joshkaufman comments on Help with a (potentially Bayesian) statistics / set theory problem? - Less Wrong

2 Post author: joshkaufman 10 November 2011 10:30PM

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Comment author: joshkaufman 10 November 2011 11:57:38PM *  0 points [-]

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?

Comment author: dlthomas 11 November 2011 12:37:00AM 1 point [-]

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.

Comment author: joshkaufman 11 November 2011 12:40:53AM 0 points [-]

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.