Forged Invariant
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Here are the relevant quotes:
- Gather proposals for a hundred RCTs ...
- Randomly pick 5% of the proposed projects, fund them as written, and pay off the investors who correctly predicted what would happen.
- Take the other 95% of the proposed projects, give the investors their money back, and use the SWEET PREDICTIVE KNOWLEDGE [to take useful actions]
Other than the difference in the portion of the markets you run (1/20 vs 1/1000), this is equivalent.
(It does not discuss liquidity costs, just the the randomization as a way to avoid having to take many random actions.)
Just for the record, Dynomight proposed this back in 2022: https://dynomight.net/prediction-market-causation/#commit-to-randomization. (I assume that the idea has been around for longer.)
(Also I would phrase it as being able to use the same money to trade on all 1000 of the markets at once. I think that is equivalent to your free loan.)
I was able to deduce them by
making a scatter-plot of Colleen vs Liboulen's predictions. You can see that this plot has the points on a "flattened prism" in 3 directions, and manually count the shifts and see that each of the underlying components has 10 possible values.
Once you have that structure, you can pick out points on the extremes and use their slopes to calculate some of the relevant slopes. Finally, I brought in Bella's info and used that to work out the remaining stats. (I used chatGPT for some help throwing together some linear regressions, but they needed a good bit of tweaking to be functional, and mostly agreed with the slopes that I had calculated by just looking at the scatterplots.)
At this point I am throwing everything that I found in a linear regression, because I ran out of time. My pick is:
Candidate 11, with an estimated 0.91 chance of success.
Candidates 19 and 7 would be my next choices, with 0.87 and 0.85 estimated chances of success respectively.
If I had had more time to work on this, I would have like to look at:
A summary of some interesting results. I am leaving how I found some of this out for now, for brevity's sake.
I have manage to extract 6 integer variables that range from 1-10.
3 of them are from the components of (Coleen, Linestra, Liboulen, Bella), the other 3 are from (Fizz, Ister, Ziqual).
Each of them has a very similar histogram, sort of like a truncated normal distribution. A linear regression of them with Holly gives approximately 1 as their coefficient, except for 1 variable (which I am calling X2 for now) which has a coefficient of roughly -1.
All of these underlying variables have a magnitude of correlation with the candidate succeeding, between 0.11 and
A few miscellaneous observations:
Ister, Ziqual and Fizz seem to have some pretty deterministic structure connecting them.
Ister always predicts an integer between 51 and 60 inclusive.
Ziqual's prediction is equal to (Ister - 50) * (Integer from 1 to 10) - (one of 0, 1). Multipliers in the 5 to 7 range are most common.
Fizz's prediction is less than or equal to (Ister's prediction + 10). Fizz's prediction is greater than or equal to 44.
Separately, a scatterplot of Liboulen and Colleen's predictions has a lot of structure: [Scatterplot removed since it seems to show up through the spoiler. Message me if you want to see it.]
Note that each of the 3 "axies" of this
Story of a mostly homeless guy who scammed Isaac King out of $300. Isaac sued in small claims court on principle, did all the things, and none of it mattered.
This link goes to Sarah's tweet, not to Isaac's story.
This is not what the article says. It says that BC is re-criminalizing hard drugs.
I am in BC, and have not heard anything about decriminalizing marijuana. I get the sense that it being legal is generally popular. Complaints about drug users are common here, but they are usually not talking about weed.
I would expect that player 2 would be able to win almost all of the time for most normal hash functions, as they could just play randomly for the first 39 turns, and then choose one of the 2^8 available moves. It is very unlikely that all of those hashes are zero. (For commonly used hashes, player 2 could just play randomly the whole game and likely win, since the hash of any value is almost never 0.)
The figure you are referring to does not need to add up to 100%, since it is showing P[data | aliens] and P[data | no aliens].
P[data | aliens] and P[not data | aliens] need to add to 100%, but that is not on the graph.
As an extreme case where P[A | B] + P[A | C] != 1, consider A = coin did not land on its edge, B = the coin is ordinary, C = the coin is weighted to land heads twice as often as tails.
Then P[A | B] = 0.9999 and P[A | C] = 0.9999 would be reasonable values.