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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