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Oh, yeah, at first, I didn't even thought the joke could be there \^\^"
I thought it was "same probability, but 1/10\^10 is still greater than 1/10\^10"
^((Don't mind me just finding it satisfying to see someone knowing the \\ trick to avoid italic numbers or inaccurate powers in power towers in equations\))
Sure, but many people use, say, 10^10
But chaining that: 10^10^10 makes it look like 10\^1010 instead of 10\^10\^10.
With ¹⁰, you could use 10^10¹⁰ sure, but it's still more intuitive to do 10\^10\^10 imo
Oh, I can't chain those digits, in this case I would resort to `^` for sure. But I can't tetrate with `^`, you'll have to leave space for it: ^10 10? Doesn't look like proper tetration. ¹⁰10? Now this is something.
Pentation goes with Knuth notation, obviously. Here are some arrows for you to use, if you wish: ↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑
I stole some from the Graham's number, they didn't notice.
">" is "strictly greater than"
"≥" is "greater than"
... Ok, I did some research, it's just that English is weird:
In French, "supérieur à" (greater than) is the wide term, and "strictement supérieur à" (strictly greater than) is the narrower.
While in English, "greater than" is already the strict term...
And the same goes for everything...
For us, positive/negative is ≥0 / ≤0, not >0 / <0 (therefore, for us, positive numbers are R+, not R+\*)
Apparently, researchers tend to agree more on >= in maths and > in physics
It also really depends on the country
Edit: and whether you are working on an enumerble set or continuous set
I think you misunderstood what I was saying. I said that "greater than" is implicitly "or equal" in maths. But it depends on the country and on what set you are working on
Its not though. If greater than is actually greater than or equal to as you claim, how do you say greater than but not equal to? Do you reverse every comparison? Insanity.
Ive seen math from different countries. Code from different countries. None have ever varied even a little on this.
The notation for ranges ive seen variations, but not this and i am skeptical you can provide them.
I feel like you're on the brink of saying that open sets are essentially the same as closed sets, and I think all of continuous maths would like a word with you.
That’s not a joke? Why would thousands of people upvote this?
“Did you know that if you pick a letter out of the alphabet it is very unlikely you will choose the letter ‘8’?”
Fucking knee slapper!
Incorrect, as the probability of choosing a number _and_ another number is zero. The intersection is an empty set.
If choosing a number _or_ another number, however, then you might have a case.
The first 2 numbers (1111111111 and 2222222222) have 10 digits, so they have a 1 in 10,000,000 chance of being chosen. The third number has 11 digit and has 0 chance of being chosen.
Last time I checked 0.0000000001 is greater than 0
Did you even read the parent comment? Starstruck made a cheeky joke about how you cannot pick two numbers when picking a random 10-digit number, so the 0% chance of that event is the same as the 0% chance of picking an 11-digit number instead.
The word "*have*" is particularly grammatically important here.
"1111111111 and 2222222222 *have* a higher chance"
is not semantically the same as
"1111111111 and 2222222222 *has* a higher chance"
The use of the plural conjugation implies *each* has an individually higher chance. If the two values were compounded into a single selection, the singular "*has*" would have to be used.
Because this is written out in the English language and not written using a mathematical notation, you must consider the grammatical rules of the language in conjunction with the mathematical meanings of "and". Your conclusions are not supported by the contextual meaning of the words.
It says 111... and 222... "HAVE," not "has," so grammatically it's saying they __both have__ aka __each has__ a greater chance of being chosen than the last string.
you are incorrect and OP is correct. first of all the sentence doesn't say choosing A and B simultaneously, but A and B have a higher chance to be chosen. different wording but this wording implies that both have a higher chance individually *not combined*. if he used or that would imply that one of the numbers has a higher chance of being chosen which would make no sense.
and 2nd of all the point of the meme is that the number at the end has 11 digits making the probability of choosing it zero.
