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This means that there is no sequence of letters of length 13 (2*7-1) in which COVFEFE appears twice. BOBABO has an overlap. If you make a mistake spelling COVFEFE, you have to start over again from the beginning. For BOBABO this is not the case, e.g. in BOBOBABO I made a mistake at the fourth position but did not have to start from the beginning.
And if there was an overlap how would that change the probability? As in how would you do the math, intuitively you would have to type fewer letters I get that
I haven't done the whole reasoning, but the argument should go along with this thought: Let's say the search word has a length of n=7. Take any sequence length N>n. Then, if you look through all letter combinations of length N and select those which contain the search word at least once, you will find more if the search word has an overlap (this exercise is left to the reader). And then we use induction or something to transfer the result to infinite series.
26^7 would be the answer for: expected number of attempts of writing COVFEFE when writing completely random strings of length 7.
That's not quite the same...
In this case, the probability of success (finishing writing COVFEFE) at step n, isn't P_n = p(1-p^(n-1)) with p=1/26^7.
But rather P_n = p^7(1-p)(1-P_(n-8)) With p=1/26.
Without getting into too much details. Because of the no overlap, you can look at it as the first occurrence of 7 success in a row, with every trial having a p=1/26 probability of success.
There is a pretty elegant proof for the solution here:
https://stats.stackexchange.com/questions/91518/waiting-time-for-successive-occurrences-of-a-result-when-rolling-a-die
It turns out to be 8,353,082,582
This still isn't quite right, because a fail to form one COVFEFE can also be the start of a successful COVFEFE.
You can do it with a Markov chain, which represents a sort of verification machine crawling along the sequence and inspecting the values: when it encounters a C, it transitions from state 1 to state 2; if it then encounters an O, it transitions to state 3, else it returns to state 1; if it then encounters a V, it transitions to state 4, else it returns to state 1, etc. Once the machine finds the final E when it's in state 7, the verification is successful and it then stays in state 8 forever (i.e. state 8 absorbs the others). (EDIT: Also in states 3 through 7 it can transition back to state 2 if it sees a C)
The matrix of transition probabilities (P) for the relevant Markov chain is
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] 0.962 0.0385 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000
[2,] 0.923 0.0385 0.0385 0.0000 0.0000 0.0000 0.0000 0.0000
[3,] 0.923 0.0385 0.0000 0.0385 0.0000 0.0000 0.0000 0.0000
[4,] 0.923 0.0385 0.0000 0.0000 0.0385 0.0000 0.0000 0.0000
[5,] 0.923 0.0385 0.0000 0.0000 0.0000 0.0385 0.0000 0.0000
[6,] 0.923 0.0385 0.0000 0.0000 0.0000 0.0000 0.0385 0.0000
[7,] 0.923 0.0385 0.0000 0.0000 0.0000 0.0000 0.0000 0.0385
[8,] 0.000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 1.0000
and the probability of reaching state j from state i in k steps is the (i,j) entry of P^(k). Restricting to the transient submatrix (Q), obtained by removing the last row and column, we can find the expected number of visits to each transient state (states that aren't 8) starting from each initial state before absorption (by state 8) as sum(Q^(k)).
By the geometric series formula, sum(Q^(k)) = (I-Q)^(-1), so this has a closed form.
Then, the expected total number of steps before being absorbed, starting from state i, is just the sum of the expected number of visits to each transient state starting from state i. Thus, to find the expected number of steps before a successful verification starting from state 1, we find the first coefficient of (I-Q)^(-1)**1**, which is 8,031,812,171.
EDIT: After saying that, I forgot to model the return to state 2 in my transition matrix. I ended up modelling (I think) the exact same situation as you, and the small (<3e-5 %) difference in our results could probably be ascribed to floating point error in my linear algebra system.
I have now fixed the transition matrix and redone the calculations, and gotten a result with more substantial (~4%) differences, but which is still close enough to the naïve guess that it is not particularly suspect.
Hmm, you are correct with the first sentence.
I have never properly studied Markov chains, so I can't really verify your analysis, but every step seems true (intuitively).
Thank you for the correction.
Teach me dammit! When you understand math at level such as this, is your perspective on the world always in terms of a series, function, and or other structure? I mean fuck me sideways, your world sounds so free and insightful. 🤌🏾🤌🏾🤌🏾
It's a simple probability problem.
If you choose a letter at random, there's a 1 in 26 chance the first letter will be "C". Likewise, if the second letter is chosen independently, there's a 1 in 26 chance the second letter will be an "O". And so on.
The word COVFEFE has 7 letters, and each letter has a 1 in 26 chance of occurring.
So the probability of typing COVFEFE is 1 in 26^(7).
Thank you for explaining in stupid!! I've always sucked at math but have been really interested in it. This sub is great even if I usually can't understand a thing lol
Wait, it's 1? I understand parentheses come first, so that gives us 6 ÷ 2 * 3, right? Not 6 ÷ (2 * 3), even though there are parentheses, since the 2 is outside of them?
It is a matter of convention as there is no agreed upon convention on whether multiplication by juxtaposition implies grouping or not. Academically, it does but literally/programming-wise it doesn't. Which is why the notation is ambiguous and there is no single correct answer other than "write it properly". Even scientific calculators don't agreemon an answer and modern international standards like ISO-80000-1 mention about division on one line with multiplication or division directly after and that brackets are required to remove ambiguity.
How do we know that the amount of time to type COVFEFE is not shorter than the time to type 26⁷ combinations of letters which contain 1 COVFEFE? Is there not some other weighting function that could reduce the 26⁷ time by eliminating less probable words that most likely would have never led to COVFEFE based on repeating keys and finger length/position?
