Update README.md

README.md

@@ -75,11 +75,40 @@ ID (int64): 367 - association ID.
 
 ### Data Splits
 There is a single TEST split. In the accompanied paper and code we sample it to create different training sets, but the intended use is to use winogavil as a test set.
-There are different number of candidates, which creates different difficulty levels: 
-  -- With 5 candidates, random chance for success is 38%.  
-  -- With 6 candidates, random chance for success is 34%.     
-  -- With 10 candidates, random chance for success is 24%. 
-  -- With 12 candidates, random chance for success is 19%.     
+There are different number of candidates, which creates different difficulty levels:   
+  -- With 5 candidates, random model expected score is 38%.  
+  -- With 6 candidates, random model expected score is 34%.  
+  -- With 10 candidates, random model expected score is 24%.  
+  -- With 12 candidates, random model expected score is 19%.  
+
+<details>
+  <summary>Why random chance for success with 5 candidates is 38%?</summary>
+  
+  It is a binomial distribution probability calculation. 
+  
+  Assuming N=5 candidates, and K=2 associations, there could be three events:   
+  (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. 
+  (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.  
+  (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.  
+  * Together, when K=2, the expected score is 0+0.198+0.1 = 0.298.   
+  
+  To calculate (1), the first guess needs to be wrong. There are 3 "wrong" guesses and 5 candidates, so the probability for it is 3/5. The next guess should also be wrong. Now there are only 2 "wrong" guesses, and 4 candidates, so the probability for it is 2/4. Multiplying 3/5 * 2/4 = 0.3. 
+  Same goes for (2) and (3). 
+  
+  Now we can perform the same calculation with K=3 associations. 
+    Assuming N=5 candidates, and K=3 associations, there could be four events:   
+  (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. 
+  (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.  
+  (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.    
+  (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.    
+  * Together, when K=3, the expected score is 0+0.06+0.3+0.1 = 0.46. 
+  
+Taking the average of 0.298 and 0.46 we reach 0.379. 
+
+Same process can be recalculated with 6 candidates (and K=2,3,4), 10 candidates (and K=2,3,4,5) and 123 candidates (and K=2,3,4,5,6). 
+
+</details>
+
 
 ## Dataset Creation