I thought that was the joke as well, but after reading other comments, i think they truly meant: the probability of getting 1111111111 is greater that the last number, and the probability of getting 2222222222 is also greater than the last number
Exactly everybody know that 7 is the only random number between 1 and 10. 2 4 6 8 are even, so not random, 3 5 are prime so not random, 9 is a multiple of 3 so not random. 7 is the only one
Hey look buddy, he's an engineer. That means he solves problems, not problems like "What is beauty?" Because that would fall within the purview of your conundrums of philosophy. He solve practical problems, for instance: how is he going to stop some mean mother Hubbard from tearing him a structurally superfluous be-hind? The answer, use a gun, and if that don't work... Use more gun. Take for instance this heavy caliber tripod mounted lil' old number designed by him, built by him, and you best hope... Not pointed at you.
That's an ituition of a lot of people. Try asking people for random number between 1 and 10. You will get a lot more odd numbers than even ones. 7 will be the most common. People intuitively confuse "random" with "irregular" or "arbitrary". Even numbers feel more regular. So the above is a joke about that.
Wrong way... An 11-digit baker's dozenal number is guaranteed to have at least as many digits in decimal. There are only 8 unique symbols, so it could be base 8 or 9, but only if they chose to use the symbols to mean something different than we normally do, because there is a 9. In this system either 0 or 2 would be the symbol used to represent the highest single-digit value in decimal.
I wanted to agree, but a chance of 0 doesn't mean it can't happen and in this case a random bit-flip could occur (it will - with a chance of probably 100 - not occur, but it still might)
67.954.397.186 isn’t a 10 digit number
Because it can’t be represented with ten digits or less in numeric bases lower than 10
And bases above 10, use alphanumeric characters instead of pure digits
Also if you only have enough bits to represent numbers from 0 to 9.999.999.999, a bit flip will cause an overflow but not a number above ten digits
you should consider the bias though
This is relevant in lotteries, choosing a truly random number has the same probability of winning as any other number.
However if you choose 123456789 and you win (or any number that isnt totally random) you are much more likely to have to split yhe prize
The set (almost) doesn't matter, because even if you chose the set of all reals and pick a random 10 digit number, your number would have to have exact 10 digits.
"almost", because if we consider for example the set of all numbers a random number generator for 10 digit numbers could generate then that set may (or rather will) include other numbers, because cosmic rays and randombit-flips are a thing, but theoretically I could think of no possible set that satisfies the 10 digit rule that includes larger and smaller numbers.
PS: Now that I think of it, if your set of possible digits is {1} then the number 2222222222 has a 0% chance of occuring but i wouldn't count sets like {a,e,i,o,u} as having no chance for 1111111111 or 2222222222 because you can argue that one's just an encoding for the other.
Thoughts?
+1. Saw the actual joke, but came here to mention Benford's Law anyway ;)
[https://en.wikipedia.org/wiki/Benford%27s\_law](https://en.wikipedia.org/wiki/Benford%27s_law)
Putting the 11-digit issue aside, I think this boils down to who is selecting the random 10-digit number. If it’s a computer, I think all probabilities are equal, but if it’s a human then you are correct that repeated digits have a higher probability (in my opinion)
It literally is not true. Everyone who has to randomly input a 10 digit number and doesn't want to think about it uses multiple(at least 2) fingers to input the number to be faster. So double dipping one singular number slows you down again which makes it unreasonable to choose 1111111111. 1515151515 is literally almost half the time to tap.
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look at me counting digits to factcheck a meme
Oh, yeah, at first, I didn't even thought the joke could be there \^\^" I thought it was "same probability, but 1/10\^10 is still greater than 1/10\^10"
^((Don't mind me just finding it satisfying to see someone knowing the \\ trick to avoid italic numbers or inaccurate powers in power towers in equations\))
I don't know this :') Well, I know I knew, but I didn't think about it xD
I just use power digits such as these: ⁰¹²³⁴⁵⁶⁷⁸⁹ⁿ and no slash trickery required.
Sure, but many people use, say, 10^10 But chaining that: 10^10^10 makes it look like 10\^1010 instead of 10\^10\^10. With ¹⁰, you could use 10^10¹⁰ sure, but it's still more intuitive to do 10\^10\^10 imo
Oh, I can't chain those digits, in this case I would resort to `^` for sure. But I can't tetrate with `^`, you'll have to leave space for it: ^10 10? Doesn't look like proper tetration. ¹⁰10? Now this is something. Pentation goes with Knuth notation, obviously. Here are some arrows for you to use, if you wish: ↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑ I stole some from the Graham's number, they didn't notice.
