Well?

>letters are independent
>using a Markov chain

No, call the time it takes [math]T[/math] then
[eqn]E[T] = \sum_{k=0}^\infty P(T \geq k) [/eqn]

You can find a simple recursive formula for those probabilites.

The sum should start at k=1 obviously.

Oh right I am retarded.

How do you find a recursion?

Funny until covfefe

Butthurt.

actually you do have to use a markov chain. the subsequent state of the chain of letters is dependent upon the current state in the chain of letters since having a chain of xxxc gives a higher probability of obtaining covfefe since the current last letter is a c.

well, there may be another strategy to solve it, but the markov chain approach isnt wrong

so you can look at the probability of transitioning from any given state to the "covfefe" state and sum those all up

"Covfefe" is pretty fucking famous in NLP circles too representing noise in twitter datasets

>Probability of xxx turning into xxxC
>Probability of xxxC turning into xxxCO
>Probability of xxxCO turning into xxxCOV
>Probability of xxxCOV turning into xxxCOVF
>Probability of xxxCOVF turning into xxxCOVFE
>Probability of xxxCOVFE turning into xxxCOVFEF
>Probability of xxxCOVFEF turning into xxxCOVFEFE

Since the starting character C is not contained anywhere later in the string, it is straightforward since any failure results in having to restart completely, except for a C-failure

Why are people that say "Drumpf" so retarded?

Why do people who suck trumps cock so much, so retarded?

Never. The word COVFEFE can only be written if Trump presses shift or uses capslock. The scenario never specified whether trump actually does that. So the word COVFEFE as written would not exist in the scenario as presented.

>nobody has solved it yet

solve it then if you are so smart Mr.Smartypants

Of course he would be typing in all caps, as befits his dominant nature.

I have already solved it, but I won't post it.

Twitter has character limits
for now

probability of that sequence is (1/26)^7

mean of a negative binomial distribution with probability of success (1/26)^7 is 8031810175

so it takes 8031810175 key presses on average before he types COVFEFE

or it just takes one because he's not a retard and knows how to type COVFEFE

That's plain wrong.

looks right to me

Success does not occur in distinct groups of 7, you cannot use a model based on independent bernoulli trials.
Your model would treat "COCKCOV" "FEFEASS" as a failure.

It is so wrong I don't even know where to start.
The negative binomial has two parameters: the probability of success and the number of successes.

also this

i know what you're saying, but it still seems right.

FUCKING FOUND IT:
en.wikipedia.org/wiki/Martingale_(probability_theory)
Google ABRACADABRA problem.

>I know I'm wrong but I still think I'm right

it's that dunning kruger effect kicking in for sure.

matches numerical results (had to use a restricted alphabet and smaller word because laptop is slow)

*crickets*

You realize that guy was objectively retarded for implying that the question should be answered in any other metric other than "# of keystrokes"?

it doesn't

anyway solve the equations, Ax = b with b = (1.0, ..., 1.0) and A =

0.0769 -0.0769 0.0000 0.0000 0.0000 0.0000 0.0000
-0.9231 0.9615 -0.0385 0.0000 0.0000 0.0000 0.0000
-0.9231 -0.0385 1.0000 -0.0385 0.0000 0.0000 0.0000
-0.9231 -0.0385 0.0000 1.0000 -0.0385 0.0000 0.0000
-0.9231 -0.0385 0.0000 0.0000 1.0000 -0.0385 0.0000
-0.9231 -0.0385 0.0000 0.0000 0.0000 1.0000 -0.0385
-0.9231 -0.0385 0.0000 0.0000 0.0000 0.0000 1.0000

I'm getting x =

4176542234.0734
4176542221.0734
4176541883.0733
4176533095.0714
4176304607.0198
4170363917.6783
4015905994.8014

so expected value looks like 4176542234.0734

>literally posts image of numerical results
>>it doesn't
ok

everyone told you why your nonsense is wrong. now you posted a very low resolution simulation with a superimposed curve that doesn't fit it. you don't deserve anything more than "it doesn't" for that.

it's a markov chain. solve the linear system and that's it.

the answer was already posted in the reddit thread, fellow pajeet

link?

reddit.com/r/india/comments/7h1ozu/covfefe_got_trumped_by_indian_statistical/

jesus christ, nobody even got close to the answer, it's a fucking trainwreck

thanks!

imagine being this much of a brainlet

daily E-minder that Reddit is superior to Veeky Forums.

