hEllo

What’s up? Not much

[exeggcute]

had a minor crisis when 12ft.io went down yesterday and thankfully it's back now but this seems like a good opportunity to compile a list of similar paywall-evading tools in case 12ft ever gets canned for real:

12ft.io: the legend himself. definitely my favorite of the bunch by virtue of being the easiest to use (and the easiest url to remember), but it's configured to disable paywall evasion for a handful of popular sites like the new york times, so you'll have to go elsewhere for those.

printfriendly: works great; never had any issues with removing paywalls, even on domains that don't work with 12ft.io. since this site is literally designed to make sites print-friendly, it might simplify the overall formatting of the page you're trying to access, which can be a good or bad thing. my only real issue is that the "element zapper" (which lets you remove content blocks from the print-friendly preview) is a little sensitive if you're browsing on a touchscreen device, which means you might accidentally delete a paragraph when you're just trying to scroll. but if that happens you can reload the page and it'll revert everything back to its original state.

fifteen feet: basically a 12ft clone, minus 12ft's restrictions. haven't used it much since I only discovered it yesterday in the wake of 12ft's 451 error but it seems to do the trick.

archive.today: an archival tool very similar to the wayback machine, but it also works as a de facto paywall removal tool. (the wayback machine seems to remove paywalls as well, but archive.today has better UX imo and is way faster to use.)

and an honorable mention for sci-hub: only works for scientific/academic journals, not random news articles, but the other sites listed above only work for random news articles and not academic publications so you gotta have this one in your toolbelt for full coverage. pubmed is your oyster.

[electron-transport-chain]

hello again Marilyn. If u happen to check this account u will see this

[electron-transport-chain]

hi marilyn

[writing-prompt-s]

Two months into your new, anonymous pen pal relationship, you begin to realize your pen pal is actually William Shakespeare.

[beben-eleben]

For future reference.

[mollimassacre-deactivated201408]

Thank you.

[fuckyeahifightlikeagirl]

For those who would ever need it. -C

[rapeculturerealities]

reblogging here because i can see this being relevant to anyone who’s ever tried to get out of an abusive relationship

[directorlazard-deactivated20171]

Reblogging because that last comment made me reread the whole thing in a new light and realize this could be vital information. So, putting it out there for everyone, and hoping no one ever really needs it.

[thescienceofdeductionandmore]

Math “Fun” with Covid-19

You’ve got to make assumptions to create a math problem, so let’s make up these. (Warning: These assumptions are made up.) 

100 million Americans will contract coronavirus.

There are 330 million Americans.

Living in a city boosts one’s risk of contracting coronavirus by 70% compared to an American chosen at random.

I live in a city.

Based on my age and risk factors, if I contract coronavirus there is a 1.3% chance I will die from it.

Question: What is the probability I will die of coronavirus.

[thescienceofdeductionandmore]

The Greatest Logic Puzzle Ever -- The Island of the Blue Eyed People

There is an island on which live 100 people all of whom have blue eyes. But here’s the thing. There are no mirrors on the island or anything else that reflects, so nobody can tell their own eye color. They all meet in the town square at noon every day and see each other’s faces. So they all know everyone else’s eye color, just not their own. Nobody on the island ever communicates with anyone else, so they don’t tell each other their eye color. And one more thing: Everyone on the island has perfect logic, and they all know that everyone else has perfect logic. One final thing. If anyone finds out that they have blue eyes, they must leave the island the following day at noon. And they all will follow this rule obediently and know that everyone else will too.

Okay, that’s the set up. Now everything goes on boringly until one day this guru from another island who always tells the truth and who everyone believes lands on the island, shows up at the town square, looks at everyone and says, “Hey. Guess what. At least one of you has blue eyes.”

Question: What happens? Think a little before you read on.

Your first reaction will probably be, “nothing happens”. After all, the guru didn’t provide any information they didn’t already have. Everyone knew that at least one of them had blue eyes. They could each see 99 other people with blue eyes.

That reaction would be wrong. Try to figure out the answer, and ask if you want help.

Edit: Here’s the solution.

[thescienceofdeductionandmore]

This is the Greatest Logic Puzzle Ever – The Island of the Blue Eyed People

[thescienceofdeductionandmore]

Solution to the Girls and Boys problem

To remind you, here’s the problem:

There’s a country in which every family wants a girl. Every couple has children until they have a girl. Then they stop.  

Assuming a 50/50 likelihood of having a girl or a boy with each pregnancy, what is the expected ratio of girls and boys in this country?

Solution: Every child is an independent event. I.e., the probability of any child being a girl or a boy is not affected by the sex of any other child. This is just like every coin flip is independent, and getting heads or tails on previous flips doesn’t affect the outcome of any other flip. If this were not so, you could develop a system to win at roulette (or a system to have more girls than boys), which you can’t.

To see, mathematically, why this is true, let’s play it out.

A couple has their first kid. The odds are 50/50 that it’s a girl or a boy.  If it’s a girl, they stop. This will happen 50% of the time. If it’s a boy, they’ll have a second child. This will happen 50% of the time. After one child, the odds are 50/50 that it’s a girl or a boy.

