal Final Standings)

1) Victoria Groce, 15 points, 5 wins, 172 correct responses, $126,000 Coryat Score, $121,000 earned before Final Jeopardy
2) Yogesh Raut, 12 points, 3 wins, 106 correct responses, $69,000 Coryat Score, $69,800 earned before Final Jeopardy
3) Paolo Pasco, 8 points, 1 win, 94 correct responses, $62,400 Coryat Score, $71,400 earned before Final Jeopardy
4) Andrew He, 5 points, 1 win, 89 correct responses, $53,000 Coryat Score, $50,400 earned before Final Jeopardy
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All Time Jeopardy! Winnings, Regular Play Only:

1. Ken Jennings $2,520,700
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5. Jamie Ding $882,605
6. Cris Pannullo $748,286
7. Mattea Roach $560,983
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All Time Jeopardy! Winnings, Including Tournaments (and Consolation Prizes):

1. Brad Rutter $4,968,436
2. Ken Jennings $4,372,700
3. James Holzhauer $3,614,216
4. Matt Amodio $1,964,601
5. Amy Schneider $1,864,800
6. Yogesh Raut $1,098,403
7. Mattea Roach $897,983
8. Jamie Ding $885,605
9. David Madden $785,733
10. Victoria Groce $773,801
11. Cris Pannullo $754,286
12. Roger Craig $706,200
13. Larissa Kelly $671,930
14. Matt Jackson $623,612
15. Jason Zuffranieri $549,496
16. Andrew He $534,365
17. Scott Riccardi $533,000
18. Jerome Vered $499,102
19. Julia Collins $495,767
20. Austin Rogers $493,000
21. Juveria Zaheer $491,000
22. Ben Ingram $449,201
23. Paolo Pasco $448,717
24. Adriana Harmeyer $446,600
25. Buzzy Cohen $441,603

Back in December 2015, Keith Williams on the Final Wager spoke of predicting Jeopardy! wild cards using combinatorics.

To make my predictions going forward here on The Jeopardy! Fan, my prediction model will use combinatorics in the same way. The one change that I have made to my model is its input percentages.

For my prediction model, I have taken every College Tournament since the doubling of dollar values (2002 and later), and calculated both the median score ($10,200) and the standard deviation ($6,226). I have thus assumed that non-winning scores are normally distributed based on that, and will use the Z-score of each non-winning total to determine the percentage chance that any one future score will be lower than the player’s current score.

From there, given the score, current wild card place, and number of non-winning scores to come, it is possible to calculate the % chance a player has of advancing.

As an example, take a $13,000 score of a second-placed player on Monday’s quarterfinal. My model gives that score a 67.4% chance of beating a random future score. Holding down the 1 seed with 8 non-winning scores to come, that score’s chances of advancing are:

As the 1 seed: 0.674^8 x 0.326^0 x 1 = 4.237%
As the 2 seed: 0.674^7 x 0.326^1 x 8 = 16.427%
As the 3 seed: 0.674^6 x 0.326^2 x 28 = 27.864%
As the 4 seed: 0.674^5 x 0.326^3 x 56 = 27.007%
Total: 75.534%

If that same $13,000 score finished 3rd, however, its chances of advancing would only be the sum of the first three rows above: 48.528%.

This same model can also be used to predict what score has a better than 50% chance of qualifying for the semifinals.

As it stands going into the 2017 College Championship, per this model:
A second-place score of $11,137 has a 50% chance of qualifying.
A third-place score of $13,103 has a 50% chance of qualifying.

Throughout this tournament, I will use this model each day to predict each wild card’s chances of advancing.


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