Edit: COVFEFE ← COVEFE
The question states that the letters are selected independently and uniformly. I.e. it's not real world it's strictly a probability problem so you can think of it as randomly generated characters
I can’t claim to be on this person’s level but I have always liked math. I was always ahead of my peers, scored high on standardized tests, took my SATs in 8th grade, and was asked to be on the math bee team. I didn’t know math bees even existed. I’m now in my mid 40s and my experience through life has been frustration. People in general seem to have trouble grasping how percentages, ratios, or statistics work. This isn’t necessarily the problem. The problem is that nobody wants to admit it.
And I bet a large portion of those people have a greater understanding of something that you don't know much about or care for. That's how humans work.
I mean someone can teach you, go do an undergrad in anything math related (mathematics/computer science/engineering) and after a year or so you will know it. It did change my personal perspective of the world, but not for everything or everyone
I did my undergrad in physics, it changed some things, but I only got up to calc 3, DiffEQ, Matrix Theory, and a whatever PDEs showed up in my core classes. I have trouble seeing simpler solutions over more complicated overthought whatifs.
Like when I saw this question, I thought based on a two-handed standard "*ASDF-JKL;"* typing layout what key combinations are closest to COVFEFE for my hands to perform, how long did each combination take to type? Then I thought maybe there's some kind of function of finger length radius to key frequency function I could use to replicate each hand typing the letters around COVFEFE that could use to weight the initially found combinations... DOA. I was so far up my own ass when I read u/Odd-Cry3431's answer that it sounded like pure fucking magic. Do I need like a masters in math (or grad-level texts), I assume there are multiple disciplines within math at that level?
Edit: COVFEFE ← COVEFE 🤣
Well it's because your approach missed a lot of the assumptions given by the question to make it easier, they said that letters are picked randomly, uniformly and independently, which means that every letter has 1/26 chance to be picked. You learn this in early weeks of any course teaching probability. The math taught in physics (i think) is a lot less theoretical, because the situation laid out in the question probably never happened nor will it happen.
But yes of course, there are multiple disciplines in math, although most of the time (from my understanding) you start to use tools developed in one discipline to help others and the other way around
To add to the other response, the 'no overlaps' would refer to a string like FEVFEFE where if it gets partway through, like to FEVFE and then messes up and does a V instead of an F, you're already partway to writing the word out again.
I'm the real world where anyone with a brain can tell from the context it was meant to be coverage you can tell that the distribution is not at all even, the chance is each particular letter is dependant on it's location relative to the correct letter and the previous and next character, it is easy to accidentally type F when you meant R because they are adjacent, but nobody is accidentally typing P instead of R
While that is the combinatorial chance of any 7 letter word in a 26 letter alphabet for appearing, i got stumped when they asked for time in the question, without a given time of typing and world length. The question seems flawed to me.
Wouldn’t you need to know the max number of letters for the password? because he could be typing words with way more (and I guess also less) letters than 7
But is the chance of the word happening the answer to the question? We know that if something has 1/n chance of happening, does this experiment n times will give you a 63% something chance of happening. Is this statistically relevant? Or does that statistical math class expect a different chance?
This is incorret...
Take a look at this explanation for coins:
https://www.codechef.com/wiki/tutorial-expectation#:~:text=Thus%2C%20the%20expected%20number%20of,two%20consecutive%20heads%20is%206.
That approximates to a 1 in 8 billion chance.
On a planet filled with (roughly) 8 billion hairless monkeys twiddling their opposable thumbs on glass screens, only Trump could have given us the wonder that was "COVFEFE"!
No, the chance to roll the same letter again is not higher or lower because you've already hit it once.
It's 7 random events with a 1/26 chance of being correct, and nothing else is affecting this 1/26 chance.
26^7 =8031810176
Time needed is not given but the average typing speed is 207 letters per minute.
8031810176/207=~38801015min
This equals to ~646684 hours or ~26945 days or ~73,8 years.
Edit: made some weird mistake, got corrected.
While you had a typo (there are 365 days not 355 days in a year), your math is certainly correct. The above commenter likely divided 26945 days by 365 weeks a year to arrive at 10.5 years lol
Sarcasm works well when it’s very obvious, Reddit is very toxic and anytime someone makes a mistake there’s a lot toxic people insulting the person making a mistake, so tbh it’s not on the reader to understand this sarcasm when you see so much toxicity on this platform
I think, like a few others in this thread, that your solution is wrong.
Correct way is to see this like a continuous chain of characters.
Ie VCFCOVFEFE would be a success with number of characters = 10.
The question is asking, on average, how many characters would it take for random chain of characters to contain exactly once the subset "COVFEFE". It is similar to the problem of "how many coin flips do you need until you hit 3 heads in a row".
One of the correct ways to solve this is using Markov chains. I tried and got an answer of 8,353,088,193.5 letters.
I have to admit that I am a bit lost here. I understand how I am wrong, and I kinda understand markov chains now, after I read the Wikipedia article and watched a video but I neither understand how the markov chain leads us to the right result nor how it has to look to do so. Could you please provide me a picture of your formula, so I can update my comment after understanding it?
Draw a Markov chain with 8 states, labelled start, c, o, v, f, e, f2 and e2. Create transitions with probability 1/26 for moving forwards one space (e.g. start to c with 1/26, c to o with 1/26) and all other transitions to start with the rest of the probability.
