very nerdy and fun! Thank you
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More math …
If the probability of an event is p, and there are N=1/p opportunities where the event might occur, then as p get smaller and smaller, the probability the event will occur at least once in those N trials is about 63.3%.
More formally, this probability converges to 1-1/e.
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I’ve thought that a really cool fact too! Or to put it in another way some might find more relatable … to buy 10 lotto tickets when each has a 1/10th chance of being a winner … or 100 tickets if each has a 1/100th chance, or buying a million if each has a millionth chance … as we keep increasing that number, the chances of your large stash of tickets including at least one winner approaches the value you’re referring to: 1-1/e (or about 63%). That has interesting implications. It means that I can choose any threshold of confidence I wish to achieve, and compute a finite number of tickets I need to buy to achieve it. Do I want to reduce chances of not getting a winning ticket down to below 2%? Then just take the 1/p number (N) and quadruple it. Buy 400 lotto tickets if each has a 1/100th chance, and you have better than 98% chance of at least one winner. Or make it higher yet to get as close to certainty as you wish. This means that in any infinitely long sequence of non-patterned numbers (like pi or any other irrational number), you can be guaranteed with all the certainty you want that the entire Bible, complete works of Shakespeare, Darwin’s “Origin of Species” is included in there somewhere (keeping in mind that the number of digits you need to get such a guarantee would be beyond imagination or any simple expression in scientific notation). You have to just bank on infinity to reach for such a conceptuality. But hey … it’s “there”!
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Yes, I’ve always thought that the “almost” certainty of finding a given pattern, like the works of Shakespeare, in a random sequence of sufficiently large length, a good way to think about infinity – mind boggling.
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If we are talking about thinking probabilistically, how about the “Monty Hall problem” (spoiler alert: don’t look at the reference if you want to think about it) Monty Hall problem - Wikipedia
Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?
That is, is the best strategy to switch your choice, or does it not matter whether you switch or not?
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Insufficient data 
See also XKCD, as usual.
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When there are 3 doors you have a 1 in 3 chance of picking correctly and a 2 in 3 chance of picking incorrectly. So chances are you picked incorrectly for the first choice. If given a second chance to pick from 2 doors you now have a better chance of picking correctly, and it was already established that the first door you chose is probably incorrect.
Another counterintuitive problem is the Birthday Problem. How many people do you need in order to have a 50% chance of two people sharing the same birthday?
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27?
That’s where my intuition lands; I have no idea why.
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My dad loved the Birthday Problem. He taught highschool math, and at the beginning of each year he’d get the birthdays of everyone in his classes. Class sizes are just around where you have even odds of a birthday twin, so he’d gather some real-world data to go with his foolscap solutions.
Then, he took it next level by sometimes taking attendance – by birthday. He had a knack for memorizing dates that I definitely didn’t inherit. Later on he substitute-taught in high schools in two school districts, and he’d still often ask students their birthdays. A few years later in a different class, he’d surprise a student by greeting them by their birthday instead of their name. Some were weirded out and others impressed, but I heard the story from quite a few people. Everyone knew the birthday substitute.
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Mourning Geckos (Lepidodactylus lugubris) are parthenogenetic. There are no males (or if one is rarely born it is sterile).
They also chirp to each other at night.
Also, Dart Frogs aren’t poisonous in captivity if you don’t feed them poisonous things.
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Tetrodotoxin is found in a few animal species (e.g. pufferfish), but it isn’t produced by the animals themselves. Instead, the toxin is made by symbiotic bacteria.
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Very good. One way to think about the Monty Hall problem is that your initial selection has a 1/3 probability of being the car. Monty always has a door with a goat to open, so that 1/3 probability doesn’t change after he opens a door. But there are only two doors remaining, with one containing the car. You can switch or not switch. So, by switching you have 1 - 1/3 = 2/3 probability of selecting the car.
The story goes that the great mathematician Paul Erdős (who published around 1500 mathematical papers) at first refused to accept that switching doors was the better strategy. It was only after demonstrating to him, by simulating the game several times, that he accepted the conclusion.
This is yet another illustration where our intuition, or “common sense”, can fail us.
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Part of the difficulty is that the problem is not always clearly stated. If you aren’t given the chance to change, then your odds of picking the right one are not changed by the revelation of one of the other options. I.e., in the two goats and a car version, if the narrative were instead “See, one of the ones you didn’t pick had a goat. Don’t you feel better about your choice now?”, the statistically correct answer is “no”.
A different take on choosing the right door is the short story “The Lady and the Tiger”.
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Another way to more easily understand why you should switch doors is to imagine an extreme version of the same problem. Say there are a thousand doors, and still only one of them is the winner. You choose one, which is of course only 1/1000 probability of being the good one, also meaning that it’s a 999/1000 chance the winner is among the 999 left. And then the host (just as in the other scenario) opens 998 of the other doors that are all losing ones! So of the two doors left, your original pick still has only the dismal chance of 1/1000, and the other door then has the 999/1000 chance of being the right one! Only a fool would refuse to switch if the chance was offered.
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We just did that in my statistics class!
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It’s easier to understand if you use a pack of cards instead of three doors.
Spread the cards out on a table face down, and ask some-one to pick out the ace of spades. Then look at all the other cards and turn them all over except one, which you giggle at and leave face down.
Ask them the probability that the card they picked is the ace of spades.
If necessary, repeat with one card fewer each time until there are only three left.
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This is superb. The only way to overcome our natural bias is be empirical. A lesson for all here.
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Mythbusters tested it and confirmed the expected results if anyone is interested in digging up that episode.
I’m nerdy enough to attempt a quick calculation of the birthday problem myself.
I am getting 20.
My reasoning was that the chance of one person matching another was 1/365. Then for a group of n people the combination of n people taken 2 at a time is n(n-1)/2 which I am aiming to get to 365/2 = 183. Then n=19 gives n(n-1)/2 = 171 and n=20 gives n(n-1)/2 = 190.
I can see reasons why this argument might be flawed. But that is the reason for my quick guess of 20 people on the problem.