This story reminded me of a story written by a Czech biologist who studied animals in Papua-New Guinea and went to a hunt with a group of local tribesmen. The dusk was approaching, they were still in the forest and he proposed that they could sleep under a tree. The hunters were adamant in their refusal: no, this is dangerous, a tree might fall on you in your sleep and kill you . He relented, but silently considered…
Heuristics that almost always work
531–540 of 541 posts
Re: Heuristics that almost always work
#532Earlier quoted context omitted.
I came up with the Goldilocks (meta?-)heuristic[1] for that: Only trust someone to say X is too high if they can also tell you when X would be too low. A corollary of which would be e.g. "Don't trust a skeptic that says 'X won't Change The World' unless they can tell you which developments would Change The World." [1] Or Scylla-Charybdis Heuristic if you prefer: http://blog.tyrannyofthemouse.com/2015/12/the-scylla-ch…
That is actually really useful.
Re: Heuristics that almost always work
#533Earlier quoted context omitted.
The "slippery slope" principle applies here though: N+1 enables N+2, which enables N+3 and so on.
Slippery slope is a fallacy, not a principle. Just because you took N steps, that doesn't necessarily mean you will take N+1 steps. It's a convincing fallacy because sometimes you do take N+1 steps. But just like in the article, heuristics aren't always right.
A fallacy should be a incorrect shape of an argument, a incorrect reasoning, not just a false statement.
Re: Heuristics that almost always work
#534Earlier quoted context omitted.
"There are bold X, and old X, but no old, bold X." Replace X with any practitioners subject to sufficient risk as a result of their practice. I first heard it in the context of mushroom foraging.
The risks are highest when learners are at a beginner to intermediate stage. They know the basics, and have gained some confidence, but don't know enough to get themselves out of trouble. This is called Stage 1 in the Gordon Model of learning: unconscious incompetence.
Re: Heuristics that almost always work
#535There is also society game level above all of that. You can persuade more and more people that stealing things just generates additional costs. Seems we have to allocate resources(private and public) on protecting property and its beneficial for all if we stop playing this game (where reasonable people have to spend money to continously fix unreasonable people's deeds).
Re: Heuristics that almost always work
#536Earlier quoted context omitted.
What am I missing? 6 x 6 X 6 = 216 or about 7 months.
Dice (typically) do not have a memory, so whatever happened yesterday will not influence what happens today. If you roll it daily, your chance of surviving at least N days is (215/216)^N, for the specific case of "rolling three 6 on three 6-sided dice" that puts you at ~50% at 149 days and at ~10% at 496 days. At sufficient scale, even incredibly unlikely things become quite probable.
runs summary(x)
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.0 62.0 149.0 216.2 300.0 1902.0
> quantile(x, c(0.5, 0.8, 0.9))
50% 80% 90%
149 350 495
A simple simulation. Run 10K times. Count the number of times it takes for three dice to add up 18.The numbers very much agree with you. The median is 149. The 90th is 495 in the simulation, which is close enough to 496. There is very much a long tail in the data. So, the median and the average will not be the same. Is it a coincidence that mean is a 216?
Re: Heuristics that almost always work
#537Earlier quoted context omitted.
What am I missing? 6 x 6 X 6 = 216 or about 7 months.
RandomSwede's comment is accurate, but maybe the below can help add some 'flesh' to their response. Basically, the problem is that you can't just multiply it all together. (1/6) ^ 3 is correct, and the probability of rolling 3 sixes is indeed 1/216 today , but if you repeat independent events, you don't just add up the probability. Imagine instead of dice it's coins, and it's only two. Your odds of getting HH today a…
runs summary(x)
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.0 62.0 149.0 216.2 300.0 1902.0
> quantile(x, c(0.5, 0.8, 0.9))
50% 80% 90%
149 350 495
A simple simulation. Run 10K times. Count the number of times it takes for three dice to add up 18.The numbers very much agree with you. The median is 149. The 90th is 495 in the simulation, which is close enough to 496. There is very much a long tail in the data. So, the median and the average will not be the same. Is it a coincidence that mean is a 216?
Re: Heuristics that almost always work
#538Earlier quoted context omitted.
Dice (typically) do not have a memory, so whatever happened yesterday will not influence what happens today. If you roll it daily, your chance of surviving at least N days is (215/216)^N, for the specific case of "rolling three 6 on three 6-sided dice" that puts you at ~50% at 149 days and at ~10% at 496 days. At sufficient scale, even incredibly unlikely things become quite probable.
runs summary(x) Min. 1st Qu. Median Mean 3rd Qu. Max. 0.0 62.0 149.0 216.2 300.0 1902.0 > quantile(x, c(0.5, 0.8, 0.9)) 50% 80% 90% 149 350 495 A simple simulation. Run 10K times. Count the number of times it takes for three dice to add up 18. The numbers very much agree with you. The median is 149. The 90th is 495 in the simulation, which is close enough to 496. There is very much a long tail in the data. So, the me…
Iteration counts gathered with Python and a (manual) binary search (actually faster than writing code).
Re: Heuristics that almost always work
#539Earlier quoted context omitted.
RandomSwede's comment is accurate, but maybe the below can help add some 'flesh' to their response. Basically, the problem is that you can't just multiply it all together. (1/6) ^ 3 is correct, and the probability of rolling 3 sixes is indeed 1/216 today , but if you repeat independent events, you don't just add up the probability. Imagine instead of dice it's coins, and it's only two. Your odds of getting HH today a…
runs summary(x) Min. 1st Qu. Median Mean 3rd Qu. Max. 0.0 62.0 149.0 216.2 300.0 1902.0 > quantile(x, c(0.5, 0.8, 0.9)) 50% 80% 90% 149 350 495 A simple simulation. Run 10K times. Count the number of times it takes for three dice to add up 18. The numbers very much agree with you. The median is 149. The 90th is 495 in the simulation, which is close enough to 496. There is very much a long tail in the data. So, the me…
Thinking about it doesn't make me feel like I'm solving a maths problem. I start stacking ideas and concepts in a way which makes me feel like I'm overlaying them in a way which is incorrect.
It makes me feel like I'm solving a riddle, which hints to me that maybe it's actually a question of semantics and definitions rather than a maths problem.
Re: Heuristics that almost always work
#540This is what Nassim Nicholas Taleb has been writing books about. He calls them black swan events, because if you took a sample of 1000 swans, chances are you'd conclude that all swans are white, but it just isn't so. People tend to round down the probability of very rare events to zero, even when the upside of them is small and the downside is catastrophically bad. Examples: the 2008 housing crisis, Fukushima, and ou…