Continuing my interest in tracking real-world predictions, I notice that the recent acquittal of Knox & Sollecito offers an interesting opportunity - specifically, many LessWrongers gave probabilities for guilt back in 2009 in komponisto’s 2 articles:
- “You Be the Jury: Survey on a Current Event”
- “The Amanda Knox Test: How an Hour on the Internet Beats a Year in the Courtroom”
Both were interesting exercises, and it’s time to do a followup. Specifically, there are at least 3 new pieces of evidence to consider:
- the failure of any damning or especially relevant evidence to surface in the ~2 years since (see also: the hope function)
- the independent experts’ report on the DNA evidence
- the freeing of Knox & Sollecito, and continued imprisonment of Rudy Guede (with reduced sentence)
Point 2 particularly struck me (the press attributes much of the acquittal to the expert report, an acquittal I had not expected to succeed), but other people may find the other 2 points or unmentioned news more weighty.
2 Probabilities
I was curious how the consensus has changed, and so, in some spare time, I summoned all the Conscientiousness I could and compiled the following list of 54 entries based on those 2 articles’ comments (sometimes inferring specific probabilities and possibly missing probabilities given in hidden subthreads), where people listed probabilities for Knox’s guilt, Sollecito’s guilt, and Guede’s guilt:
Knox | Sollecito | Guede | LWer |
---|---|---|---|
.20 | .20 | .70 | badger |
.05 | .10 | .90 | mattnewport |
.20 | .25 | .90 | AngryParsley |
.05 | .05 | .95 | tut |
.05 | .05 | .95 | bentarm |
.85 | .60 | .20 | bgrah449 |
.01 | .01 | .99 | kodos96 |
.01 | .01 | .99 | Daniel_Burfoot |
.40 | .40 | .90 | nerzhin |
.45 | .45 | .60 | Matt_Simpson |
.33 | .33 | .90 | Cyan |
.50 | .50 | .95 | jimmy |
.05 | .05 | .99 | Psychohistorian |
.40 | .40 | .90 | Threads |
.50 | .50 | .80 | Morendil |
.15 | — | — | Eliezer_Yudkowsky |
.20 | .35 | .98 | LauraABJ |
.10 | .10 | .90 | curious |
.20 | .20 | .96 | jpet |
.06 | .06 | .70 | saliency |
.80 | .60 | .95 | Mario |
.20 | .20 | .95 | Yvain |
.70 | — | — | Shalmanese |
.05 | .05 | .95 | gelisam |
.05 | .05 | .90 | Mononofu |
.90 | .90 | .90 | lordweiner27 (changed mind) |
.50 | .50 | .99 | GreenRoot |
.99 | .99 | .99 | dilaudid |
.13 | .15 | .97 | Jack |
.05 | .05 | .90 | wedrifid |
.01 | .01 | .90 | Nanani |
.35 | .35 | .95 | imaxwell |
.01 | .01 | .99 | jenmarie |
.25 | .25 | .75 | Jawaka |
.41 | .38 | .99 | magfrump |
.40 | .20 | .60 | gwern |
.08 | .10 | .95 | loqi |
.25 | .25 | .50 | JamesAndrix |
.90 | .85 | .99 | Unknowns |
.35 | .35 | .90 | Sebastian_Hagen |
.90 | .90 | .99 | brazil84 |
.30 | .30 | .40 | ChrisHibbert |
.02 | .02 | .98 | wnoise |
.50 | .40 | .90 | John_Maxwell_IV |
.10 | .10 | — | k3nt |
.01 | .01 | .99 | Sinai |
.00 | .00 | 1.0 | KayPea |
.00 | .00 | .60 | MerleRideout |
.15 | .10 | .80 | TheRev |
.01 | .01 | .99 | komponisto |
.30 | — | — | pete22 |
.01 | — | — | SforSingularity |
.00 | .00 | .90 | AnnaGilmour |
.05 | .05 | .95 | Seth_Goldin |
.60 | .60 | .95 | bigjeff5 |
It’s interesting how many people assign a high-probability to Knox being guilty; I had remembered LW as being a hive of Amanda fans, but either I’m succumbing to hindsight bias or people updated significantly after those articles. (For example, Eliezer says .15 is too high, but doesn’t seem otherwise especially convinced; and later one reads in Methods of Rationality that "[Hagrid] is the most blatantly innocent bystander to be convicted by the magical British legal system since Grindelwald's Confunding of Neville Chamberlain was pinned on Amanda Knox.")
EDIT: Jack graphed the probability against karma:
2.1 Outliers
If we look just at >41% (chosen to keep contacts manageable), we find 12 entries out of 54:
Knox | Sollecito | Guede | LWer |
---|---|---|---|
.45 | .45 | .60 | Matt_Simpson |
.50 | .40 | .90 | John_Maxwell_IV |
.50 | .50 | .80 | Morendil |
.50 | .50 | .95 | jimmy |
.50 | .50 | .99 | GreenRoot |
.60 | .60 | .95 | bigjeff5 |
.70 | — | — | Shalmanese |
.80 | .60 | .95 | Mario |
.85 | .60 | .20 | bgrah449 |
.90 | .85 | .99 | Unknowns |
.90 | .90 | .90 | lordweiner27 |
.90 | .90 | .99 | brazil84 |
.99 | .99 | .99 | dilaudid |
I have messaged each of them, asking them to comment here, describing if and how they have since updated, and any other thoughts they might have. (I have also messaged the first 12 commenters or so, chronologically, with <41% confidence in Knox’s guilt, with the same message.) The commenters:
AngryParsley / Cyan / Daniel_Burfoot / Eliezer_Yudkowsky / GreenRoot / John_Maxwell_IV / LauraABJ / Mario / Matt_Simpson / Morendil / Psychohistorian / Shalmanese / Threads / Unknowns / badger / bentarm / bgrah449 / bigjeff5 / brazil84 / dilaudid / jimmy / kodos96 / lordweiner27 / mattnewport / nerzhin / tut
I look forward to seeing their retrospectives, or indeed, anyone's retrospectives on the matter.
- Allknowing and most merciful Bayes;
- We have erred, and strayed from thy ways like biased sheep.
- We have followed too much the devices and desires of our own hearts.
- We have offended against thy axiomatic laws.
- We have left undone those updates which we ought to have done;
- And we have done those updates which we ought not to have done;
- And there is no calibration in us.
- But thou, O Bayes, have mercy upon us, miserable wannabes.
- Spare thou them, O Bayes, who confess their faults.
It is a little over the chance that if you are dealt two cards from a standard deck of cards that one of them will be the ace of spaces. It is a little under the chance that if you are dealt three cards from a standard deck that one them will be the ace of spades.
It is roughly the chance that if you pick three random members of the US House of Representatives that at least two of the three will not be reelected.
If It is about half as likely as the chance that a given US soldier in Iraq over the last decade will have been killed or too badly injured to return to duty (generally estimated to be around 9%). ETA: This number is wildly off. Disregard.
It is slightly less likely than your expectation for Schrodinger's cat to be alive if you run the experiment 5 times.
It is a bit under the chance that if you put your money on two numbers on a roulette wheel that one of them will turn up.
It is slightly over the chance that if you meet two random South Koreans that their last names will both be "Kim".
ETA: Here's a depressing one: It is around the chance that if you pick two children with childhood leukemia that they will both survive five years.
Who exactly?
... (read more)