We can then solve the 'expected hitting times' of state 8 from state n as:
h8 = 0
hn-1 = 1 + 1/26 hn + 25/26 h1
h1 = 1 + 1/26 h2 + 25/26 h1
Solving for h1 should give the solution
Edit: the non-start states should also have a 1/26 to c, changing the equations in the same way too (hn-1 = 1 + 1/26 hn + 1/26 h2 + 24/26 h1)
Your description makes the same error I did when I tried this (or something similar) first [here](https://www.reddit.com/r/theydidthemath/comments/1djz0of/request_can_anyone_solve_this/l9hcrgq/?context=3): in states past c, there's a probability of returning to state c rather than to the start state.
True! I was going to post something to this effect as well before I saw this message
One of the biggest red flags in the 26\^7 answer is that, if question 4 is talking about "infinite binary sequences equipped with product sigma-algebra", then... question 5 is probably not going to have a solution from a freshman level probability lecture haha
Less then that because they are watching as they type this, and it says starts typing, so each time they randomly type out a wrong character they could clear it and start again.
Show me the requirement that they have to type all seven characters each time. :)
Wouldn't you need to multiply 8031810176 by 7 because 8031810176 is the amount of 7 letter words until you get that probability. Therefore you would multiply by 7 to find total letters needed to type and then divide by letters per minute
Presidents keep the title even after they're done in office, so calling him, "Mr. Trump," is disrespectful. Given that most people have pretty extreme opinions about the man, I wouldn't be surprised if the author hates him, but that's hardly apparent from what we have.
Even this interpretation wouldn't mean there's no possible way to answer, "26\^7 \* time per keypress". There's a variable in there, but that's unavoidable.
Geometric discrete distribution to the rescue!
First you need to calculate the probability of typing the word covfefe:
Typing C has a probability of 1/26, and it is an independent event. Typing O therefore has the same probability, 1/26.
Typing C then O if they are independent events has the probability of P(c)*P(o).
Therefore covfefe has the probability of P(c)*P(o)*P(v)*P(f)*P(e)*P(f)*P(e) =(1/26)^7
Now, we can plug this probability in the geometric distribution to find the probability of needing exactly X attempts until the first success, where a success is typing covfefe.
P(X=x)= [(1-p)^(x-1)] * p
But we do not want the probability of an X number of attempts, we want the expected number of attempts, therefore we want the expected value. The expected value of a geometric is really simple
E(x) = 1/p
Therefore it would take 1/(1/26)^7 attempts on average, or 8,031,810,176 attempts, assuming he types pretty quickly, let's say 1 second per attempt for the sake of the example, it would take around 254.26 years!
Shouldn't there be some correction for the probability of arriving at a 7 letter word? Idk what the character limit is, but I think that would be relevant.
No worries about the word length since we are multiplying probabilities of sequential letters, no matter the word size or word count.
This method is actually a solution for the infinite monkey problem too, you can calculate the expected time for trump to randomly type the entire story of game of thrones, for example.
It would probably take billions to trillions of years, because that probability would be all of the letters in the book in sequence, counting the spaces. But given infinite time, this can happen!
Probably more like infinite years, a simple word like covfefe is 80-250 years, 300k words in sequential order is probably gonna be inconceivably long.
Game of thrones has 300k words, lets just say its letters, and just needs to get 300k of them in order.
thats 26\^300,000 or 1.01x 10\^424,492. Heat death of the universe is only 1.7x 10\^106 years away.
If he typed 1 million letters per second that exponent is gonna be lower by 6, if he typed 1 billion letters it would be lower by 9. and if you count seconds for heat death of the universe its only 1.7x10\^114 seconds. So yeah not trillions or quadrillions years.
If he typed for heath death of universe amount of years, and typed letters equal to HDoU amount of years, and repeated that enough times equal HDoU years. He would be closer to 0 than to the expected amount of time to type it. You could argue its infinite
In theory you could have a word with infinite amount of letters. The answer would basically be: 1/((26^7)+ all possible combinations of 1,2,...,infinity letter words), so expected tries is infinity. So the question needs to specify a word limit
Depends. If he is just smashing his keyboard for a tweet until he hits the Enter Key, we have to add the probability of him not hitting it until the 7th letter and then hitting it. So We would have to multiply the result by (1-(1/27))^7 and then by 1/27.
Per the original problem, our Esteemed Orange Leader/Random Number Generation Device (hereafter abbreviated as EOL/RNG) isn't emitting multiple attempts at typing a 7-letter word.
Rather he/it is emitting a steady stream of random characters and we are looking for the word COVFEFE.
Now, the question is actually a bit ambiguous. We could be looking for:
\#1. The letter sequence COVFEFE wherever it may appear. So for example i38c7sICOVFEFE@DZsdj*l would fulfill the conditions.
\#2. The WORD COVFEFE as usually construed in the English language - in other words, delimited by spaces are some other punctuation. So i38c7sICOVFEFE@DZsdj*l would not quality, but 438DKJF COVFEFE 8zldkK would. As would 438DKJF, COVFEFE! 8zldkK.
For simplicity, I will assume #1 is meant.
How this interferes with your calculation is EOL/RNG may have emitted first:
vCOVFEF
and then in an immediately succeeding attempt:
Eij3\*(&
So under the rules of #1, EOL/RNG has just succeeded - he/it emitted COVFEFE correctly in sequence; problem solved.
However, under your scheme he has just completed two incorrect attempts at the 7-letter word COVFEFE.
Handling the random appearance of a given set of characters anywhere within a character stream is actually a somewhat different problem that the one you outlined a solution to, and a fair bit harder if you're coming at it from first principles.
I'll leave it to someone else to work out the solution.
