The alternate path would be to assume that there is a preference ordering from 1 to 7, with some sort of noise applied. This feels clunky to me, though.
While clunky, this seems to be a perfectly workable approach for 7 objects. There are 5040 permutations. For each piece of evidence, add probability mass to those that correspond and remove it from those that differ (perhaps weighted by how much they correspond or differ?). The probability of your object being most preferred, then, is the sum of the probabilities of those permutations in which it occupies the highest point.
Okay, if A is preferred from { A , [B-G] }, that should add probability mass to [A, [B,...,G] ], where [A, [B,...,G] ] is a ranked set of objects where the first slot is most preferred. That would represent 720 (6!) sets out of 5040.
All other sets (7! − 6! = 4,320) should either stay the same probability or have probability mass removed.
Then, the probability of A being “most preferred” = the sum of the probability mass of all 720 sets that have A as the highest ranked member. Likewise for B through G. Highest total probability mass wins.
We see a new piece of evidence—one of the people prefers C to E
C will be preferred to E in half the lists. Those lists become more probable, the other half become less probable. How much more/less probable depends on how much error you expect to see and of what type.
Repeat on all the data.
You only actually look at the first member when asking the odds that a particular object is there—at which point, yes, you sum up the probability of those 720 sets.
While clunky, this seems to be a perfectly workable approach for 7 objects. There are 5040 permutations. For each piece of evidence, add probability mass to those that correspond and remove it from those that differ (perhaps weighted by how much they correspond or differ?). The probability of your object being most preferred, then, is the sum of the probabilities of those permutations in which it occupies the highest point.
Okay, if A is preferred from { A , [B-G] }, that should add probability mass to [A, [B,...,G] ], where [A, [B,...,G] ] is a ranked set of objects where the first slot is most preferred. That would represent 720 (6!) sets out of 5040.
All other sets (7! − 6! = 4,320) should either stay the same probability or have probability mass removed.
Then, the probability of A being “most preferred” = the sum of the probability mass of all 720 sets that have A as the highest ranked member. Likewise for B through G. Highest total probability mass wins.
Am I understanding that correctly?
I don’t think I’m following you.
We see a new piece of evidence—one of the people prefers C to E
C will be preferred to E in half the lists. Those lists become more probable, the other half become less probable. How much more/less probable depends on how much error you expect to see and of what type.
Repeat on all the data.
You only actually look at the first member when asking the odds that a particular object is there—at which point, yes, you sum up the probability of those 720 sets.
Ah, I see. Instead of updating half the lists, I was updating the 720 sets where C is the #1 preference. Thanks for the clarification.