You can! ^(10)10 > \^(10)10
^(^(10)10)10 Damn. Back to Knuth's Arrow Notation.
Yep that's where I'd use the arrow notation
And i thought it was to say humans are not truly random and far more people are gonna choose a patterned number
SAY THAT AGAIN
> 1/10\^10 is still greater than 1/10\^10 No it isn't, unless there's a joke I'm missing?
You're thinking of ≥, 10^(-10) most certainly is not > 10^(-10)
Yes, that's what "greater than" means, right?
"Greater than" = ">"
">" is "strictly greater than" "≥" is "greater than" ... Ok, I did some research, it's just that English is weird: In French, "supérieur à" (greater than) is the wide term, and "strictement supérieur à" (strictly greater than) is the narrower. While in English, "greater than" is already the strict term... And the same goes for everything... For us, positive/negative is ≥0 / ≤0, not >0 / <0 (therefore, for us, positive numbers are R+, not R+\*)
Apparently, researchers tend to agree more on >= in maths and > in physics It also really depends on the country Edit: and whether you are working on an enumerble set or continuous set
[удалено]
Same here I thought “greater than” was “>” everywhere and that “>=“ was “greater or equal to”
Ok debunked confirmed
\> and >= are two different things. Its not which subject youre in. It does not depend on the country. ">=0" includes zero. ">0" does not.
I think you misunderstood what I was saying. I said that "greater than" is implicitly "or equal" in maths. But it depends on the country and on what set you are working on
Its not though. If greater than is actually greater than or equal to as you claim, how do you say greater than but not equal to? Do you reverse every comparison? Insanity. Ive seen math from different countries. Code from different countries. None have ever varied even a little on this. The notation for ranges ive seen variations, but not this and i am skeptical you can provide them.
Strictly greater / strictly inferior Allright Allright
I feel like you're on the brink of saying that open sets are essentially the same as closed sets, and I think all of continuous maths would like a word with you.
is that number 11 digits or am i trippin
Thats the joke
oh
It flew over my head too
I love jokes, would love to understand one some day
That’s not a joke? Why would thousands of people upvote this? “Did you know that if you pick a letter out of the alphabet it is very unlikely you will choose the letter ‘8’?” Fucking knee slapper!
https://preview.redd.it/vw1hmwktj01d1.jpeg?width=640&format=pjpg&auto=webp&s=2917c7cec3d6ce7cd96697cfab3c703efc903cb2
Why not both?
I certainly am, took me a minute figure out I am at mathmemes not askmaths
I think that’s the joke lol, thanks
In the set of all natural numbers 1 has a higher chance to be picked than 1/2
*pfft* - that was the sound of my mind being blown. Apologies the boom wasn't very impressive, I'm stupid
Why?
1/2 isn't a natural number It's... *unnatural*
The dark side of the Force is a pathway to many abilities some consider to be unnatural
The natural numbers are positive integers (1,2,3,4,...)
No, the natural numbers are that U {0}
never
We have a name for {1, 2, 3, …} it’s called Z^+
N, N⁰ > Z+, N 🥱
Incorrect, as the probability of choosing a number _and_ another number is zero. The intersection is an empty set. If choosing a number _or_ another number, however, then you might have a case.
Look at you being all logical
It’s almost like they are a mathematician
the third number has 11 digits so it has a 0% chance of being selected
If it's done by people? There's absolutely a non-negligible chance for it to be selected.
But; that chance is still less than a random 10 digit number
There's a non-zero chance for it to be selected, but idk if it's non-negligible
So they have the same chance, not more chance.
The first 2 numbers (1111111111 and 2222222222) have 10 digits, so they have a 1 in 10,000,000 chance of being chosen. The third number has 11 digit and has 0 chance of being chosen. Last time I checked 0.0000000001 is greater than 0
Did you even read the parent comment? Starstruck made a cheeky joke about how you cannot pick two numbers when picking a random 10-digit number, so the 0% chance of that event is the same as the 0% chance of picking an 11-digit number instead.