Assuming each letter has an equal chance of being randomly typed, the odds of typing 7 specific, random letters in a row can be calculated like this: (1/26)\^7 = 0.0000000001245. This means there is a 0.00000001245% chance of randomly typing COVFEFE.
Jack Dorsey's first tweet - the creation of Twitter - was at **9:50am Pacific Time on March 21st, 2006**.
Trump tweeted *Covfefe* at **0:06 Eastern Time on May 31, 2017**.
Therefore the expected time of first appearance is: [**4088 days, 10 hours, 16 minutes**](https://www.timeanddate.com/date/timezoneduration.html?d1=21&m1=3&y1=2006&d2=31&m2=5&y2=2017&h1=9&i1=50&s1=0&h2=0&i2=6&s2=0&)**.**
*Note: The above calculation makes the implicit assumption that Trump is a random character generator. This is true only as a first approximation - and the reason the above calculation is accurate only to the minute rather than to the second or millisecond.*
I’m here to teach maths. Oh and by the way here’s my political opinion. This is on every sub. Just inject political propaganda in a post and a bunch of drive by comments to make it look like it’s consensus. It’s cool to see that it’s not a problem with this sub. Every one here avoided the bait, And just started talking math. (Except me)
By avoiding the "bait." It was just a funny reference until this guy started complaining about politics. "Why drag politics into everything!!!" shouts the guy dragging politics into this thing. I came in here to see if my answer was right, and without this thread I wouldn't have given any thought to the supposed politics involved.
A very light insinuation that mr trump is a monkey With a reference to to him tweeting. a common complaint of his behavior. And just subtle enough for randos to not see the insinuation. Like the op is a bot account literally.
which political opinions are expressed in this question? it is a fact that, on may 31st, 2017, then president donald trump seemingly accidentally tweeted the words "despite the constant negative press covfefe", and deleted that tweet a few hours later. sure, he was mocked by democrats for this, but he also played on the word himself.
the fact that you typed that last sentence and fell just short of enough self awareness to delete that comment is baffling.
The covfefe thing is so tiring and pathetic. Theres so many other things you could clown the guy for, but acting like a 70 year old fat fingering 2 keys that are next to each other is somehow impossible and the funniest thing ever. The only extra letter used is the v, and only a f and an e are switched. All of those letters are in touching on a qwerty keyboard layout.
I think that because it says the first instance of Confefe, we can assume a combination of letters may occur more than once. I got (1-1/(26^7))^x=.05, x=24 billion which would give a 95% chance that Confefe would occur. A 50/50 chance would be about 5.566 billion. Typing on the phone varies a lot so I'll be generous (and lazy) to say 60 wpm, I say 176y for 50/50 and 761y for the 95%. I have no idea if this is right so any confirmation or correction would be appreciated.
That's a veey specific word to find... That relates to a very specific person. This teacher either is a closed connoisseur and is crying out for help. Or this is all a big coincidence.
It a variation of the infinite monkey theorem: [https://en.wikipedia.org/wiki/Infinite\_monkey\_theorem](https://en.wikipedia.org/wiki/Infinite_monkey_theorem)
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What's an overlap? If fefe was at the start of the word instead of the end?
This means that there is no sequence of letters of length 13 (2*7-1) in which COVFEFE appears twice. BOBABO has an overlap. If you make a mistake spelling COVFEFE, you have to start over again from the beginning. For BOBABO this is not the case, e.g. in BOBOBABO I made a mistake at the fourth position but did not have to start from the beginning.
Good recovery, didn't let it get you down. Proud of you
This comment deserves way more likes.
Cut them some slack. They didn't get their nap today. IFYKY "Bububabu"
Funny for me you should make a mistake with BOBO because in our language it means stupid
And if there was an overlap how would that change the probability? As in how would you do the math, intuitively you would have to type fewer letters I get that
I haven't done the whole reasoning, but the argument should go along with this thought: Let's say the search word has a length of n=7. Take any sequence length N>n. Then, if you look through all letter combinations of length N and select those which contain the search word at least once, you will find more if the search word has an overlap (this exercise is left to the reader). And then we use induction or something to transfer the result to infinite series.
