FAQ/ranksum102013-03-08 10:17:10localhostconverted to 1.6 markup92007-10-25 08:46:01PeterWatson82007-10-25 08:45:21PeterWatson72007-10-25 08:44:48PeterWatson62007-10-25 08:39:03PeterWatson52007-10-24 16:09:47PeterWatson42007-10-24 16:07:41PeterWatson32007-10-24 16:02:52PeterWatson22007-10-24 15:46:12PeterWatson12007-10-24 15:43:48PeterWatsonWhat is the expected total discrepancy score in a R choice task?Suppose we have R possible choices and each of these is equally likely to be the true one. If we consider a discrepancy as the difference between the true choice and the one given by a subject then The expected total discrepancy of the absolute value of discrepancies equals $$\sum_{k=1}^R (k=1)k $$, $$1 \leq k \leq R $$ with the average sum of the absolute values of discrepancies per rating equal to $$\frac{\sum_{k=1}^R (k=1)k}{R}$$. For example the table below gives all the abs(discrepancies) for the case where R = 4. k=1 k=2 k=3 k=4 True Rank

0 1 2 3 1

1 0 1 2 2

2 1 0 1 3

3 2 1 0 4

Expected total score assuming random guesses at true rank = 2(1+2+3)+2(1+1+2)= 20 = 1x2 + 2x3 + 3x4 = $$\sum_{k=1}^4 (k=1)k $$. The average sum of abs(discrepancies) per rating = 20/4 = 5.