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Interesting probability question 
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#1
Sep907, 07:16 AM

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The problem statement, all variables and given/known data
Firstly, I'd like to point out that this question is a little left of field...but here it is anyway. Here is what Puck has to say at the end of A Midsummer Nights Dream 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. 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, you always end up with the same word, “restore”. In fact, even if you had started anywhere else in A Midsummer Nights Dream you would have wound up with the same word. Is this a secret message left by Shakespeare, is it coincidence, or is there another easy explanation? The attempt at a solution I really don't have any ideas. At first I thought it may be Shakespeare's style of writing, but after some research it seems iambic pentameter doesn't have a lot to do with it. Anybody think there is any mathematical basis in this question? 


#2
Sep907, 06:14 PM

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To make this trick work you only have to pay attention to last few lines and make sure all paths lead to 'restore'. The rest of the play is irrelevant. I think it actually is a pretty interesting question. I would think the odds a random piece of prose would have this property is low but hardly infinitesimal, but then probability "doth make fools of us all". I do think the odds are infinitesimal it's a message from Shakespeare. What's the message? Predicting the "Restoration"? Hah. The correct thing to do is write a computer program to skip through prose in this way and see what happens.



#3
Sep907, 06:51 PM

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Do you always get the initial r of restore? Or end up anywhere within that word?



#4
Sep907, 07:26 PM

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Interesting probability question



#5
Sep1007, 12:18 AM

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I think you always end up at one particular word, though I can't actually prove it. But here's how to analyze it.
Make a list of the number of letters in each word, starting with the last word and going backwards through the prose. Call the n^{th} number in this list w_{n}. For Midsummer Night's Dream, this list begins w_{1}=6, w_{2}=7, w_{3}=5, w_{4}=5, w_{5}=3, ... Now make a new list of numbers, r_{1}, r_{2}, ..., according to the following rule: starting with n=1, if nw_{n} is negative or zero, r_{n}=n. If nw_{n} is positive, r_{n}=r_{nwn}. The meaning of the r list is this: if you start at the n^{th} word in the prose (counting backwards from the end), and move forwards according to the rule given by t_n_p, r_{n} gives you the number of the word where you end up. For Midsummer Night's Dream, the r list is r_{1}=1, r_{2}=2, r_{3}=3, r_{4}=4, r_{5}=2, r_{6}=6, r_{7}=2, r_{8}=6, r_{9}=2, and then r_{n}=2 for every n>9. Now, I ran this on Mathematica a couple dozen times, using a list of 100 word lengths chosen randomly between 1 and 11 letters (not an ideal simulation of English, I grant you). Every time, the r list eventually was a single repeating number. 


#6
Sep1007, 12:45 AM

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Interesting concept avodyne, how did you actually figure that out!
Another person also told me something along the lines of finding the mean number of letters in a word and so on and so forth. Can't exactly remember how they explained it though.. 


#7
Sep1007, 01:05 AM

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The idea of the r list is to label the n^{th} word with n if it's a word you end on, or with the label of the word you go to if it's not a word you end on. By doing this backwards through the prose, the labels propagate, so that if one word goes to another and then to another (where it stops), they all get the label of that last word. Somehow I don't think that's very clear but it's the best I can do right now ... 


#8
Sep1007, 10:08 AM

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I realized that it is possible to construct a counterexample: suppose all words have an even number of letters. Then, if you start at any oddnumbered word, you end on an oddnumbered word, and ditto for even. More generally, you need a pattern of word lengths that divides the prose into two or more sets so that if you start on a word in a particular set, you always stay in that set. But my randomnumber experiments indicate such patterns of word lengths are pretty rare ...



#9
Sep1007, 10:31 AM

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So my guess such behavior is just uncommon is really wrong. It's exceedingly common. Makes sense, any sequence of ten or so words can be a bottleneck, any path coming in leaves on the same path. And as the text gets longer the odds of having a bottleneck get greater. This is a great tool to generate 'messages' from some of the great minds of the past! Well done!



#10
Sep1007, 03:57 PM

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#11
Sep1007, 04:02 PM

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Indeed, means the longer the text, the more likely there's a unique destination word.



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