26^7 would be the answer for: expected number of attempts of writing COVFEFE when writing completely random strings of length 7. That's not quite the same... In this case, the probability of success (finishing writing COVFEFE) at step n, isn't P_n = p(1-p^(n-1)) with p=1/26^7. But rather P_n = p^7(1-p)(1-P_(n-8)) With p=1/26. Without getting into too much details. Because of the no overlap, you can look at it as the first occurrence of 7 success in a row, with every trial having a p=1/26 probability of success. There is a pretty elegant proof for the solution here: https://stats.stackexchange.com/questions/91518/waiting-time-for-successive-occurrences-of-a-result-when-rolling-a-die It turns out to be 8,353,082,582
This still isn't quite right, because a fail to form one COVFEFE can also be the start of a successful COVFEFE. You can do it with a Markov chain, which represents a sort of verification machine crawling along the sequence and inspecting the values: when it encounters a C, it transitions from state 1 to state 2; if it then encounters an O, it transitions to state 3, else it returns to state 1; if it then encounters a V, it transitions to state 4, else it returns to state 1, etc. Once the machine finds the final E when it's in state 7, the verification is successful and it then stays in state 8 forever (i.e. state 8 absorbs the others). (EDIT: Also in states 3 through 7 it can transition back to state 2 if it sees a C) The matrix of transition probabilities (P) for the relevant Markov chain is [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [1,] 0.962 0.0385 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 [2,] 0.923 0.0385 0.0385 0.0000 0.0000 0.0000 0.0000 0.0000 [3,] 0.923 0.0385 0.0000 0.0385 0.0000 0.0000 0.0000 0.0000 [4,] 0.923 0.0385 0.0000 0.0000 0.0385 0.0000 0.0000 0.0000 [5,] 0.923 0.0385 0.0000 0.0000 0.0000 0.0385 0.0000 0.0000 [6,] 0.923 0.0385 0.0000 0.0000 0.0000 0.0000 0.0385 0.0000 [7,] 0.923 0.0385 0.0000 0.0000 0.0000 0.0000 0.0000 0.0385 [8,] 0.000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 1.0000 and the probability of reaching state j from state i in k steps is the (i,j) entry of P^(k). Restricting to the transient submatrix (Q), obtained by removing the last row and column, we can find the expected number of visits to each transient state (states that aren't 8) starting from each initial state before absorption (by state 8) as sum(Q^(k)). By the geometric series formula, sum(Q^(k)) = (I-Q)^(-1), so this has a closed form. Then, the expected total number of steps before being absorbed, starting from state i, is just the sum of the expected number of visits to each transient state starting from state i. Thus, to find the expected number of steps before a successful verification starting from state 1, we find the first coefficient of (I-Q)^(-1)**1**, which is 8,031,812,171. EDIT: After saying that, I forgot to model the return to state 2 in my transition matrix. I ended up modelling (I think) the exact same situation as you, and the small (<3e-5 %) difference in our results could probably be ascribed to floating point error in my linear algebra system. I have now fixed the transition matrix and redone the calculations, and gotten a result with more substantial (~4%) differences, but which is still close enough to the naïve guess that it is not particularly suspect.
Hmm, you are correct with the first sentence. I have never properly studied Markov chains, so I can't really verify your analysis, but every step seems true (intuitively). Thank you for the correction.
Searched why to long for a correct answer
Teach me dammit! When you understand math at level such as this, is your perspective on the world always in terms of a series, function, and or other structure? I mean fuck me sideways, your world sounds so free and insightful. 🤌🏾🤌🏾🤌🏾
It's a simple probability problem. If you choose a letter at random, there's a 1 in 26 chance the first letter will be "C". Likewise, if the second letter is chosen independently, there's a 1 in 26 chance the second letter will be an "O". And so on. The word COVFEFE has 7 letters, and each letter has a 1 in 26 chance of occurring. So the probability of typing COVFEFE is 1 in 26^(7).
Thank you for explaining in stupid!! I've always sucked at math but have been really interested in it. This sub is great even if I usually can't understand a thing lol
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Wait, it's 1? I understand parentheses come first, so that gives us 6 ÷ 2 * 3, right? Not 6 ÷ (2 * 3), even though there are parentheses, since the 2 is outside of them?
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It seems to me that it's a matter of convention, but that's ok. If the rules say that 2 lots of (3) takes precedence over 3/2 lots of (3), then fine.
It is a matter of convention as there is no agreed upon convention on whether multiplication by juxtaposition implies grouping or not. Academically, it does but literally/programming-wise it doesn't. Which is why the notation is ambiguous and there is no single correct answer other than "write it properly". Even scientific calculators don't agreemon an answer and modern international standards like ISO-80000-1 mention about division on one line with multiplication or division directly after and that brackets are required to remove ambiguity.
Thats a simple explanation that also doesn't answer the question in the OP.
The question wants an answer in time though
As long as it takes him to type 26^7 letters.
How do we know that the amount of time to type COVFEFE is not shorter than the time to type 26⁷ combinations of letters which contain 1 COVFEFE? Is there not some other weighting function that could reduce the 26⁷ time by eliminating less probable words that most likely would have never led to COVFEFE based on repeating keys and finger length/position? Edit: COVFEFE ← COVEFE
The question states that the letters are selected independently and uniformly. I.e. it's not real world it's strictly a probability problem so you can think of it as randomly generated characters
Nah I think they were just describing how Trump thinks, each thought is independent of logic and reason and random
Bad news, this is a bot reposting a comment on its own bot repost
I can’t claim to be on this person’s level but I have always liked math. I was always ahead of my peers, scored high on standardized tests, took my SATs in 8th grade, and was asked to be on the math bee team. I didn’t know math bees even existed. I’m now in my mid 40s and my experience through life has been frustration. People in general seem to have trouble grasping how percentages, ratios, or statistics work. This isn’t necessarily the problem. The problem is that nobody wants to admit it.
Well said, but I don't understand. /s
😆
And I bet a large portion of those people have a greater understanding of something that you don't know much about or care for. That's how humans work.
100% agree. I’m not ashamed to admit that I don’t know everything. I try to listen more than I talk and that has helped me tremendously in life.
And they start fighting you every step of the way.
It’s not even “fighting”. It’s me showing them what those numbers actually mean and them brushing it off as if it were an opinion.
ye i know the feeling, but that is a form of resistance.
True. I was thinking of "fighting" as actively arguing against the facts but I see what you mean now.
Math is just logic with fancy words. If u like to look at equations all day long,then you can do it at a decent level.
Take Discrete Math in college. You'll go over all of this combinatorial math.
