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yonatanbitton commited on
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Update README.md

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@@ -87,7 +87,7 @@ There are different number of candidates, which creates different difficulty lev
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  It is a binomial distribution probability calculation.
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  Assuming N=5 candidates, and K=2 associations, there could be three events:
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- (1) The probability for a random guess is correct in 0 associations is 0.3 (elaborate below), and the Jaccard index is 0 (there is no intersection between the correct labels and the wrong guesses). Therefore the expected random score is 0.
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  (2) The probability for a random guess is correct in 1 associations is 0.6, and the Jaccard index is 0.33 (intersection=1, union=3, one of the correct guesses, and one of the wrong guesses). Therefore the expected random score is 0.6*0.33 = 0.198.
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  (3) The probability for a random guess is correct in 2 associations is 0.1, and the Jaccard index is 1 (intersection=2, union=2). Therefore the expected random score is 0.1*1 = 0.1.
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  * Together, when K=2, the expected score is 0+0.198+0.1 = 0.298.
@@ -97,7 +97,7 @@ There are different number of candidates, which creates different difficulty lev
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  Now we can perform the same calculation with K=3 associations.
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  Assuming N=5 candidates, and K=3 associations, there could be four events:
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- (4) The probability for a random guess is correct in 0 associations is 0, and the Jaccard index is 0. Therefore the expected random score is 0.
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  (5) The probability for a random guess is correct in 1 associations is 0.3, and the Jaccard index is 0.2 (intersection=1, union=4). Therefore the expected random score is 0.3*0.2 = 0.06.
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  (6) The probability for a random guess is correct in 2 associations is 0.6, and the Jaccard index is 0.5 (intersection=2, union=4). Therefore the expected random score is 0.6*5 = 0.3.
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  (7) The probability for a random guess is correct in 3 associations is 0.1, and the Jaccard index is 1 (intersection=3, union=3). Therefore the expected random score is 0.1*1 = 0.1.
 
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  It is a binomial distribution probability calculation.
88
 
89
  Assuming N=5 candidates, and K=2 associations, there could be three events:
90
+ (1) The probability for a random guess is correct in 0 associations is 0.3 (elaborate below), and the Jaccard index is 0 (there is no intersection between the correct labels and the wrong guesses). Therefore the expected random score is 0.
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  (2) The probability for a random guess is correct in 1 associations is 0.6, and the Jaccard index is 0.33 (intersection=1, union=3, one of the correct guesses, and one of the wrong guesses). Therefore the expected random score is 0.6*0.33 = 0.198.
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  (3) The probability for a random guess is correct in 2 associations is 0.1, and the Jaccard index is 1 (intersection=2, union=2). Therefore the expected random score is 0.1*1 = 0.1.
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  * Together, when K=2, the expected score is 0+0.198+0.1 = 0.298.
 
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  Now we can perform the same calculation with K=3 associations.
99
  Assuming N=5 candidates, and K=3 associations, there could be four events:
100
+ (4) The probability for a random guess is correct in 0 associations is 0, and the Jaccard index is 0. Therefore the expected random score is 0.
101
  (5) The probability for a random guess is correct in 1 associations is 0.3, and the Jaccard index is 0.2 (intersection=1, union=4). Therefore the expected random score is 0.3*0.2 = 0.06.
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  (6) The probability for a random guess is correct in 2 associations is 0.6, and the Jaccard index is 0.5 (intersection=2, union=4). Therefore the expected random score is 0.6*5 = 0.3.
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  (7) The probability for a random guess is correct in 3 associations is 0.1, and the Jaccard index is 1 (intersection=3, union=3). Therefore the expected random score is 0.1*1 = 0.1.