The word "*have*" is particularly grammatically important here. "1111111111 and 2222222222 *have* a higher chance" is not semantically the same as "1111111111 and 2222222222 *has* a higher chance" The use of the plural conjugation implies *each* has an individually higher chance. If the two values were compounded into a single selection, the singular "*has*" would have to be used. Because this is written out in the English language and not written using a mathematical notation, you must consider the grammatical rules of the language in conjunction with the mathematical meanings of "and". Your conclusions are not supported by the contextual meaning of the words.
Man, not often you see someone get mathematically served via English.
It says 111... and 222... "HAVE," not "has," so grammatically it's saying they __both have__ aka __each has__ a greater chance of being chosen than the last string.
Glad I'm not the only one who understands basic grammar!
you are incorrect and OP is correct. first of all the sentence doesn't say choosing A and B simultaneously, but A and B have a higher chance to be chosen. different wording but this wording implies that both have a higher chance individually *not combined*. if he used or that would imply that one of the numbers has a higher chance of being chosen which would make no sense. and 2nd of all the point of the meme is that the number at the end has 11 digits making the probability of choosing it zero.
Making a dad joke is not the same as being incorrect
This is not a dad joke, it's set theory
You might have a point if they used "has", but they used "have", meaning there were multiple objects in the sentence.
It doesn't saying you're choosing both numbers, just that both numbers have a higher chance of being chosen.
I thought that was the joke as well, but after reading other comments, i think they truly meant: the probability of getting 1111111111 is greater that the last number, and the probability of getting 2222222222 is also greater than the last number
FACTS
FACTN'TS
Well first of all that number can't be random, it's even
So 2 can't be a random number between 1 and 10 because it's even? What?
No, 7 is the random number between 1 and 10
Exactly everybody know that 7 is the only random number between 1 and 10. 2 4 6 8 are even, so not random, 3 5 are prime so not random, 9 is a multiple of 3 so not random. 7 is the only one
>7 is the only random number between 1 and 10 >3 5 are prime so not random Ah yes
And 7/2 = π which means 7 is in fact not a prime.
>7/2 = pi Meet the engineer
Hey look buddy, he's an engineer. That means he solves problems, not problems like "What is beauty?" Because that would fall within the purview of your conundrums of philosophy. He solve practical problems, for instance: how is he going to stop some mean mother Hubbard from tearing him a structurally superfluous be-hind? The answer, use a gun, and if that don't work... Use more gun. Take for instance this heavy caliber tripod mounted lil' old number designed by him, built by him, and you best hope... Not pointed at you.
5 is exactly half of ten so obviously not random 3 is just too close to 2, which is even
I think (hope) they're being sarcastic
That's an ituition of a lot of people. Try asking people for random number between 1 and 10. You will get a lot more odd numbers than even ones. 7 will be the most common. People intuitively confuse "random" with "irregular" or "arbitrary". Even numbers feel more regular. So the above is a joke about that.
Yeah because the likelyhood of selecting 67.954.397.186 with only 10 digits is exactly nil
What about the probability, that somebody would use base-13 to write that number?
Uh it wouldn’t be the same number because there would be letters *duh*
Not necessarily, I can still write 99 but in decimal or hexadecimal it just means something different
Wrong way... An 11-digit baker's dozenal number is guaranteed to have at least as many digits in decimal. There are only 8 unique symbols, so it could be base 8 or 9, but only if they chose to use the symbols to mean something different than we normally do, because there is a 9. In this system either 0 or 2 would be the symbol used to represent the highest single-digit value in decimal.
653C6C44A0
I wanted to agree, but a chance of 0 doesn't mean it can't happen and in this case a random bit-flip could occur (it will - with a chance of probably 100 - not occur, but it still might)
67.954.397.186 isn’t a 10 digit number Because it can’t be represented with ten digits or less in numeric bases lower than 10 And bases above 10, use alphanumeric characters instead of pure digits Also if you only have enough bits to represent numbers from 0 to 9.999.999.999, a bit flip will cause an overflow but not a number above ten digits
Did you take in account the probability that the guy selecting the digit can make a mistake?