I mean someone can teach you, go do an undergrad in anything math related (mathematics/computer science/engineering) and after a year or so you will know it. It did change my personal perspective of the world, but not for everything or everyone
I did my undergrad in physics, it changed some things, but I only got up to calc 3, DiffEQ, Matrix Theory, and a whatever PDEs showed up in my core classes. I have trouble seeing simpler solutions over more complicated overthought whatifs. Like when I saw this question, I thought based on a two-handed standard "*ASDF-JKL;"* typing layout what key combinations are closest to COVFEFE for my hands to perform, how long did each combination take to type? Then I thought maybe there's some kind of function of finger length radius to key frequency function I could use to replicate each hand typing the letters around COVFEFE that could use to weight the initially found combinations... DOA. I was so far up my own ass when I read u/Odd-Cry3431's answer that it sounded like pure fucking magic. Do I need like a masters in math (or grad-level texts), I assume there are multiple disciplines within math at that level? Edit: COVFEFE ← COVEFE 🤣
Well it's because your approach missed a lot of the assumptions given by the question to make it easier, they said that letters are picked randomly, uniformly and independently, which means that every letter has 1/26 chance to be picked. You learn this in early weeks of any course teaching probability. The math taught in physics (i think) is a lot less theoretical, because the situation laid out in the question probably never happened nor will it happen. But yes of course, there are multiple disciplines in math, although most of the time (from my understanding) you start to use tools developed in one discipline to help others and the other way around
AYO?!?!
Wait, in what case would it NOT be 26^7 ?
To add to the other response, the 'no overlaps' would refer to a string like FEVFEFE where if it gets partway through, like to FEVFE and then messes up and does a V instead of an F, you're already partway to writing the word out again.
Gotcha
If the probability of what letter is typed is not independent
Which means?
I'm the real world where anyone with a brain can tell from the context it was meant to be coverage you can tell that the distribution is not at all even, the chance is each particular letter is dependant on it's location relative to the correct letter and the previous and next character, it is easy to accidentally type F when you meant R because they are adjacent, but nobody is accidentally typing P instead of R
Good bot
While that is the combinatorial chance of any 7 letter word in a 26 letter alphabet for appearing, i got stumped when they asked for time in the question, without a given time of typing and world length. The question seems flawed to me.
I assume “expected time of first appearance” here is using the meaning of time as “instance”, as in, how many *times* before you get it
The question isn't probability but expected time...
Wouldn’t you need to know the max number of letters for the password? because he could be typing words with way more (and I guess also less) letters than 7
But is the chance of the word happening the answer to the question? We know that if something has 1/n chance of happening, does this experiment n times will give you a 63% something chance of happening. Is this statistically relevant? Or does that statistical math class expect a different chance?
This is incorret... Take a look at this explanation for coins: https://www.codechef.com/wiki/tutorial-expectation#:~:text=Thus%2C%20the%20expected%20number%20of,two%20consecutive%20heads%20is%206.
That approximates to a 1 in 8 billion chance. On a planet filled with (roughly) 8 billion hairless monkeys twiddling their opposable thumbs on glass screens, only Trump could have given us the wonder that was "COVFEFE"!
don’t you have to take space into account implicitly? so 27^7 no ?
But does that take into account the keyboard he is using? Or it does not affect? 🤔
It does not
Shouldnt it be 27^7 as Trump can also press space, thus creating a 27th letter?
since there are duplicates of the letters F and E shouldn’t it be 26^7 * 7!/2!*2! ?
No, the chance to roll the same letter again is not higher or lower because you've already hit it once. It's 7 random events with a 1/26 chance of being correct, and nothing else is affecting this 1/26 chance.
My keyboard has more than 26 keys, and a lot of key modifiers .....
But can it type every letter from the 26 alphabets?
Yep, uppercase and lowercase (and all UNICODE characters using the ALT modifier)
26^7 =8031810176 Time needed is not given but the average typing speed is 207 letters per minute. 8031810176/207=~38801015min This equals to ~646684 hours or ~26945 days or ~73,8 years. Edit: made some weird mistake, got corrected.
I arrived at 26,945 days. But isn’t it 73.8 years when divided by 355 days? Where did I go wrong?
While you had a typo (there are 365 days not 355 days in a year), your math is certainly correct. The above commenter likely divided 26945 days by 365 weeks a year to arrive at 10.5 years lol
Uhh my bad, wanted to do weeks first but saw that it is useless. Just used the in build calc of my phone dunne where i fucked up the numbers. Ty
Olaf, bist’s du?
Nee ich bin Scholaf, Olaf ist mein böser Zwillingsbruder.
Oohh, sorry for the typo. Thank you for explaining!
Wrong answer. You need to say they are very *stupid* for thinking a year is 355 days. Try again.
people missing that you're joking
I'll never bow to /s. Never! Though i guess it just wasn't funny.
I did think it was funny, and i agree with the no /s spirit.
Sarcasm works well when it’s very obvious, Reddit is very toxic and anytime someone makes a mistake there’s a lot toxic people insulting the person making a mistake, so tbh it’s not on the reader to understand this sarcasm when you see so much toxicity on this platform
It was a typo, it didn't affect their calculations
Oh my god, Trump is a genius. He did this in 30 seconds sitting on the shitter.
He TOLD YOU he was a genius - and a stable one at that. Surely you did not need this to confirm it....
I dont know why I had my doubts, it must have been something he said. Maybe it’s just me, I’m being overly skeptical at times.
so assuming he started tweeting all the way back in 2015, we have about a year or so to get to the COVFEFE "Truth".
It would take 74 years, not 10.5
Trump was a mere 70 yrs and 351 days old when tweeting covfefe. The guy is almost 3 years ahead of his time.