Yet nobody in the reddit thread got the correct solution either.

>implying

How do I learn to code such stuff?

Give me a quick rundown on mathematica.

Read SICP.

wolfram.com/mathematica/resources/
lab.open.wolframcloud.com/objects/wpl/GetStarted.nb

i've never read sicp

this might not be a bad approximation since [math]P(S_{t+1} | F_t) = P(S_{t+1})/P(F_t) \approx P(S_{t+1}) [/math] and we are only concerned with the first instance of "failure". don't think it's exactly right now that i think about it more

so what i'm saying is that "success" (as in not covfefe) for characters t to t+7 does not give much information about whether characters t+1 to t+8 spell out "covfefe", and since we stop at the first failure, they can basically be treated as independent events even though they aren't.

someone correct me if i'm wrong.

The effect is most clearly seen at step 1. The probability that letters 2-9 spell "covfefe", given that letters 1-8 do not spell "covfefe", is increased, because the chance that letters 2-8 read "ovfefe" is reduced.

The Markov chain calculations handle this exactly.

(So I can't count how many letters there are in "covfefe".)

to see why you're wrong, write out the linear equations for the expected time based on the markov chain. realize it's a different result.

oh sorry I thought you were saying it's the right answer. I think you're right? being further down the path doesn't really decrease the expectancy much at all. see where x is the expectancy when having 0, 1, ..., 6 correct letters

26^7

i'm not the user who originally made the graph. yes, it's not exactly correct, but since [math] P(S_t|F_t) \approx P(S_t) [/math] you can probably assume independence without being off by much, if any at all.

(1/26)**7/(1-(1/26)**7) - (1/26)**7 = 1.550149392621445e-20

we're talking about a very small difference here. i'm not even sure that you'd get a different mean when you round it to the nearest step.

looks like you're right about it not being negative binomial

i yield

I would expect the effect to be visible when the target word is "ab", and the alphabet is {a,b}. I'll accept that the effect is negligible for "covfefe" over {a,...,z}.

yeah it makes sense considering the pdf of the negative binomial is non-zero for n before the length of the target string. didn't think of that.

i think the reddit dude got it right. but anyway, you'll probably introduce more numerical errors solving matrices of floats than just assuming independence

But that's the wrong Markov chain. At each step, there are three options:
1) The correct next letter (p=1/26) Go to state I+1
2) The letter "c" (p=1/26) Go to state 1
3) Any other letter (p=24/26) Go to state 0

Your Markov chain overestimates the number of steps needed.

welp

you're almost there, see

no you won't. it's 7x7. it's very few operations. even if it was a shitload of operations, mathematica works with arbitrary precision. even if it didn't, those aren't floats, they're rationals.

is that correct? I got like half with my program, did I fuck up?

really confused as to why the two have the same mean.

i didn't say anything about mathematica.

WHOOPS. here's the corrected linear system. A =
0.0385 -0.0385 0.0000 0.0000 0.0000 0.0000 0.0000
-0.9231 0.9615 -0.0385 0.0000 0.0000 0.0000 0.0000
-0.9231 -0.0385 1.0000 -0.0385 0.0000 0.0000 0.0000
-0.9231 -0.0385 0.0000 1.0000 -0.0385 0.0000 0.0000
-0.9231 -0.0385 0.0000 0.0000 1.0000 -0.0385 0.0000
-0.9231 -0.0385 0.0000 0.0000 0.0000 1.0000 -0.0385
-0.9231 -0.0385 0.0000 0.0000 0.0000 0.0000 1.0000

b = (1, 1, ..., 1)

x = (expectancy when starting at 0, 1, 2, ..., 6)
8031811989.6027
8031811963.6027
8031811313.6026
8031794413.5988
8031355013.4996
8019930610.9199
7722896143.8488

you didn't say anything at all then

It just turns out that the approximation in is a very good one. I would estimate a relative error of about 26^{-7} between the conditioned and unconditioned probabilities.

>a group of pajeets is smarter than Veeky Forums
Why do I even come here? It's very clear that all the superiority complexes people on this board have are misplaced.

oh just realized they are actually different by 1 gotcha.

but nobody in there has the right answer.

i'd like to see his solution properly worked out, and not the elipicklemorty version

Here is the solution of almost the same problem just a different word:

jeremykun.com/2014/03/03/martingales-and-the-optional-stopping-theorem/

Thanks guys for letting me know that youre retards
t. Sociology student