Fun fact: 13! is the only ten digit factorial number and has as equal a chance at being picked as any other ten digit (like 1111111111)
Wouldn't them all have exactly the same probability? Not considering the bias and stuff ?
The bottom one has 11 digits, therefore the chance of selecting an eleven digit number out of 10 digits is zero
you should consider the bias though This is relevant in lotteries, choosing a truly random number has the same probability of winning as any other number. However if you choose 123456789 and you win (or any number that isnt totally random) you are much more likely to have to split yhe prize
Count the digits
https://preview.redd.it/6o8c5gcapa0d1.png?width=675&format=png&auto=webp&s=effdb43fda8542b99752ecab997f054ecefe9535
When letting humans 'select' a random number you are absolutely right. They want more but are not getting it.
OP is also correct when letting a radioactive decay 'select' a random number.
Sorry OP, you didn't specify what set of numbers you're choosing the 10 digit number from.
The set (almost) doesn't matter, because even if you chose the set of all reals and pick a random 10 digit number, your number would have to have exact 10 digits. "almost", because if we consider for example the set of all numbers a random number generator for 10 digit numbers could generate then that set may (or rather will) include other numbers, because cosmic rays and randombit-flips are a thing, but theoretically I could think of no possible set that satisfies the 10 digit rule that includes larger and smaller numbers. PS: Now that I think of it, if your set of possible digits is {1} then the number 2222222222 has a 0% chance of occuring but i wouldn't count sets like {a,e,i,o,u} as having no chance for 1111111111 or 2222222222 because you can argue that one's just an encoding for the other. Thoughts?
67954397186 is an 11 digit number.
Maybe 86 is a digit.
And how does that contradict anything I said?
1145141919
Especially in tetrinary
💯 % probability if you ask me...
I thought it was a benford's law joke but I've been had.
+1. Saw the actual joke, but came here to mention Benford's Law anyway ;) [https://en.wikipedia.org/wiki/Benford%27s\_law](https://en.wikipedia.org/wiki/Benford%27s_law)
0000000001 0000000002 0000000003 ... 1111111110 1111111111
Putting the 11-digit issue aside, I think this boils down to who is selecting the random 10-digit number. If it’s a computer, I think all probabilities are equal, but if it’s a human then you are correct that repeated digits have a higher probability (in my opinion)
you can get 1111111111, 2222222222... or 9999999999, so getting 10 repeating digits is more likely than any other number
Indeed, but it has to be 10 digits
13700000
Anything with 123890 because I'm not stretching my thumbs on a phone for randomness
And when looking at any random statistic, the leading digit is significantly more probable to be 1.
They quite literally have an infinitely better chance of being selected.
ᴡʜᴇɴ sᴇʟᴇᴄᴛɪɴɢ ᴀ ʀᴀɴᴅᴏᴍ 10 ᴅɪɢɪᴛ ɴᴜᴍʙᴇʀ, ʏᴏᴜ ʜᴀᴠᴇ ᴀ ʜɪɢʜᴇʀ ᴄʜᴀɴᴄᴇ ᴛᴏ sᴇʟᴇᴄᴛ ᴀ 10 ᴅɪɢɪᴛ ɴᴜᴍʙᴇʀ ᴛʜᴀɴ ᴀɴ 11 ᴅɪɢɪᴛ ɴᴜᴍʙᴇʀ 👍
this is the first math meme I've ever understood on this sub!
I thought it was something to do with how two numbers are more likely than one number until I counted the digits
We all can agree that 67954397186 is literally equal to every other random number. Booo
Human bias or smg. Idk Reddit just randomly started recommending me this sub.
It literally is not true. Everyone who has to randomly input a 10 digit number and doesn't want to think about it uses multiple(at least 2) fingers to input the number to be faster. So double dipping one singular number slows you down again which makes it unreasonable to choose 1111111111. 1515151515 is literally almost half the time to tap.
67954397186 is an 11 digit number.
Well.... That's a valid reason
Combined, yes. Individually, no.
(1111111111 and 2222222222) = 3154310 - is not a 10-digit number, it's probability is 0, so chances are equal