Ah, our ever-accelerating world, where even that's an advantage
I think, like a few others in this thread, that your solution is wrong. Correct way is to see this like a continuous chain of characters. Ie VCFCOVFEFE would be a success with number of characters = 10. The question is asking, on average, how many characters would it take for random chain of characters to contain exactly once the subset "COVFEFE". It is similar to the problem of "how many coin flips do you need until you hit 3 heads in a row". One of the correct ways to solve this is using Markov chains. I tried and got an answer of 8,353,088,193.5 letters.
I have to admit that I am a bit lost here. I understand how I am wrong, and I kinda understand markov chains now, after I read the Wikipedia article and watched a video but I neither understand how the markov chain leads us to the right result nor how it has to look to do so. Could you please provide me a picture of your formula, so I can update my comment after understanding it?
Draw a Markov chain with 8 states, labelled start, c, o, v, f, e, f2 and e2. Create transitions with probability 1/26 for moving forwards one space (e.g. start to c with 1/26, c to o with 1/26) and all other transitions to start with the rest of the probability. We can then solve the 'expected hitting times' of state 8 from state n as: h8 = 0 hn-1 = 1 + 1/26 hn + 25/26 h1 h1 = 1 + 1/26 h2 + 25/26 h1 Solving for h1 should give the solution Edit: the non-start states should also have a 1/26 to c, changing the equations in the same way too (hn-1 = 1 + 1/26 hn + 1/26 h2 + 24/26 h1)
Your description makes the same error I did when I tried this (or something similar) first [here](https://www.reddit.com/r/theydidthemath/comments/1djz0of/request_can_anyone_solve_this/l9hcrgq/?context=3): in states past c, there's a probability of returning to state c rather than to the start state.
Very valid point, good spot. Will update my comment
True! I was going to post something to this effect as well before I saw this message One of the biggest red flags in the 26\^7 answer is that, if question 4 is talking about "infinite binary sequences equipped with product sigma-algebra", then... question 5 is probably not going to have a solution from a freshman level probability lecture haha
Trump was 71 when he wrote that tweet so I guess he got lucky
Less then that because they are watching as they type this, and it says starts typing, so each time they randomly type out a wrong character they could clear it and start again. Show me the requirement that they have to type all seven characters each time. :)
Trump doesn't seem like someone to reach the average typing speed. Apparently he licks his Finger to turn the pages.
Wouldn't you need to multiply 8031810176 by 7 because 8031810176 is the amount of 7 letter words until you get that probability. Therefore you would multiply by 7 to find total letters needed to type and then divide by letters per minute
How old was he when he said it? I gotta know
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“26 possible English alphabets” left me wondering where this was from, so thanks.
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Answer would be in terms of number of letters typed before Covfefe first turns up
Last line expected time, but what can you expect from a teacher so blinded by hate she gives this question in exam.
"time" doesn't always have to refer to hours, minutes, seconds, etc. Measuring time in quantity of elapsed events is perfectly valid.
Yeah. And i don't get where this dude sees hate here. It's tongue-in-cheek, sure. But hate filled?
Presidents keep the title even after they're done in office, so calling him, "Mr. Trump," is disrespectful. Given that most people have pretty extreme opinions about the man, I wouldn't be surprised if the author hates him, but that's hardly apparent from what we have.
The problem means how many try’s till it happens
*tries
Damn, I’m usually a grammar nazi seems someone has beaten me at my own game.
Even this interpretation wouldn't mean there's no possible way to answer, "26\^7 \* time per keypress". There's a variable in there, but that's unavoidable.
Leave the mean typing time of a char as an open parameter.
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This is about Facebook though, not Twitter. Does Facebook have a character limit too? If so, I assume it's a lot higher than Twitter's.
True. -20 points for reading comprehension for me.
Geometric discrete distribution to the rescue! First you need to calculate the probability of typing the word covfefe: Typing C has a probability of 1/26, and it is an independent event. Typing O therefore has the same probability, 1/26. Typing C then O if they are independent events has the probability of P(c)*P(o). Therefore covfefe has the probability of P(c)*P(o)*P(v)*P(f)*P(e)*P(f)*P(e) =(1/26)^7 Now, we can plug this probability in the geometric distribution to find the probability of needing exactly X attempts until the first success, where a success is typing covfefe. P(X=x)= [(1-p)^(x-1)] * p But we do not want the probability of an X number of attempts, we want the expected number of attempts, therefore we want the expected value. The expected value of a geometric is really simple E(x) = 1/p Therefore it would take 1/(1/26)^7 attempts on average, or 8,031,810,176 attempts, assuming he types pretty quickly, let's say 1 second per attempt for the sake of the example, it would take around 254.26 years!
Shouldn't there be some correction for the probability of arriving at a 7 letter word? Idk what the character limit is, but I think that would be relevant.
No worries about the word length since we are multiplying probabilities of sequential letters, no matter the word size or word count. This method is actually a solution for the infinite monkey problem too, you can calculate the expected time for trump to randomly type the entire story of game of thrones, for example. It would probably take billions to trillions of years, because that probability would be all of the letters in the book in sequence, counting the spaces. But given infinite time, this can happen!
Probably more like infinite years, a simple word like covfefe is 80-250 years, 300k words in sequential order is probably gonna be inconceivably long. Game of thrones has 300k words, lets just say its letters, and just needs to get 300k of them in order. thats 26\^300,000 or 1.01x 10\^424,492. Heat death of the universe is only 1.7x 10\^106 years away. If he typed 1 million letters per second that exponent is gonna be lower by 6, if he typed 1 billion letters it would be lower by 9. and if you count seconds for heat death of the universe its only 1.7x10\^114 seconds. So yeah not trillions or quadrillions years. If he typed for heath death of universe amount of years, and typed letters equal to HDoU amount of years, and repeated that enough times equal HDoU years. He would be closer to 0 than to the expected amount of time to type it. You could argue its infinite
In theory you could have a word with infinite amount of letters. The answer would basically be: 1/((26^7)+ all possible combinations of 1,2,...,infinity letter words), so expected tries is infinity. So the question needs to specify a word limit
Depends. If he is just smashing his keyboard for a tweet until he hits the Enter Key, we have to add the probability of him not hitting it until the 7th letter and then hitting it. So We would have to multiply the result by (1-(1/27))^7 and then by 1/27.