The 50% of the couples whose first child is a boy have a second child. There’s a 50/50 chance the second child will be a girl or a boy. So ½ of the 50% of the couples who have a second child (25% of couples overall) will have a girl and 25% of them will have a second child who’s a boy.

Let’s add up what we know so far:

50% of the couples have just 1 girl.

25% of the couples have a boy and then a girl

25% of the couples have a boy and then a boy.

The distribution of children after up to two children over, say, 100 couples is.

1 girl - 50

1 girl and 1 boy 25

2 boys - 25

That adds up to a total of 75 girls and 75 boys for these 100 couples.

But 25% of them (the ones with 2 boys), will have a 3rd kid. Let’s do one more round to be sure it keeps working this way.

The 25% of the couples with 2 boys will have another child, which has a 50/50 shot of being a girl or a boy. So, after up to 3 kids, 12.5% of the couples will have 2 boys and a girl, and 12.5% of the couples will have 3 boys.

Distribution among 100 couples after up to 3 kids:

1 girl - 50

1 girl and 1 boy - 25

2 boys and a girl -12.5

3 boys - 12.5

That adds up to 87.5 girls and 87.5 boys.

The ones with 3 boys will have a 4th kid, and so on, but the distribution of kids after each round will stay the same because there’s always a 50/50 chance that any pregnancy will lead to a girl or a boy.

Solution: The country will be half girl and half boy.

[thescienceofdeductionandmore]

The Girls and Boys problem

There’s a country in which every family wants a girl. Every couple has children until they have a girl. Then they stop.

Assuming a 50/50 likelihood of having a girl or a boy with each pregnancy, what is the expected ratio of girls and boys in this country?

[odinsblog]

Have a great day today.

[csnews]

‘Days or months left’ for porpoise species Vaquita as it nears extinction

Catrin Rutland - August 5, 2019

It could be just days or weeks until the world’s smallest porpoise becomes extinct, experts have warned. The Vaquita species was only discovered in 1958, but within a few decades numbers have hit critical levels.

“We are down to 6-19 individuals remaining, by now these numbers have probably halved,” Professor Len Thomas, a statistician at the University of St Andrews, wrote in the British journal Royal Society Open Science.

“We are in the final few days or months for this species.”

The plight of the rare mammal has been tracked extensively, and since 2011 researchers have reported a 98.6% drop in numbers. The porpoise only lives in a small area within the Gulf of California. It has already been classified as critically endangered by the International Union for Conservation of Nature.

Keep reading

[gayfools-deactivated20200607]

if a plant is sad do u photosympathise with it

[melaninboy]

[thescienceofdeductionandmore]

Solution to the So Many Friends and Not a Single Birthday problem

To remind you, here’s the question:

Mary has 1,143 Facebook friends. Every morning, she checks to see which of her friends are having birthdays. Today, not one of her Facebook friends had a birthday. Assuming birth dates are evenly distributed throughout the days of the year (and assuming 365 days in a year - forget leap years), what were the chances of that?

Solution:

We’re trying to calculate the odds that none of the 1,143 people have their birthday on a given day. Any individual has a 1/365 chance of having their birthday on a given day. So they have a 364/365 chance of not having their birthday on a given day.

The rule we follow is this: The odds of two independent events both occurring is equal to the product of the odds of each of the events occurring.

To put that into simple terms: Events are called “independent” if the odds of one event occurring are not affected by the odds of the other event occurring. For example, every flip of a coin has a 50/50 chance, and the outcome of one flip doesn’t affect the outcome of another flip. So we call coin flips “independent” events.

(An example of dependent events. Today is snowy. Today is also cold. Are being snowy and being cold independent? No. It is more likely to be snowy on a cold day than on a warm day, so they are not independent events.) 

Getting back to the coin flips, which are independent.  If I flip a coin, the odds are ½ of getting tails. If I flip it again, the odds are ½ of getting tails. Since the two events are independent, the odds I will get tails on BOTH flips is the product of the two odds - ½ x ½ - or ¼.

That’s the rule we follow to solve this problem.

To simplify the solution, let’s imagine I have only 2 Facebook friends, not 1,143.  Assuming there are only 2 friends, and knowing that their birthdays are independent of each other: Since the odds of Friend #1 not having their birthday on that day is 364/365, and the odds of Friend #2 not having their birthday on that day is 364/365, the odds of both of them not having their birthday on that day is 364/365 x 364/365 or about 99.5%.  So I know if I have 2 Facebook friends, on any given day there is a 99.5% chance that neither of them will have their birthday.

To solve it for 3 friends, we multiply 364/365 x 364/365 x 364/365. The odds are still very high that none of them have birthdays on any given day. But they went down a drop - from around 99.5% to around 99.1%

To solve for 1,143 friends, we need to multiply 364/365 by itself 1,143 times. Stated mathematically, that’s 364/365 ^1,143 (i.e., to the 1,143 power).  And the answer is approximately 4.35%.

Answer: 4.35%

[anexperimentallife]

[mulletlove-deactivated20200811]

if she’s ur girl why is she covering me in glaze and carefully placing me in the kiln so i may become impervious and hold liquid :/

[belgaer]

I just read The Secret History and, being me, was immediately taken with Francis. 

(click for better quality)