Probability gets interesting?

  • Thread starter dj023102
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  • #1
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Probability gets interesting???

Hey guys thought this was an interesting problem that our lecturer purposed:

Pick any word in the first few lines and do the following: Let’s say your choice is the
word “shadows”. This word has seven letters. The seventh word following “shadows”
is “all”. This word has three letters. The third word following “all” is “That”. This
word has four letters, etc. proceed until you come across the word “restore” in the last
line, from where you cannot move any further. Why is it that no matter which word
you choose in the first few lines, you always end up with the same word, “restore”. In
fact, even if you had started somewhere in the middle of A Midsummer Nights Dream
you would have wound up with the same word. And, even stranger, every play by
Shakespeare contains a special word like “restore”. Are these words secret messages left
by Shakespeare, is all this coincidence, or is there another easy explanation?

If we shadows have offended,
Think but this, and all is mended,
That you have but slumber’d here
While these visions did appear.
And this weak and idle theme,
No more yielding but a dream,
Gentles, do not reprehend:
if you pardon, we will mend:
And, as I am an honest Puck,
If we have unearned luck
Now to ’scape the serpent’s tongue,
We will make amends ere long;
Else the Puck a liar call;
So, good night unto you all.
Give me your hands, if we be friends,
And Robin shall restore amends.

Anyone have any ideas to how that is, apparently it also occurs with The Harry Potter books no matter what word u start with in the book you will always end up with the same word in the end of the book? Any ideas?
 

Answers and Replies

  • #2
CompuChip
Science Advisor
Homework Helper
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Hey, that's funny.
I think it's a general phenomenon. If I apply the procedure you described to the first paragraph of your post ("Pick any word ... another easy explanation") I end up on the word "there" 138 out of 147 times (6/147: Shakespeare, 1/147: coincidence, 2/147: another).
 
  • #3
6
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Hey any chance your doing maths 1030 at monash clayton. I'm doing an assignment now exact same wording. Tell us if you find the answer plz. I reckon its just a matter of convergence. No matter where you start in wat text if you continue through enough lines they all end up on the same path. Like in that shakespear quote its sumtin like 5 or six lines till they all on same word... now how do i explain that in mathematics lol
 
  • #4
15
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No idea, hence i am asking for some help. But I don't think it is any probability that he has taught during the lecture, i just don't know how to apply binomial, poisson or normal distribution to work it out.
 
  • #5
6
0


oi i worked it out... you hav to add up the distribution of word length for the text then use that to find the expected word length

it turns out to be 4... then starting from the first word you end up at the magic word

using a binomial distribution with probabilty of success equal to prob of gettin a word 4 letters long solve mean=n*p with the mean set the the expected value of the distribution and solve for n thats the expected number of trials before the average word length reaches 4

then to prove it you anylse the text and count the amount of words you have to go through before each word in the first line reach the same path as moving four words at a time then average that it works out to be the same number

are you actualy doing 1030 tho at monash??
 
  • #6
15
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So basically divide the number of letters with thenumber of words which apprioxmately equals 3.95 so letters per words.Then starting at the first word (if) count every fourth word so we have(if), (offended), (and) and keep going until you reach (restore).
Then count the number of 4letter words in the text divide the number of words in text gives you the probability of getting a 4letter word which is 25/94 = 0.266
mean = 4 so therefore n=15.04
Is this right?
And can u give an example of ur proof?
and yes am doing MTH1030 at Monash. Hating this probability assignment atm
 
Last edited:
  • #7
6
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well it wasn't exactly a proof more of a justification using the text...

highlight the path of moving four words starting from the first word...
then starting from each word in the first sentence count how many words it takes till they are on the same path as moving 4 words at a time...
average this and it turns out to b 14.6 or sumthin and use the actual mean not 4 it is about 14,8 or sumtin for your second calulation
 
  • #8
2,257
7


lines (beginning with different words) converge but never split apart. eventually it is bound to end on one single line.
 
  • #9


well it wasn't exactly a proof more of a justification using the text...

highlight the path of moving four words starting from the first word...
then starting from each word in the first sentence count how many words it takes till they are on the same path as moving 4 words at a time...
average this and it turns out to b 14.6 or sumthin and use the actual mean not 4 it is about 14,8 or sumtin for your second calulation

Can anyone further elaborate on this?(bold writing)
 
  • #11
7
0


Can anyone clarify what Stanaz meant? I'd like to prove this problem using some kind of probability/binomial method. I don't understand his method or reasoning at all.
 

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