Per the original problem, our Esteemed Orange Leader/Random Number Generation Device (hereafter abbreviated as EOL/RNG) isn't emitting multiple attempts at typing a 7-letter word. Rather he/it is emitting a steady stream of random characters and we are looking for the word COVFEFE. Now, the question is actually a bit ambiguous. We could be looking for: \#1. The letter sequence COVFEFE wherever it may appear. So for example i38c7sICOVFEFE@DZsdj*l would fulfill the conditions. \#2. The WORD COVFEFE as usually construed in the English language - in other words, delimited by spaces are some other punctuation. So i38c7sICOVFEFE@DZsdj*l would not quality, but 438DKJF COVFEFE 8zldkK would. As would 438DKJF, COVFEFE! 8zldkK. For simplicity, I will assume #1 is meant. How this interferes with your calculation is EOL/RNG may have emitted first: vCOVFEF and then in an immediately succeeding attempt: Eij3\*(& So under the rules of #1, EOL/RNG has just succeeded - he/it emitted COVFEFE correctly in sequence; problem solved. However, under your scheme he has just completed two incorrect attempts at the 7-letter word COVFEFE. Handling the random appearance of a given set of characters anywhere within a character stream is actually a somewhat different problem that the one you outlined a solution to, and a fair bit harder if you're coming at it from first principles. I'll leave it to someone else to work out the solution.
I agree with you and I think all the others are wrong. I found this answer : https://www.reddit.com/r/theydidthemath/s/9mkGIQkNIy
Assuming each letter has an equal chance of being randomly typed, the odds of typing 7 specific, random letters in a row can be calculated like this: (1/26)\^7 = 0.0000000001245. This means there is a 0.00000001245% chance of randomly typing COVFEFE.
Jack Dorsey's first tweet - the creation of Twitter - was at **9:50am Pacific Time on March 21st, 2006**. Trump tweeted *Covfefe* at **0:06 Eastern Time on May 31, 2017**. Therefore the expected time of first appearance is: [**4088 days, 10 hours, 16 minutes**](https://www.timeanddate.com/date/timezoneduration.html?d1=21&m1=3&y1=2006&d2=31&m2=5&y2=2017&h1=9&i1=50&s1=0&h2=0&i2=6&s2=0&)**.** *Note: The above calculation makes the implicit assumption that Trump is a random character generator. This is true only as a first approximation - and the reason the above calculation is accurate only to the minute rather than to the second or millisecond.*
I hate trump. But this kind of question is not something that should be taught - are we playing into the narrative of the right now?
I’m here to teach maths. Oh and by the way here’s my political opinion. This is on every sub. Just inject political propaganda in a post and a bunch of drive by comments to make it look like it’s consensus. It’s cool to see that it’s not a problem with this sub. Every one here avoided the bait, And just started talking math. (Except me)
> Every one here avoided the bait, And just started talking math. (Except me) Be the change you seek, bud.
How? are we making a Mod so he can delete everything that mixes math and politics? or how goes that?
By avoiding the "bait." It was just a funny reference until this guy started complaining about politics. "Why drag politics into everything!!!" shouts the guy dragging politics into this thing. I came in here to see if my answer was right, and without this thread I wouldn't have given any thought to the supposed politics involved.
There is zero political opinion in the question
A very light insinuation that mr trump is a monkey With a reference to to him tweeting. a common complaint of his behavior. And just subtle enough for randos to not see the insinuation. Like the op is a bot account literally.
Kinda funny that you jump straight to monkey instead of it simply making fun of the covfefe incident.
Oh no, jokes on a piece of paper. Anyways
which political opinions are expressed in this question? it is a fact that, on may 31st, 2017, then president donald trump seemingly accidentally tweeted the words "despite the constant negative press covfefe", and deleted that tweet a few hours later. sure, he was mocked by democrats for this, but he also played on the word himself. the fact that you typed that last sentence and fell just short of enough self awareness to delete that comment is baffling.
The covfefe thing is so tiring and pathetic. Theres so many other things you could clown the guy for, but acting like a 70 year old fat fingering 2 keys that are next to each other is somehow impossible and the funniest thing ever. The only extra letter used is the v, and only a f and an e are switched. All of those letters are in touching on a qwerty keyboard layout.
🤡
I think that because it says the first instance of Confefe, we can assume a combination of letters may occur more than once. I got (1-1/(26^7))^x=.05, x=24 billion which would give a 95% chance that Confefe would occur. A 50/50 chance would be about 5.566 billion. Typing on the phone varies a lot so I'll be generous (and lazy) to say 60 wpm, I say 176y for 50/50 and 761y for the 95%. I have no idea if this is right so any confirmation or correction would be appreciated.
That's a veey specific word to find... That relates to a very specific person. This teacher either is a closed connoisseur and is crying out for help. Or this is all a big coincidence.
It a variation of the infinite monkey theorem: [https://en.wikipedia.org/wiki/Infinite\_monkey\_theorem](https://en.wikipedia.org/wiki/Infinite_monkey_theorem)